Length of longest subarray in which elements greater than K are more than elements not greater than K

Given an array arr[] of length N. The task is to find the length of the longest subarray in which elements greater than a given number K are more than elements not greater than K.

Examples:

Input : N = 5, K = 2, arr[]={ 1, 2, 3, 4, 1 }
Output : 3
The subarray [2, 3, 4] or [3, 4, 1] satisfy the given condition, and there is no subarray of
length 4 or 5 which will hold the given condition, so the answer is 3.



Input : N = 4, K = 2, arr[]={ 6, 5, 3, 4 }
Output : 4

Approach:

  • Idea is to use the concept of binary search over Partial Sum .
  • First, replace the elements which are greater than K by 1 and other elements by -1 and calculate the prefix sum over it. Now if a subarray has a sum greater than 0, it implies that it holds more elements greater than K than elements which are less than K.
  • To find the answer use binary search over the answer. In each step of binary search, check every subarray of that length and then decide whether to go for larger length or not.

Below is the implementation of above Approach:

C++

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#include <bits/stdc++.h>
using namespace std;
// C++ implementation of above approach
  
// Function to find the length of a 
// longest subarray in which elements 
// greater than K are more than 
// elements not greater than K
int LongestSubarray(int a[], int n, int k)
{
  
    int pre[n] = { 0 };
  
    // Create a new array in which we store 1
    // if a[i] > k otherwise we store -1.
    for (int i = 0; i < n; i++) {
        if (a[i] > k)
            pre[i] = 1;
        else
            pre[i] = -1;
    }
  
    // Taking prefix sum over it
    for (int i = 1; i < n; i++) 
        pre[i] = pre[i - 1] + pre[i];
  
    // len will store maximum
    // length of subarray
    int len = 0; 
  
    int lo = 1, hi = n;
  
    while (lo <= hi) {
        int mid = (lo + hi) / 2;
  
        // This indicate there is at least one
        // subarray of length mid that has sum > 0
        bool ok = false;
  
        // Check every subarray of length mid if 
        // it has sum > 0 or not if sum > 0 then it
        // will satisfy our required condition
        for (int i = mid - 1; i < n; i++) {
  
            // x will store the sum of 
            // subarray of length mid
            int x = pre[i];
            if (i - mid >= 0)
                x -= pre[i - mid];
  
            // Satisfy our given condition
            if (x > 0) { 
                ok = true;
                break;
            }
        }
  
        // Check for higher length as we
        // get length mid
        if (ok == true) {
            len = mid;
            lo = mid + 1;
        }
        // Check for lower length as we
        // did not get length mid
        else
            hi = mid - 1;
    }
  
    return len;
}
  
// Driver code
int main()
{
    int a[] = { 2, 3, 4, 5, 3, 7 };
    int k = 3;
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << LongestSubarray(a, n, k);
  
    return 0;
}

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Java














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// Java implementation of above approach 


class GFG  


{ 


  


// Function to find the length of a  


// longest subarray in which elements  


// greater than K are more than  


// elements not greater than K 


static int LongestSubarray(int a[], int n, int k) 


{ 


  


    int []pre = new int[n]; 


  


    // Create a new array in which we store 1 


    // if a[i] > k otherwise we store -1. 


    for (int i = 0; i < n; i++) 


    { 


        if (a[i] > k) 


            pre[i] = 1; 


        else


            pre[i] = -1; 


    } 


  


    // Taking prefix sum over it 


    for (int i = 1; i < n; i++)  


        pre[i] = pre[i - 1] + pre[i]; 


  


    // len will store maximum 


    // length of subarray 


    int len = 0;  


  


    int lo = 1, hi = n; 


  


    while (lo <= hi) 


    { 


        int mid = (lo + hi) / 2; 


  


        // This indicate there is at least one 


        // subarray of length mid that has sum > 0 


        boolean ok = false; 


  


        // Check every subarray of length mid if  


        // it has sum > 0 or not if sum > 0 then it 


        // will satisfy our required condition 


        for (int i = mid - 1; i < n; i++) 


        { 


  


            // x will store the sum of  


            // subarray of length mid 


            int x = pre[i]; 


            if (i - mid >= 0) 


                x -= pre[i - mid]; 


  


            // Satisfy our given condition 


            if (x > 0)  


            {  


                ok = true; 


                break; 


            } 


        } 


  


        // Check for higher length as we 


        // get length mid 


        if (ok == true)  


        { 


            len = mid; 


            lo = mid + 1; 


        } 


          


        // Check for lower length as we 


        // did not get length mid 


        else


            hi = mid - 1; 


    } 


  


    return len; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int a[] = { 2, 3, 4, 5, 3, 7 }; 


    int k = 3; 


    int n = a.length; 


  


    System.out.println(LongestSubarray(a, n, k)); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















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Python3














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# Python3 implementation of above approach 


  


# Function to find the Length of a 


# longest subarray in which elements 


# greater than K are more than 


# elements not greater than K 


def LongestSubarray(a, n, k): 


  


    pre = [0 for i in range(n)] 


  


    # Create a new array in which we store 1 


    # if a[i] > k otherwise we store -1. 


    for i in range(n): 


        if (a[i] > k): 


            pre[i] = 1


        else: 


            pre[i] = -1


  


    # Taking prefix sum over it 


    for i in range(1, n): 


        pre[i] = pre[i - 1] + pre[i] 


  


    # Len will store maximum 


    # Length of subarray 


    Len = 0


  


    lo = 1


    hi = n 


  


    while (lo <= hi): 


        mid = (lo + hi) // 2


  


        # This indicate there is at least one 


        # subarray of Length mid that has sum > 0 


        ok = False


  


        # Check every subarray of Length mid if 


        # it has sum > 0 or not if sum > 0 then it 


        # will satisfy our required condition 


        for i in range(mid - 1, n): 


  


            # x will store the sum of 


            # subarray of Length mid 


            x = pre[i] 


            if (i - mid >= 0): 


                x -= pre[i - mid] 


  


            # Satisfy our given condition 


            if (x > 0): 


                ok = True


                break


  


        # Check for higher Length as we 


        # get Length mid 


        if (ok == True): 


            Len = mid 


            lo = mid + 1


              


        # Check for lower Length as we 


        # did not get Length mid 


        else: 


            hi = mid - 1


  


    return Len


  


# Driver code 


a = [2, 3, 4, 5, 3, 7] 


k = 3


n = len(a) 


  


print(LongestSubarray(a, n, k)) 


  


# This code is contributed by Mohit Kumar 



















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C#














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// C# implementation of above approach 


using System; 


  


class GFG  


{ 


  


// Function to find the length of a  


// longest subarray in which elements  


// greater than K are more than  


// elements not greater than K 


static int LongestSubarray(int[] a, int n, int k) 


{ 


    int []pre = new int[n]; 


  


    // Create a new array in which we store 1 


    // if a[i] > k otherwise we store -1. 


    for (int i = 0; i < n; i++) 


    { 


        if (a[i] > k) 


            pre[i] = 1; 


        else


            pre[i] = -1; 


    } 


  


    // Taking prefix sum over it 


    for (int i = 1; i < n; i++)  


        pre[i] = pre[i - 1] + pre[i]; 


  


    // len will store maximum 


    // length of subarray 


    int len = 0;  


  


    int lo = 1, hi = n; 


  


    while (lo <= hi) 


    { 


        int mid = (lo + hi) / 2; 


  


        // This indicate there is at least one 


        // subarray of length mid that has sum > 0 


        bool ok = false; 


  


        // Check every subarray of length mid if  


        // it has sum > 0 or not if sum > 0 then it 


        // will satisfy our required condition 


        for (int i = mid - 1; i < n; i++) 


        { 


  


            // x will store the sum of  


            // subarray of length mid 


            int x = pre[i]; 


            if (i - mid >= 0) 


                x -= pre[i - mid]; 


  


            // Satisfy our given condition 


            if (x > 0)  


            {  


                ok = true; 


                break; 


            } 


        } 


  


        // Check for higher length as we 


        // get length mid 


        if (ok == true)  


        { 


            len = mid; 


            lo = mid + 1; 


        } 


          


        // Check for lower length as we 


        // did not get length mid 


        else


            hi = mid - 1; 


    } 


    return len; 


} 


  


// Driver code 


public static void Main() 


{ 


    int[] a = { 2, 3, 4, 5, 3, 7 }; 


    int k = 3; 


    int n = a.Length; 


  


    Console.WriteLine(LongestSubarray(a, n, k)); 


} 


} 


  


// This code is contributed by Code_Mech 



















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Output:


5



Time Complexity: O(N*logN)



          
          
          
          

            

    

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