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

Given an array arr[], find the maximum j – i such that arr[j] > arr[i].

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: -1 


Method 1 (Simple but Inefficient)
Run two loops. In the outer loop, pick elements one by one from left. In the inner loop, compare the picked element with the elements starting from 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++

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#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[] = {9, 2, 3, 4, 5, 6, 7, 8, 18, 0};
    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)

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C

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#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;
}

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Java

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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);
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output :

8

Time Complexity : O(n^2)

Method 2 (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 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 RMa[] 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.

C

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#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;
}

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Java

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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);
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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

Asked in: AmazonGoogleVMWare

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



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