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

// 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

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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Output:

5

Time Complexity: O(N*logN)



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