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Given an array arr[], find the maximum j – i such that arr[i] <= arr[j]

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Given an array arr[] of N positive integers. The task is to find the maximum of j – i subjected to the constraint of arr[i] <= arr[j].

Examples : 

  Input: {34, 8, 10, 3, 2, 80, 30, 33, 1}
  Output: 6  (j = 7, i = 1)

  Input: {9, 2, 3, 4, 5, 6, 7, 8, 18, 0}
  Output: 8 ( j = 8, i = 0)

  Input:  {1, 2, 3, 4, 5, 6}
  Output: 5  (j = 5, i = 0)

  Input:  {6, 5, 4, 3, 2, 1}
  Output: 0
Recommended Practice

Method 1 (Simple but Inefficient): Run two loops. In the outer loop, pick elements one by one from the left. In the inner loop, compare the picked element with the elements starting from the right side. Stop the inner loop when you see an element greater than the picked element and keep updating the maximum j-i so far. 

 

C++




// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
/* For a given array arr[],
   returns the maximum j – i such
   that arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
    int maxDiff = -1;
    int i, j;
 
    for (i = 0; i < n; ++i) {
        for (j = n - 1; j > i; --j) {
            if (arr[j] >= arr[i] && maxDiff < (j - i))
                maxDiff = j - i;
        }
    }
 
    return maxDiff;
}
 
int main()
{
    int arr[] = { 1,1,1,1,1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    cout << "\n" << maxDiff;
    return 0;
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)


C




// C program for the above approach
#include <stdio.h>
/* For a given array arr[],
   returns the maximum j – i such
   that arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
    int maxDiff = -1;
    int i, j;
 
    for (i = 0; i < n; ++i) {
        for (j = n - 1; j > i; --j) {
            if (arr[j] > arr[i] && maxDiff < (j - i))
                maxDiff = j - i;
        }
    }
 
    return maxDiff;
}
 
int main()
{
    int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    printf("\n %d", maxDiff);
    getchar();
    return 0;
}


Java




// Java program for the above approach
public class FindMaximum {
    /* For a given array arr[],
        returns the maximum j-i such
       that arr[j] > arr[i] */
    int maxIndexDiff(int arr[], int n)
    {
        int maxDiff = -1;
        int i, j;
 
        for (i = 0; i < n; ++i) {
            for (j = n - 1; j > i; --j) {
                if (arr[j] > arr[i] && maxDiff < (j - i))
                    maxDiff = j - i;
            }
        }
 
        return maxDiff;
    }
 
    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        FindMaximum max = new FindMaximum();
        int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
        int n = arr.length;
        int maxDiff = max.maxIndexDiff(arr, n);
        System.out.println(maxDiff);
    }
}


Python3




# Python3 program to find the maximum
# j – i such that arr[j] > arr[i]
 
# For a given array arr[], returns
# the maximum j – i such that
# arr[j] > arr[i]
 
 
def maxIndexDiff(arr, n):
    maxDiff = -1
    for i in range(0, n):
        j = n - 1
        while(j > i):
            if arr[j] > arr[i] and maxDiff < (j - i):
                maxDiff = j - i
            j -= 1
 
    return maxDiff
 
 
# driver code
arr = [9, 2, 3, 4, 5, 6, 7, 8, 18, 0]
n = len(arr)
maxDiff = maxIndexDiff(arr, n)
print(maxDiff)
 
# This article is contributed by Smitha Dinesh Semwal


C#




// C# program to find the maximum
// j – i such that arr[j] > arr[i]
using System;
class GFG {
    // For a given array arr[], returns
    // the maximum j-i such that arr[j] > arr[i]
    static int maxIndexDiff(int[] arr, int n)
    {
        int maxDiff = -1;
        int i, j;
 
        for (i = 0; i < n; ++i) {
            for (j = n - 1; j > i; --j) {
                if (arr[j] > arr[i] && maxDiff < (j - i))
                    maxDiff = j - i;
            }
        }
 
        return maxDiff;
    }
 
    // Driver program
    public static void Main()
    {
 
        int[] arr = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
        int n = arr.Length;
        int maxDiff = maxIndexDiff(arr, n);
        Console.Write(maxDiff);
    }
}
// This Code is Contributed by Sam007


PHP




<?php
// PHP program to find the maximum
// j – i such that arr[j] > arr[i]
 
// For a given array arr[], returns
// the maximum j – i such that
// arr[j] > arr[i]
function maxIndexDiff($arr, $n)
{
    $maxDiff = -1;
     
    for ($i = 0; $i < $n; ++$i)
    {
        for ($j = $n - 1; $j > $i; --$j)
        {
            if($arr[$j] > $arr[$i] &&
               $maxDiff < ($j - $i))
                $maxDiff = $j - $i;
        }
    }
 
    return $maxDiff;
}
 
// Driver Code
$arr = array(9, 2, 3, 4, 5,
             6, 7, 8, 18, 0);
$n = count($arr);
$maxDiff = maxIndexDiff($arr, $n);
echo $maxDiff ;
 
// This code is contributed by Sam007
?>


Javascript




<script>
// JavaScript program for the above approach
 
/* For a given array arr[],
returns the maximum j – i such
that arr[j] > arr[i] */
function maxIndexDiff(arr, n)
{
    let maxDiff = -1;
    let i, j;
 
    for (i = 0; i < n; ++i)
    {
        for (j = n - 1; j > i; --j)
        {
            if (arr[j] > arr[i] && maxDiff < (j - i))
                maxDiff = j - i;
        }
    }
 
    return maxDiff;
}
 
    // Driver code
    let arr = [ 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 ];
    let n = arr.length;
    let maxDiff = maxIndexDiff(arr, n);
    document.write(maxDiff);
 
// This code is contributed by Manoj.
</script>


Output

8

Time Complexity: O(n2)
Auxiliary Space: O(1)

Method 2: Improvising the Brute Force Algorithm and looking for BUD, i.e Bottlenecks, unnecessary and duplicated works. A quick observation actually shows that we have been looking to find the first greatest element traversing from the end of the array to the current index. We can see that we are trying to find the first greatest element again and again for each element in the array. Let’s say we have an array with us for example [1, 5, 12, 4, 9] now we know that 9 is the element that is greater than 1, 5, and 4 but why do we need to find that again and again. We can actually keep a track of the maximum number moving from the end to the start of the array. The approach will help us understand better and also this improvisation is great to come up with in an interview. 

Approach :  

  1. Traverse the array from the end and keep a track of the maximum number to the right of the current index including self
  2. Now we have a monotonous decreasing array, and we know we can use binary search to find the index of the rightmost greater element
  3. Now we will just use binary search for each of the elements in the array and store the maximum difference of the indices and that’s it we are done.

C++




/* For a given array arr[],
   calculates the maximum j – i
   such that arr[j] > arr[i] */
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    vector<long long int> v{
        34, 8, 10, 3, 2, 80, 30, 33, 1
    };
    int n = v.size();
    vector<long long int> maxFromEnd(n + 1, INT_MIN);
 
    // create an array maxfromEnd
    for (int i = v.size() - 1; i >= 0; i--) {
        maxFromEnd[i] = max(maxFromEnd[i + 1], v[i]);
    }
 
    int result = 0;
 
    for (int i = 0; i < v.size(); i++) {
        int low = i + 1, high = v.size() - 1, ans = i;
 
        while (low <= high) {
            int mid = (low + high) / 2;
 
            if (v[i] <= maxFromEnd[mid]) {
               
                // We store this as current answer and look
                // for further larger number to the right
                // side
                ans = max(ans, mid);
                low = mid + 1;
            }
            else {
                high = mid - 1;
            }
        }
        // keeping a track of the
        // maximum difference in indices
        result = max(result, ans - i);
    }
    cout << result << endl;
}


C




/* C program to implement
the above approach */
 
/* For a given array arr[],
calculates the maximum j – i
such that arr[j] > arr[i] */
#include <limits.h>
#include <stdio.h>
 
/* Function for maximum of
two numbers in C */
int max(int num1, int num2)
{
    return (num1 > num2 ) ? num1 : num2;
}
 
int main()
{
    int v[] = { 34, 8, 10, 3, 2, 80, 30, 33, 1 };
    int n = sizeof(v) / sizeof(v[0]);
    int maxFromEnd[n+1];
    for (int i = 0; i < n+1; i++) {
        maxFromEnd[i] = INT_MIN;
    }
    // create an array maxfromEnd
    for (int i = n - 1; i >= 0; i--) {
        maxFromEnd[i] = max(maxFromEnd[i + 1], v[i]);
    }
 
    int result = 0;
 
    for (int i = 0; i < n; i++) {
        int low = i + 1, high = n - 1, ans = i;
 
        while (low <= high) {
            int mid = (low + high) / 2;
 
            if (v[i] <= maxFromEnd[mid]) {
             
                // We store this as current answer and look
                // for further larger number to the right
                // side
                ans = max(ans, mid);
                low = mid + 1;
            }
            else {
                high = mid - 1;
            }
        }
        // keeping a track of the
        // maximum difference in indices
        result = max(result, ans - i);
    }
    printf("\n %d", result);
}
 
/* This code is contributed by Pushpesh Raj */


Java




// Java program to implement
// the above approach
 
// For a given array arr[],
// calculates the maximum j – i
// such that arr[j] > arr[i]
import java.util.*;
class GFG{
 
public static void main(String[] args)
{
  int []v = {34, 8, 10, 3, 2,
             80, 30, 33, 1};
  int n = v.length;
  int []maxFromEnd = new int[n + 1];
  Arrays.fill(maxFromEnd, Integer.MIN_VALUE);
 
  // Create an array maxfromEnd
  for (int i = v.length - 1; i >= 0; i--)
  {
    maxFromEnd[i] = Math.max(maxFromEnd[i + 1],
                             v[i]);
  }
 
  int result = 0;
 
  for (int i = 0; i < v.length; i++)
  {
    int low = i + 1, high = v.length - 1,
        ans = i;
 
    while (low <= high)
    {
      int mid = (low + high) / 2;
 
      if (v[i] <= maxFromEnd[mid])
      {
        // We store this as current
        // answer and look for further
        // larger number to the right side
        ans = Math.max(ans, mid);
        low = mid + 1;
      }
      else
      {
        high = mid - 1;
      }
    }
     
    // Keeping a track of the
    // maximum difference in indices
    result = Math.max(result, ans - i);
  }
  System.out.print(result + "\n");
}
}
 
// This code is contributed by shikhasingrajput


Python3




# Python3 program to implement
# the above approach
 
# For a given array arr,
# calculates the maximum j – i
# such that arr[j] > arr[i]
 
# Driver code
if __name__ == '__main__':
   
    v = [34, 8, 10, 3,
         2, 80, 30, 33, 1];
    n = len(v);
    maxFromEnd = [-38749432] * (n + 1);
 
    # Create an array maxfromEnd
    for i in range(n - 1, 0, -1):
        maxFromEnd[i] = max(maxFromEnd[i + 1],
                            v[i]);
 
    result = 0;
 
    for i in range(0, n):
        low = i + 1; high = n - 1; ans = i;
 
        while (low <= high):
            mid = int((low + high) / 2);
 
            if (v[i] <= maxFromEnd[mid]):
               
                # We store this as current
                # answer and look for further
                # larger number to the right side
                ans = max(ans, mid);
                low = mid + 1;
            else:
                high = mid - 1;       
 
        # Keeping a track of the
        # maximum difference in indices
        result = max(result, ans - i);
     
    print(result, end = "");
     
# This code is contributed by Rajput-Ji


C#




// C# program to implement
// the above approach
 
// For a given array []arr,
// calculates the maximum j – i
// such that arr[j] > arr[i]
using System;
class GFG{
 
public static void Main(String[] args)
{
  int []v = {34, 8, 10, 3, 2,
             80, 30, 33, 1};
  int n = v.Length;
  int []maxFromEnd = new int[n + 1];
   
  for (int i = 0;
           i < maxFromEnd.Length; i++)
    maxFromEnd[i] = int.MinValue;
 
  // Create an array maxfromEnd
  for (int i = v.Length - 1;
           i >= 0; i--)
  {
    maxFromEnd[i] = Math.Max(maxFromEnd[i + 1],
                             v[i]);
  }
 
  int result = 0;
 
  for (int i = 0; i < v.Length; i++)
  {
    int low = i + 1,
        high = v.Length - 1,
    ans = i;
 
    while (low <= high)
    {
      int mid = (low + high) / 2;
 
      if (v[i] <= maxFromEnd[mid])
      {
        // We store this as current
        // answer and look for further
        // larger number to the right side
        ans = Math.Max(ans, mid);
        low = mid + 1;
      }
      else
      {
        high = mid - 1;
      }
    }
 
    // Keeping a track of the
    // maximum difference in indices
    result = Math.Max(result, ans - i);
  }
  Console.Write(result + "\n");
}
}
 
// This code is contributed by shikhasingrajput


Javascript




<script>
 
    // Javascript program to implement
    // the above approach
       
    // For a given array []arr,
    // calculates the maximum j – i
    // such that arr[j] > arr[i]
     
    let v = [34, 8, 10, 3, 2, 80, 30, 33, 1];
    let n = v.length;
    let maxFromEnd = new Array(n + 1);
 
    for (let i = 0; i < maxFromEnd.length; i++)
      maxFromEnd[i] = Number.MIN_VALUE;
 
    // Create an array maxfromEnd
    for (let i = v.length - 1; i >= 0; i--)
    {
      maxFromEnd[i] = Math.max(maxFromEnd[i + 1], v[i]);
    }
 
    let result = 0;
 
    for (let i = 0; i < v.length; i++)
    {
      let low = i + 1, high = v.length - 1, ans = i;
 
      while (low <= high)
      {
        let mid = parseInt((low + high) / 2, 10);
 
        if (v[i] <= maxFromEnd[mid])
        {
          // We store this as current
          // answer and look for further
          // larger number to the right side
          ans = Math.max(ans, mid);
          low = mid + 1;
        }
        else
        {
          high = mid - 1;
        }
      }
 
      // Keeping a track of the
      // maximum difference in indices
      result = Math.max(result, ans - i);
    }
    document.write(result);
   
</script>


Output

6

Time complexity : O(N*log(N)) 
Space complexity: O(N)

Method 3 O(n * log n): Use hashing and sorting to solve this problem in less than quadratic complexity after taking special care of the duplicates. 
Approach :  

  1. Traverse the array and store the index of each element in a list (to handle duplicates).
  2. Sort the array.
  3. Now traverse the array and keep track of the maximum difference of i and j.
  4. For j consider the last index from the list of possible indexes of the element and for i consider the first index from the list. (As the index was appended in ascending order).
  5. Keep updating the max difference till the end of the array.

Below is the implementation of the above approach:

C++




// C++ implementation of
// the hashmap approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum
// index difference
int maxIndexDiff(vector<int>& arr, int n)
{
   
    // Initialise unordered_map
    unordered_map<int, vector<int> > hashmap;
 
    // Iterate from 0 to n - 1
    for (int i = 0; i < n; i++) {
        hashmap[arr[i]].push_back(i);
    }
 
    // Sort arr
    sort(arr.begin(), arr.end());
    int maxDiff = INT_MIN;
    int temp = n;
   
    // Iterate from 0 to n - 1
    for (int i = 0; i < n; i++) {
        if (temp > hashmap[arr[i]][0]) {
            temp = hashmap[arr[i]][0];
        }
        maxDiff = max(
            maxDiff,
            hashmap[arr[i]][hashmap[arr[i]].size() - 1]
                - temp);
    }
    return maxDiff;
}
 
// Driver Code
int main()
{
 
    int n = 9;
    vector<int> arr{ 34, 8, 10, 3, 2, 80, 30, 33, 1 };
 
    // Function Call
    int ans = maxIndexDiff(arr, n);
    cout << "The maxIndexDiff is : " << ans << endl;
 
    return 1;
}


Java




// Java implementation of
// the hashmap approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find maximum
// index difference
static int maxIndexDiff(ArrayList<Integer> arr, int n)
{
     
    // Initialise unordered_map
    Map<Integer,
    ArrayList<Integer>> hashmap = new HashMap<Integer,
                                    ArrayList<Integer>>();
         
    // Iterate from 0 to n - 1
    for(int i = 0; i < n; i++)
    {
        if(hashmap.containsKey(arr.get(i)))
        {
            hashmap.get(arr.get(i)).add(i);
        }
        else
        {
            hashmap.put(arr.get(i), new ArrayList<Integer>());
            hashmap.get(arr.get(i)).add(i);
        }
    }
     
    // Sort arr
    Collections.sort(arr);
    int maxDiff = Integer.MIN_VALUE;
    int temp = n;
     
    // Iterate from 0 to n - 1
    for(int i = 0; i < n; i++)
    {
        if (temp > hashmap.get(arr.get(i)).get(0))
        {
            temp = hashmap.get(arr.get(i)).get(0);
        }
        maxDiff = Math.max(maxDiff,
        hashmap.get(arr.get(i)).get(
            hashmap.get(arr.get(i)).size() - 1) - temp);
    }
    return maxDiff;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 9;
    ArrayList<Integer> arr = new ArrayList<Integer>(
        Arrays.asList(34, 8, 10, 3, 2, 80, 30, 33, 1));
         
    // Function Call
    int ans = maxIndexDiff(arr, n);
     
    System.out.println("The maxIndexDiff is : " + ans);
}
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 implementation of the above approach
n = 9
a = [34, 8, 10, 3, 2, 80, 30, 33, 1]
 
# To store the index of an element.
index = dict()
for i in range(n):
    if a[i] in index:
 
        # append to list (for duplicates)
        index[a[i]].append(i) 
    else:
 
        # if first occurrence
        index[a[i]] = [i]  
 
# sort the input array
a.sort()    
maxDiff = 0
 
# Temporary variable to keep track of minimum i
temp = n    
for i in range(n):
    if temp > index[a[i]][0]:
        temp = index[a[i]][0]
    maxDiff = max(maxDiff, index[a[i]][-1]-temp)
 
print(maxDiff)


C#




// C# implementation of
// the hashmap approach
 
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Function to find maximum
  // index difference
  static int maxIndexDiff(List<int> arr, int n)
  {
    Dictionary<int,List<int>> hashmap = new Dictionary<int,List<int>>();
 
    // Iterate from 0 to n - 1
    for(int i = 0; i < n; i++)
    {
      if(hashmap.ContainsKey(arr[i]))
      {
        hashmap[arr[i]].Add(i);
      }
      else
      {
        hashmap.Add(arr[i], new List<int>());
        hashmap[arr[i]].Add(i);
      }
    }
 
    // Sort arr
    arr.Sort();
 
    int maxDiff = -1;
    int temp = n;
 
    // Iterate from 0 to n - 1
    for(int i = 0; i < n; i++)
    {
      if(temp > hashmap[arr[i]][0] )
      {
        temp = hashmap[arr[i]][0];
      }
      maxDiff = Math.Max(maxDiff,hashmap[arr[i]][hashmap[arr[i]].Count - 1]- temp);
    }
    return maxDiff;
  }
   
  // Driver Code
  static public void Main (){
    int n = 9;
    List<int> arr = new List<int>();
    arr.Add(34);
    arr.Add(8);
    arr.Add(10);
    arr.Add(3);
    arr.Add(2);
    arr.Add(80);
    arr.Add(30);
    arr.Add(33);
    arr.Add(1);
     
    // Function Call
    int ans = maxIndexDiff(arr, n);
 
    Console.WriteLine("The maxIndexDiff is : " + ans );
  }
}
 
// This code is contributed by rag2127.


Javascript




<script>
// JavaScript implementation of
// the hashmap approach
  
// Function to find maximum
// index difference
function maxIndexDiff(arr,n)
{
    
    // Initialise map in JavaScript
    let hashmap = new Map()
  
    // Iterate from 0 to n - 1
    for (let i = 0; i < n; i++) {
     
        hashmap[arr[i]] = hashmap[arr[i]] || []
        hashmap[arr[i]].push(i)
     
    }
  
    // Sort arr
    arr.sort((a,b)=> (a - b))
 
    let maxDiff = 0
    let temp = n
    
    // Iterate from 0 to n - 1
    for (let i = 0; i < n; i++) {
 
        if (temp > hashmap[arr[i]][0]) {
            temp = hashmap[arr[i]][0]
        }
        maxDiff = Math.max( maxDiff,hashmap[arr[i]][hashmap[arr[i]].length - 1]- temp )
 
    }
    return maxDiff
}
  
// Driver Code
 
let n = 9
const arr = [ 34, 8, 10, 3, 2, 80, 30, 33, 1 ]
  
// Function Call
let ans = maxIndexDiff(arr, n)
document.write(`The maxIndexDiff is : ${ans}`)
  
 // This code is contributed by shinjanpatra
 </script>


Output

The maxIndexDiff is : 6

Time complexity : O(N*log(N)) 
Auxiliary Space: O(N)

Method 4 (Efficient): To solve this problem, we need to get two optimum indexes of arr[]: left index i and right index j. For an element arr[i], we do not need to consider arr[i] for left index if there is an element smaller than arr[i] on left side of arr[i]. Similarly, if there is a greater element on right side of arr[j] then we do not need to consider this j for the right index. So we construct two auxiliary arrays LMin[] and RMax[] such that LMin[i] holds the smallest element on left side of arr[i] including arr[i], and RMax[j] holds the greatest element on right side of arr[j] including arr[j]. After constructing these two auxiliary arrays, we traverse both of these arrays from left to right. While traversing LMin[] and RMax[] if we see that LMin[i] is greater than RMax[j], then we must move ahead in LMin[] (or do i++) because all elements on left of LMin[i] are greater than or equal to LMin[i]. Otherwise, we must move ahead in RMax[j] to look for a greater j – i value.

Thanks to celicom for suggesting the algorithm for this method. 

Working Example:

Lets consider any example [7 3 1 8 9 10 4 5 6]

what is maxRight ?

Filling from right side 6 is first element now 6 > 5 so again we fill 6 till we reach 10 > 6 :

[10 10 10 10 10 10 6 6 6] this is maxR

[7 3 1 1 1 1 1 1 1 ] this is minL

now we see that how to reach answer from these to and its proof !!!

lets compare first elements of the arrays now we see 10 > 7,

now we increase maxR by 1 till it becomes lesser than 7 i.e at index 5

hence answer till now is. 5-0 = 5

now we will increase minL we get 3 which is lesser than 6 so we increase maxR till it reaches last index and the answer becomes 8-1= 7

so we see how we are getting correct answer.

As we need the max difference j – i such that A[i]<= A[j], hence we do not need to consider element after the index j and element before index i.

in previous hint, make 2 arrays,

First, will store smallest occurring element before the element

Second, will store largest occurring element after the element

Traverse the Second array, till the element in second array is larger than or equal to First array, and store the index difference. And if it becomes smaller, traverse the first array till it again becomes larger.

And store the max difference of this index difference.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
/* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
    int maxDiff;
    int i, j;
 
    int* LMin = new int[(sizeof(int) * n)];
    int* RMax = new int[(sizeof(int) * n)];
 
    /* Construct LMin[] such that
    LMin[i] stores the minimum value
    from (arr[0], arr[1], ... arr[i]) */
    LMin[0] = arr[0];
    for (i = 1; i < n; ++i)
        LMin[i] = min(arr[i], LMin[i - 1]);
 
    /* Construct RMax[] such that
    RMax[j] stores the maximum value from
    (arr[j], arr[j+1], ..arr[n-1]) */
    RMax[n - 1] = arr[n - 1];
    for (j = n - 2; j >= 0; --j)
        RMax[j] = max(arr[j], RMax[j + 1]);
 
    /* Traverse both arrays from left to right
    to find optimum j - i. This process is similar to
    merge() of MergeSort */
    i = 0, j = 0, maxDiff = -1;
    while (j < n && i < n) {
        if (LMin[i] <= RMax[j]) {
            maxDiff = max(maxDiff, j - i);
            j = j + 1;
        }
        else
            i = i + 1;
    }
 
    return maxDiff;
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 2, 3, 4, 5,
                  6, 7, 8, 18, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    cout << maxDiff;
    return 0;
}
 
// This code is contributed by rathbhupendra


C




#include <stdio.h>
 
/* Utility Functions to get max and minimum of two integers */
int max(int x, int y)
{
    return x > y ? x : y;
}
 
int min(int x, int y)
{
    return x < y ? x : y;
}
 
/* For a given array arr[], returns the maximum j – i such that
    arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
    int maxDiff;
    int i, j;
 
    int* LMin = (int*)malloc(sizeof(int) * n);
    int* RMax = (int*)malloc(sizeof(int) * n);
 
    /* Construct LMin[] such that LMin[i] stores the minimum value
       from (arr[0], arr[1], ... arr[i]) */
    LMin[0] = arr[0];
    for (i = 1; i < n; ++i)
        LMin[i] = min(arr[i], LMin[i - 1]);
 
    /* Construct RMax[] such that RMax[j] stores the maximum value
       from (arr[j], arr[j+1], ..arr[n-1]) */
    RMax[n - 1] = arr[n - 1];
    for (j = n - 2; j >= 0; --j)
        RMax[j] = max(arr[j], RMax[j + 1]);
 
    /* Traverse both arrays from left to right to find optimum j - i
        This process is similar to merge() of MergeSort */
    i = 0, j = 0, maxDiff = -1;
    while (j < n && i < n) {
        if (LMin[i] <= RMax[j]) {
            maxDiff = max(maxDiff, j - i);
            j = j + 1;
        }
        else
            i = i + 1;
    }
 
    return maxDiff;
}
 
/* Driver program to test above functions */
int main()
{
    int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    printf("\n %d", maxDiff);
    getchar();
    return 0;
}


Java




public class FindMaximum {
    /* Utility Functions to get max and minimum of two integers */
    int max(int x, int y)
    {
        return x > y ? x : y;
    }
 
    int min(int x, int y)
    {
        return x < y ? x : y;
    }
 
    /* For a given array arr[], returns the maximum j-i such that
       arr[j] > arr[i] */
    int maxIndexDiff(int arr[], int n)
    {
        int maxDiff;
        int i, j;
 
        int RMax[] = new int[n];
        int LMin[] = new int[n];
 
        /* Construct LMin[] such that LMin[i] stores the minimum value
           from (arr[0], arr[1], ... arr[i]) */
        LMin[0] = arr[0];
        for (i = 1; i < n; ++i)
            LMin[i] = min(arr[i], LMin[i - 1]);
 
        /* Construct RMax[] such that RMax[j] stores the maximum value
           from (arr[j], arr[j+1], ..arr[n-1]) */
        RMax[n - 1] = arr[n - 1];
        for (j = n - 2; j >= 0; --j)
            RMax[j] = max(arr[j], RMax[j + 1]);
 
        /* Traverse both arrays from left to right to find optimum j - i
           This process is similar to merge() of MergeSort */
        i = 0;
        j = 0;
        maxDiff = -1;
        while (j < n && i < n) {
            if (LMin[i] <= RMax[j]) {
                maxDiff = max(maxDiff, j - i);
                j = j + 1;
            }
            else
                i = i + 1;
        }
 
        return maxDiff;
    }
 
    /* Driver program to test the above functions */
    public static void main(String[] args)
    {
        FindMaximum max = new FindMaximum();
        int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
        int n = arr.length;
        int maxDiff = max.maxIndexDiff(arr, n);
        System.out.println(maxDiff);
    }
}


Python3




# Utility Functions to get max
# and minimum of two integers
def max(a, b):
    if(a > b):
        return a
    else:
        return b
 
def min(a, b):
    if(a < b):
        return a
    else:
        return b
 
# For a given array arr[],
# returns the maximum j - i
# such that arr[j] > arr[i]
def maxIndexDiff(arr, n):
    maxDiff = 0;
    LMin = [0] * n
    RMax = [0] * n
 
    # Construct LMin[] such that
    # LMin[i] stores the minimum
    # value from (arr[0], arr[1],
    # ... arr[i])
    LMin[0] = arr[0]
    for i in range(1, n):
        LMin[i] = min(arr[i], LMin[i - 1])
 
    # Construct RMax[] such that
    # RMax[j] stores the maximum
    # value from (arr[j], arr[j + 1],
    # ..arr[n-1])
    RMax[n - 1] = arr[n - 1]
    for j in range(n - 2, -1, -1):
        RMax[j] = max(arr[j], RMax[j + 1]);
 
    # Traverse both arrays from left
    # to right to find optimum j - i
    # This process is similar to
    # merge() of MergeSort
    i, j = 0, 0
    maxDiff = -1
    while (j < n and i < n):
        if (LMin[i] <= RMax[j]):
            maxDiff = max(maxDiff, j - i)
            j = j + 1
        else:
            i = i + 1
 
    return maxDiff
 
# Driver Code
if(__name__ == '__main__'):
    arr = [9, 2, 3, 4, 5,
           6, 7, 8, 18, 0]
    n = len(arr)
    maxDiff = maxIndexDiff(arr, n)
    print (maxDiff)
 
# This code is contributed
# by gautam karakoti


C#




// C# program to find the maximum
// j – i such that arr[j] > arr[i]
using System;
 
class GFG {
    // Utility Functions to get max
    // and minimum of two integers
    static int max(int x, int y)
    {
        return x > y ? x : y;
    }
 
    static int min(int x, int y)
    {
        return x < y ? x : y;
    }
 
    // For a given array arr[], returns
    // the maximum j-i such thatarr[j] > arr[i]
    static int maxIndexDiff(int[] arr, int n)
    {
        int maxDiff;
        int i, j;
 
        int[] RMax = new int[n];
        int[] LMin = new int[n];
 
        // Construct LMin[] such that LMin[i]
        // stores the minimum value
        // from (arr[0], arr[1], ... arr[i])
        LMin[0] = arr[0];
        for (i = 1; i < n; ++i)
            LMin[i] = min(arr[i], LMin[i - 1]);
 
        // Construct RMax[] such that
        // RMax[j] stores the maximum value
        // from (arr[j], arr[j+1], ..arr[n-1])
        RMax[n - 1] = arr[n - 1];
        for (j = n - 2; j >= 0; --j)
            RMax[j] = max(arr[j], RMax[j + 1]);
 
        // Traverse both arrays from left
        // to right to find optimum j - i
        // This process is similar to merge()
        // of MergeSort
        i = 0;
        j = 0;
        maxDiff = -1;
        while (j < n && i < n) {
            if (LMin[i] <= RMax[j]) {
                maxDiff = max(maxDiff, j - i);
                j = j + 1;
            }
            else
                i = i + 1;
        }
 
        return maxDiff;
    }
 
    // Driver program
    public static void Main()
    {
 
        int[] arr = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
        int n = arr.Length;
        int maxDiff = maxIndexDiff(arr, n);
        Console.Write(maxDiff);
    }
}
// This Code is Contributed by Sam007


PHP




<?php
// PHP program to find the maximum
// j – i such that arr[j] > arr[i]
 
// For a given array arr[],
// returns the maximum j - i
// such that arr[j] > arr[i]
function maxIndexDiff($arr, $n)
{
    $maxDiff = 0;
    $LMin = array_fill(0, $n, NULL);
    $RMax = array_fill(0, $n, NULL);
 
    // Construct LMin[] such that
    // LMin[i] stores the minimum
    // value from (arr[0], arr[1],
    // ... arr[i])
    $LMin[0] = $arr[0];
    for($i = 1; $i < $n; $i++)
        $LMin[$i] = min($arr[$i],
                        $LMin[$i - 1]);
 
    // Construct RMax[] such that
    // RMax[j] stores the maximum
    // value from (arr[j], arr[j+1],
    // ..arr[n-1])
    $RMax[$n - 1] = $arr[$n - 1];
    for($j = $n - 2; $j >= 0; $j--)
        $RMax[$j] = max($arr[$j],
                        $RMax[$j + 1]);
 
    // Traverse both arrays from left
    // to right to find optimum j - i
    // This process is similar to
    // merge() of MergeSort
    $i = 0;
    $j = 0;
    $maxDiff = -1;
    while ($j < $n && $i < $n)
        if ($LMin[$i] <= $RMax[$j])
        {
            $maxDiff = max($maxDiff, $j - $i);
            $j = $j + 1;
        }
        else
            $i = $i + 1;
 
    return $maxDiff;
}
 
// Driver Code
$arr = array(9, 2, 3, 4, 5,
             6, 7, 8, 18, 0);
$n = sizeof($arr);
$maxDiff = maxIndexDiff($arr, $n);
echo $maxDiff;
 
// This code is contributed
// by ChitraNayal
?>


Javascript




<script>
    // Javascript program to find the maximum
    // j – i such that arr[j] > arr[i]
     
    // Utility Functions to get max
    // and minimum of two integers
    function max(x, y)
    {
        return x > y ? x : y;
    }
  
    function min(x, y)
    {
        return x < y ? x : y;
    }
  
    // For a given array arr[], returns
    // the maximum j-i such thatarr[j] > arr[i]
    function maxIndexDiff(arr, n)
    {
        let maxDiff;
        let i, j;
  
        let RMax = new Array(n);
        let LMin = new Array(n);
  
        // Construct LMin[] such that LMin[i]
        // stores the minimum value
        // from (arr[0], arr[1], ... arr[i])
        LMin[0] = arr[0];
        for (i = 1; i < n; ++i)
            LMin[i] = min(arr[i], LMin[i - 1]);
  
        // Construct RMax[] such that
        // RMax[j] stores the maximum value
        // from (arr[j], arr[j+1], ..arr[n-1])
        RMax[n - 1] = arr[n - 1];
        for (j = n - 2; j >= 0; --j)
            RMax[j] = max(arr[j], RMax[j + 1]);
  
        // Traverse both arrays from left
        // to right to find optimum j - i
        // This process is similar to merge()
        // of MergeSort
        i = 0;
        j = 0;
        maxDiff = -1;
        while (j < n && i < n) {
            if (LMin[i] <= RMax[j]) {
                maxDiff = max(maxDiff, j - i);
                j = j + 1;
            }
            else
                i = i + 1;
        }
  
        return maxDiff;
    }
     
    let arr = [ 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 ];
    let n = arr.length;
    let maxDiff = maxIndexDiff(arr, n);
    document.write(maxDiff);
</script>


Output

8

Time Complexity: O(n) 
Auxiliary Space: O(n) 

Asked in: Amazon,  Google,  VMWare

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

Another Approach: ( only using one extra array ): We consider an auxiliary array : rightMax[] , such that, rightMax[i] = max element of the subarray arr[i…(n-1)], the largest or equal element after arr[i] element Suppose (arr[i], arr[jLast] ) is a pair, such that arr[jLast] is the last greater or equal element than arr[i]. For the pairs ending with arr[jLast] :  ( arr[k], arr[jLast] ) for all k = (i+1) to jLast we don’t need to consider (jLast – k) because (jLast – i ) > (jLast – k) for all such k’s. So we can skip those pairs. Traversing from left to right of both arrays : arr[] and rightMax[]  , when we first encounter rightMax[j] < arr[i[  , we know that jLast = j-1, and we can skip the pairs (arr[k], arr[jLast]) for all k = (i+1) to jLast. And also rightMax[] is non increasing sequence , so all elements at right side of rightMax[j] is smaller than or equal to rightMax[j]. But there may be arr[x]  after arr[i] (x > i) such that arr[x] < rightMax[j] for x > i, so increment i when rightMax[j] < arr[i] is encountered.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
/* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
 
        int rightMax[n];
        rightMax[n-1]= arr[n-1];
        for(int i = n-2; i>=0; i--)
            rightMax[i] = max(rightMax[i+1] , arr[i]);
         
        //rightMax[i] = max{ arr[i...(n-1] }
 
 
         
        int maxDist = INT_MIN;
        int i = 0, j = 0;
        while(i<n && j<n)
        {
            if(rightMax[j] >= arr[i])
            {
                maxDist = max( maxDist, j-i );
                j++;
            }
            else // if(rightMax[j] < leftMin[i])
                i++;
        }
         
        return maxDist;
         
         
}
 
// Driver Code
int main()
{
    int arr[] = { 34,8,10,3,2,80,30,33,1};
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    cout << maxDiff;
    return 0;
}
 
// This code is contributed by Sourashis Mondal


Java




import java.util.*;
 
class GFG{
 
  /* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
  static int maxIndexDiff(int arr[], int n)
  {
 
    int []rightMax = new int[n];
    rightMax[n-1]= arr[n-1];
    for(int i = n-2; i>=0; i--)
      rightMax[i] = Math.max(rightMax[i+1] , arr[i]);
 
    // rightMax[i] = max{ arr[i...(n-1] }
    int maxDist = Integer.MIN_VALUE;
    int i = 0, j = 0;
    while(i < n && j < n)
    {
      if(rightMax[j] >= arr[i])
      {
        maxDist = Math.max( maxDist, j-i );
        j++;
      }
      else // if(rightMax[j] < leftMin[i])
        i++;
    }
 
    return maxDist;
 
 
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = {34, 8, 10, 3, 2, 80, 30, 33, 1};
    int n = arr.length;
    int maxDiff = maxIndexDiff(arr, n);
    System.out.print(maxDiff);
  }
}
 
// This code is contributed by Rajput-Ji


Python3




# For a given array arr[], returns the
# maximum j – i such that arr[j] > arr[i]
def maxIndexDiff(arr, n):
     
    rightMax = [0] * n
    rightMax[n - 1] = arr[n - 1]
    for i in range(n - 2, -1, -1):
        rightMax[i] = max(rightMax[i + 1], arr[i])
         
    # rightMax[i] = max arr[i...(n-1]
    maxDist = -2**31
    i = 0
    j = 0
     
    while (i < n and j < n):
        if (rightMax[j] >= arr[i]):
            maxDist = max(maxDist, j - i)
            j += 1
             
        else:
             
            # if(rightMax[j] < leftMin[i])
            i += 1
     
    return maxDist
 
# Driver Code
arr = [ 34, 8, 10, 3, 2, 80, 30, 33, 1 ]
n = len(arr)
maxDiff = maxIndexDiff(arr, n)
 
print(maxDiff)
 
# This code is contributed by Shubham Singh


C#




/* For a given array arr[], 
returns the maximum j – i such that
arr[j] > arr[i] */
using System;
 
public class GFG
{
 
  static int maxIndexDiff(int[] arr, int n)
  {
  
    int []rightMax = new int[n];
    rightMax[n - 1] = arr[n - 1];
    int i = 0, j = 0;
    for(i = n - 2; i >= 0; i--)
      rightMax[i] = Math.Max(rightMax[i+1] , arr[i]);
 
    // rightMax[i] = max{ arr[i...(n-1] }
    int maxDist = Int32.MinValue;
    i = 0;
    while(i < n && j < n)
    {
      if(rightMax[j] >= arr[i])
      {
        maxDist = Math.Max( maxDist, j - i);
        j++;
      }
      else // if(rightMax[j] < leftMin[i])
        i++;
    }
 
    return maxDist;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = {34, 8, 10, 3, 2, 80, 30, 33, 1};
    int n = arr.Length;
    int maxDiff = maxIndexDiff(arr, n);
    Console.Write(maxDiff);
  }
}
 
// This code is contributed by Shubham Singh


Javascript




<script>
 
/* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
function maxIndexDiff(arr, n)
{
    var rightMax = new Array(n).fill(0);;
    rightMax[n - 1] = arr[n - 1];
    for(var i = n - 2; i >= 0; i--){
        rightMax[i] = Math.max(rightMax[i+1] , arr[i]);
    }
     
    // rightMax[i] = max{ arr[i...(n-1] }
    var maxDist = Number.MIN_VALUE;
    var i = 0;
    var j = 0;
    while(i < n && j < n)
    {
        if(rightMax[j] >= arr[i])
        {
            maxDist = Math.max( maxDist, j-i );
            j++;
        }
        else // if(rightMax[j] < leftMin[i])
        {    i++;
        }
     }
     return maxDist;   
}
 
// Driver Code
var arr = [ 34,8,10,3,2,80,30,33,1];
var n = arr.length;
var maxDiff = maxIndexDiff(arr, n);
document.write(maxDiff);
 
// This code is contributed by Shubham Singh
 
</script>


Output

6

Time complexity: O(n), As i and j pointers are traversing at most n elements, time complexity  = O(n) + O(n) = O(n)
Auxiliary Space: O(n)

Using leftMin[]: We can also do this using leftMin[] array only , where leftMin[i] = min element of the subarray arr[0…i]

C++




#include <bits/stdc++.h>
using namespace std;
 
/* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
    int leftMin[n] ;
    leftMin[0] = arr[0];
    for(int i = 1 ; i<n; i++)
        leftMin[i] = min(leftMin[i-1], arr[i]);
 
 
    //leftMin[i] = min{ arr[0...i] }
 
 
    int maxDist = INT_MIN;
    int i = n-1, j = n-1;
 
    while(i>=0  &&  j>=0)
    {
        if(arr[j] >= leftMin[i])
        {
            maxDist = max(maxDist, j-i);
            i--;
        }
        else
            j--;
    }
 
    return maxDist;
 
         
}
 
// Driver Code
int main()
{
    int arr[] = { 34,8,10,3,2,80,30,33,1};
    int n = sizeof(arr) / sizeof(arr[0]);
    int maxDiff = maxIndexDiff(arr, n);
    cout << maxDiff;
    return 0;
}
 
// This code is contributed by Sourashis Mondal


Java




import java.util.*;
class GFG
{
 
  /* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
  static int maxIndexDiff(int arr[], int n)
  {
 
    int []leftMin = new int[n];
    leftMin[0] = arr[0];
    for(int i = 1; i < n; i++)
      leftMin[i] = Math.min(leftMin[i - 1] , arr[i]);
 
    // leftMin[i] = min{ arr[i...(n-1] }
 
    int maxDist = Integer.MIN_VALUE;
    int i = n - 1, j = n - 1;
    while(i >= 0 && j >= 0)
    {
      if(arr[j] >= leftMin[i])
      {
        maxDist = Math.max( maxDist, j - i );
        i--;
      }
      else
        j--;
    }
 
    return maxDist;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = {34, 8, 10, 3, 2, 80, 30, 33, 1};
    int n = arr.length;
    int maxDiff = maxIndexDiff(arr, n);
    System.out.print(maxDiff);
  }
}
 
// This code is contributed by Shubham Singh


Python3




# For a given array arr[], 
#   returns the maximum j – i such that
#   arr[j] > arr[i] */
def maxIndexDiff(arr, n):
     
    leftMin = [0]*n
    leftMin[0] = arr[0]
    for i in range(1,n):
        leftMin[i] = min(leftMin[i-1], arr[i])
         
    # leftMin[i] = min arr[0...i] 
    maxDist = - 2**32
    i = n-1
    j = n-1
     
    while(i>=0  and  j>=0):
         
        if(arr[j] >= leftMin[i]):
            maxDist = max(maxDist, j-i)
            i-=1
        else:
            j-=1
             
    return maxDist
 
# Driver Code
arr = [34,8,10,3,2,80,30,33,1]
n = len(arr)
maxDiff = maxIndexDiff(arr, n)
print(maxDiff)
 
# This code is contributed by Shubham Singh


C#




using System;
 
public class GFG{
 
  /* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
  static int maxIndexDiff(int[] arr, int n)
  {
 
    int []leftMin = new int[n];
    leftMin[0] = arr[0];
    int i,j;
    for( i = 1; i < n; i++)
      leftMin[i] = Math.Min(leftMin[i - 1] , arr[i]);
 
    // leftMin[i] = min{ arr[i...(n-1] }
 
    int maxDist = Int32.MinValue;
    i = n - 1;
    j = n - 1;
    while(i >= 0 && j >= 0)
    {
      if(arr[j] >= leftMin[i])
      {
        maxDist = Math.Max( maxDist, j - i );
        i--;
      }
      else
        j--;
    }
 
    return maxDist;
  }
 
  // Driver Code
  static public void Main ()
  {
    int[] arr = {34, 8, 10, 3, 2, 80, 30, 33, 1};
    int n = arr.Length;
    int maxDiff = maxIndexDiff(arr, n);
    Console.Write(maxDiff);
  }
}
 
// This code is contributed by Shubham Singh


Javascript




<script>
 
/* For a given array arr[], 
   returns the maximum j – i such that
   arr[j] > arr[i] */
function maxIndexDiff(arr, n)
{
    var leftMin = new Array(n).fill(0);;
    leftMin[0] = arr[0];
    for(var i = 1; i < n; i++){
        leftMin[i] = Math.min(leftMin[i-1] , arr[i]);
    }
     
    // leftMin[i] = min{ arr[i...(n-1] }
    var maxDist = Number.MIN_VALUE;
    var i = n-1;
    var j = n-1;
    while(i >= 0 && j >= 0)
    {
        if(arr[j] >= leftMin[i])
        {
            maxDist = Math.max( maxDist, j-i );
            i--;
        }
        else // if(rightMax[j] < leftMin[i])
        {    j--;
        }
     }
     return maxDist;   
}
 
// Driver Code
var arr = [ 34,8,10,3,2,80,30,33,1];
var n = arr.length;
var maxDiff = maxIndexDiff(arr, n);
document.write(maxDiff);
 
// This code is contributed by Shubham Singh
 
</script>


Output

6

Time Complexity: O(n) 
Auxiliary Space: O(n) 

Please suggest if someone has a better solution that is more efficient in terms of space and time.

Another approach : Using Stack

Idea is :

1) First we do one traversal from left to right of array and store only those indexes  in stack whose array values appears in decreasing order in stack.. This is because, say if any value at i and j is satisfying our condition A[ i ]<=A[ j ] and if any index k that appears before ith index  i.e.  k<i and if  A[ k ]<= A[ i ] then value at kth index will also satisfy the condition A[k]<=A[ j ] . so we can ignore ith value .

2) Now we simpaly traverse from right of array with index i and comapre it with the top of stack . 

if we find A[stack.peek()] <=A [i] we pop from stack and compare with next value else we decrement our i counter

at any point in loop if  A[stack.peek()] <=A [i]. We compute tempMax = i – stack.peeak(). And keep updating maxSofar result.

below is the code for this Approach:

C++




#include<bits/stdc++.h>
using namespace std;
 
int maxIndexDiff(int A[], int N) {
     
    stack<int> stkForIndex;
     
    //loop for storing index in stack whose value appears in decreasing order
    for(int i=0;i<N;i++){           
        if(stkForIndex.empty() || A[stkForIndex.top()]>A[i])
            stkForIndex.push(i);
    }
 
    int maxDiffSoFar = 0;
    int tempdiff;
     
    //Now we traverse from right to left.
    int i = N-1;
    while(i>=0){
         
        /*
        This will compare top value of array at index stack top.
        if it satisfy our condition we check the difference  and update out result.
        else we decrement our counter
        */
        if(!stkForIndex.empty() && A[stkForIndex.top()] <= A[i]){
            tempdiff = i - stkForIndex.top();
            stkForIndex.pop();
            if(tempdiff>maxDiffSoFar){
                maxDiffSoFar  = tempdiff;
            }
            continue;
        }           
        i--;
    }
 
    return maxDiffSoFar;
     
    //This Code is Contributed by Ashwini Chourasia
}
     
int main() {
     
    int A[] = {34,8,10,3,2,80,30,33,1};       
    int N = sizeof(A) / sizeof(int);
    cout<<"Max diff be : " << maxIndexDiff(A, N);
}


Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.Stack;
 
class GFG {
    public static void main(String[] args) {
         
        int A[] = {34,8,10,3,2,80,30,33,1};       
        int N = A.length;
        System.out.println("Max diff be : " + maxIndexDiff(A, N));
    }
     
    static int maxIndexDiff(int A[], int N) {
         
        Stack<Integer> stkForIndex = new Stack<>();
         
        //loop for storing index in stack whose value appears in decreasing order
        for(int i=0;i<N;i++){           
            if(stkForIndex.isEmpty() || A[stkForIndex.peek()]>A[i])
                stkForIndex.push(i);
        }
 
        int maxDiffSoFar = 0;
        int tempdiff;
         
        //Now we traverse from right to left.
        int i = N-1;
        while(i>=0){
             
            /*
            This will compare top value of array at index stack top.
            if it satisfy our condition we check the difference  and update out result.
            else we decrement our counter
            */
            if(!stkForIndex.isEmpty() && A[stkForIndex.peek()] <= A[i]){
                tempdiff = i - stkForIndex.pop();
                if(tempdiff>maxDiffSoFar){
                    maxDiffSoFar  = tempdiff;
                }
                continue;
            }           
            i--;
        }
 
        return maxDiffSoFar;
         
        //This Code is Contributed by Ashwini Chourasia
    }
}


Python3




def maxIndexDiff( A, N) :
    stkForIndex = list();
     
    # loop for storing index in stack whose value appears in decreasing order
    for i in range(0, N):          
        if(len(stkForIndex) == 0 or A[stkForIndex[-1]] > A[i]):
            stkForIndex.append(i);
 
    maxDiffSoFar = 0;
    tempdiff=-1;
     
    # Now we traverse from right to left.
    i = N-1;
    while(i >= 0):
         
        # This will compare top value of array at index stack top.
        # if it satisfy our condition we check the difference  and update out result.
        # else we decrement our counter
        if(len(stkForIndex) and A[stkForIndex[-1]] <= A[i]):
            tempdiff = i - stkForIndex.pop();
            if(tempdiff > maxDiffSoFar):
                maxDiffSoFar  = tempdiff;
            continue;
        i -= 1;
    return maxDiffSoFar;
     
A = [34,8,10,3,2,80,30,33,1];       
N = len(A);
print("Max diff be :" , maxIndexDiff(A, N));
 
# This code is contributed by ratiagrawal.


C#




// C# implementation
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
 
public class Gfg
{
 
  static int maxIndexDiff(int[] A, int N)
  {
 
    Stack<int> stkForIndex = new Stack<int>();
 
    // loop for storing index in stack whose value appears in decreasing order
    int i = 0;
    for(i = 0; i < N; i++){           
      if(stkForIndex.Count == 0 || A[stkForIndex.Peek()] > A[i])
        stkForIndex.Push(i);
    }
 
    int maxDiffSoFar = 0;
    int tempdiff;
 
    //Now we traverse from right to left.
    i = N-1;
    while(i >= 0){
 
      /*
            This will compare top value of array at index stack top.
            if it satisfy our condition we check the difference  and update out result.
            else we decrement our counter
            */
      if(stkForIndex.Count!=0 && A[stkForIndex.Peek()] <= A[i]){
        tempdiff = i - stkForIndex.Peek();
        stkForIndex.Pop();
        if(tempdiff > maxDiffSoFar){
          maxDiffSoFar  = tempdiff;
        }
        continue;
      }           
      i--;
    }
 
    return maxDiffSoFar;
  }
 
  public static void Main(string[] args)
  {   
    int[] A = {34,8,10,3,2,80,30,33,1};       
    int N = A.Length;
    Console.Write("Max diff be : " + maxIndexDiff(A, N));
  }
}
 
// This code is contributed by poojaagarwal2.


Javascript




// Javascript implementation
function maxIndexDiff(A, N)
{
  let stkForIndex = [];
 
  // loop for storing index in stack
  // whose value appears in decreasing order
  for(let i = 0; i < N; i++)
  {
if(stkForIndex.length === 0 || A[stkForIndex[stkForIndex.length - 1]] > A[i]) {
  stkForIndex.push(i);
}
  }
 
  let maxDiffSoFar = 0;
  let tempdiff;
 
  // Now we traverse from right to left.
  let i = N - 1;
  while(i >= 0) {
 
/*
This will compare top value of array at index stack top.
if it satisfy our condition we check the difference and update out result.
else we decrement our counter
*/
if(stkForIndex.length !== 0 && A[stkForIndex[stkForIndex.length - 1]] <= A[i]) {
  tempdiff = i - stkForIndex[stkForIndex.length - 1];
  stkForIndex.pop();
  if(tempdiff > maxDiffSoFar) {
    maxDiffSoFar = tempdiff;
  }
  continue;
}
i--;
  }
 
  return maxDiffSoFar;
}
 
let A = [34, 8, 10, 3, 2, 80, 30, 33, 1];
let N = A.length;
console.log("Max diff be : " + maxIndexDiff(A, N));
 
 
// This code is contributed by Aman Kumar.


Output

Max diff be : 6

Time complexity: O(n),  As it requires two traversal  so TC is O(n) + O(n) = O(n)
Auxiliary Space: O(n)  O(n) is its worst case complexity in case whole array is already sorted in decreasing order.Other wise it stores only those values which appears in decreasing order.



Last Updated : 04 May, 2023
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