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Count indices where the maximum in the prefix array is less than that in the suffix array
  • Last Updated : 21 Apr, 2021

Given an array arr[] of size N, the task is to find the number of prefix arrays whose maximum is less than the maximum element in the remaining suffix array.

Examples:

Input: arr[] = {2, 3, 4, 8, 1, 4}
Output: 3
Explanation:
Prefix array = {2}, {2, 3}, {2, 3, 4}, {2, 3, 4, 8}, {2, 3, 4, 8, 1} 
Respective Suffix = {3, 4, 8, 1, 4}, {4, 8, 1, 4}, {8, 1, 4}, {1, 4}, {4}
Only the first 3 the prefix arrays have maximum valued element less than the maximum value element in the respective suffix array.

Input: arr[] = {4, 4, 4, 4, 5}
Output: 4

Naive Approach: The simplest approach is to find all possible prefixes and suffixes of the given array and count the number of indices where maximum element in the prefix array is less than the maximum element in the suffix array.



Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to store the maximum value for every prefix and suffix of the array and then calculate the number of prefix arrays with a maximum value less than its corresponding suffix array. Follow the steps below to solve the problem:

  • Initialize a variable, say ans and two arrays prefix[] and suffix[] of size N.
  • Traverse the array arr[] over the range [0, N – 1] and for each ith index in prefix[], store the maximum element as prefix[i] = max(prefix[i – 1], arr[i]).
  • Traverse the given array in the reverse order over the range [N – 1, 0] and for each ith index in suffix[], store the maximum element as suffix[i] = max(suffix[i + 1], arr[i]).
  • Now, traverse the array arr[] over the range [0, N – 2] and if prefix[i] is less than suffix[i], then increment ans by 1.
  • After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
void count(int a[], int n)
{
    // If size of array is 1
    if (n == 1) {
        cout << 0;
        return;
    }
 
    // pre[]: Prefix array
    // suf[]: Suffix array
    int pre[n - 1], suf[n - 1];
    int max = a[0];
 
    // Stores the required count
    int ans = 0, i;
 
    pre[0] = a[0];
 
    // Find the maximum in prefix array
    for (i = 1; i < n - 1; i++) {
 
        if (a[i] > max)
            max = a[i];
 
        pre[i] = max;
    }
 
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
 
    // Find the maximum in suffix array
    for (i = n - 2; i >= 1; i--) {
 
        if (a[i] > max)
            max = a[i];
 
        suf[i - 1] = max;
    }
 
    // Traverse the array
    for (i = 0; i < n - 1; i++) {
 
        // If maximum in prefix array
        // is less than maximum in
        // the suffix array
        if (pre[i] < suf[i])
            ans++;
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 8, 1, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    count(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
public class gfg
{
  // Function to print the count of indices
  // in which the maximum in prefix arrays
  // is less than that in the suffix array
  static void count(int a[], int n)
  {
 
    // If size of array is 1
    if (n == 1)
    {
      System.out.print(0);
      return;
    }
 
    // pre[]: Prefix array
    // suf[]: Suffix array
    int[] pre = new int[n - 1];
    int[] suf = new int[n - 1];
    int max = a[0];
 
    // Stores the required count
    int ans = 0, i;         
    pre[0] = a[0];
 
    // Find the maximum in prefix array
    for(i = 1; i < n - 1; i++)
    {
      if (a[i] > max)
        max = a[i];
 
      pre[i] = max;
    }       
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
 
    // Find the maximum in suffix array
    for(i = n - 2; i >= 1; i--)
    {
      if (a[i] > max)
        max = a[i];                 
      suf[i - 1] = max;
    }
 
    // Traverse the array
    for(i = 0; i < n - 1; i++)
    {
 
      // If maximum in prefix array
      // is less than maximum in
      // the suffix array
      if (pre[i] < suf[i])
        ans++;
    }
 
    // Print the answer
    System.out.print(ans);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = { 2, 3, 4, 8, 1, 4 };
    int N = arr.length;
 
    // Function Call
    count(arr, N);
  }
}
 
// This code is contributed by divyesh072019.

Python3




# Python program for the above approach
 
# Function to print the count of indices
# in which the maximum in prefix arrays
# is less than that in the suffix array
def count(a, n) :
 
    # If size of array is 1
    if (n == 1) :
        print(0)
        return
 
    # pre[]: Prefix array
    # suf[]: Suffix array
    pre = [0] * (n - 1)
    suf = [0] * (n - 1)
    max = a[0]
 
    # Stores the required count
    ans = 0
 
    pre[0] = a[0]
 
    # Find the maximum in prefix array
    for i in range(n-1):
 
        if (a[i] > max):
            max = a[i]
 
        pre[i] = max
 
    max = a[n - 1]
    suf[n - 2] = a[n - 1]
 
    # Find the maximum in suffix array
    for i in range(n-2, 0, -1):
 
        if (a[i] > max):
            max = a[i]
 
        suf[i - 1] = max
 
    # Traverse the array
    for i in range(n - 1):
 
        # If maximum in prefix array
        # is less than maximum in
        # the suffix array
        if (pre[i] < suf[i]) :
            ans += 1
 
    # Print the answer
    print(ans)
     
 
# Driver Code
 
arr = [ 2, 3, 4, 8, 1, 4 ]
N = len(arr)
 
# Function Call
count(arr, N)
 
# This code is contributed by code_hunt.

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
static void count(int[] a, int n)
{
     
    // If size of array is 1
    if (n == 1)
    {
        Console.Write(0);
        return;
    }
   
    // pre[]: Prefix array
    // suf[]: Suffix array
    int[] pre = new int[n - 1];
    int[] suf = new int[n - 1];
    int max = a[0];
   
    // Stores the required count
    int ans = 0, i;
     
    pre[0] = a[0];
     
    // Find the maximum in prefix array
    for(i = 1; i < n - 1; i++)
    {
        if (a[i] > max)
            max = a[i];
             
        pre[i] = max;
    }
   
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
     
    // Find the maximum in suffix array
    for(i = n - 2; i >= 1; i--)
    {
        if (a[i] > max)
            max = a[i];
             
        suf[i - 1] = max;
    }
   
    // Traverse the array
    for(i = 0; i < n - 1; i++)
    {
         
        // If maximum in prefix array
        // is less than maximum in
        // the suffix array
        if (pre[i] < suf[i])
            ans++;
    }
     
    // Print the answer
    Console.Write(ans);
}
 
// Driver code
static void Main()
{
    int[] arr = { 2, 3, 4, 8, 1, 4 };
    int N = arr.Length;
     
    // Function Call
    count(arr, N);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
    // Javascript program for the above approach
    // Function to print the count of indices
     
    // in which the maximum in prefix arrays
    // is less than that in the suffix array
    function count(a, n)
    {
 
        // If size of array is 1
        if (n == 1)
        {
            document.write(0);
            return;
        }
 
        // pre[]: Prefix array
        // suf[]: Suffix array
        let pre = new Array(n - 1);
        let suf = new Array(n - 1);
        let max = a[0];
 
        // Stores the required count
        let ans = 0, i;
 
        pre[0] = a[0];
 
        // Find the maximum in prefix array
        for(i = 1; i < n - 1; i++)
        {
            if (a[i] > max)
                max = a[i];
 
            pre[i] = max;
        }
 
        max = a[n - 1];
        suf[n - 2] = a[n - 1];
 
        // Find the maximum in suffix array
        for(i = n - 2; i >= 1; i--)
        {
            if (a[i] > max)
                max = a[i];
 
            suf[i - 1] = max;
        }
 
        // Traverse the array
        for(i = 0; i < n - 1; i++)
        {
 
            // If maximum in prefix array
            // is less than maximum in
            // the suffix array
            if (pre[i] < suf[i])
                ans++;
        }
 
        // Print the answer
        document.write(ans);
    }
     
    let arr = [ 2, 3, 4, 8, 1, 4 ];
    let N = arr.length;
      
    // Function Call
    count(arr, N);
     
    // This code is contributed by suresh07.
</script>
Output: 
3

 

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

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