Count of Subarrays of given Array with median at least X

Last Updated : 23 Jul, 2025

Given an array arr[]of integers with length N and an integer X, the task is to calculate the number of subarrays with median greater than or equal to the given integer X.

Examples:

Input: N=4, A = [5, 2, 4, 1], X = 4
Output: 7
Explanation: For subarray [5], median is 5. (>= 4)
For subarray [5, 2], median is 5. (>= 4)
For subarray [5, 2, 4], median is 4. (>= 4)
For subarray [5, 2, 4, 1], median is 4. (>= 4)
For subarray [2, 4], median is 4. (>= 4)
For subarray [4], median is 4. (>= 4)
For subarray [4, 1], median is 4. (>= 4)

Input: N = [3, 7, 2, 0, 1, 5], X = 10
Output: 0
Explanation: There are no subarrays with median greater than or equal to X.

Approach: The problem can be solved based on the following idea.

To find a subarray with median greater or equal to X at least half of the elements should be greater than or equal to X.

Follow the below steps to implement the above idea:

Replace each element of an array with 1 if it is greater than or equal to X, else replace it with -1.
Based on the above idea, for the new array, median of any subarray to be greater than or equal to X, its sum of elements should be greater than or equal to 0.
For calculating the number of subarray with a sum greater than or equal to 0:
- Find prefix sum up to each index of the new array.
- Traverse the newly created prefix array starting from index 1 and calculate the number of elements before it with a value less than or equal to the current value.
- Add all those in the final answer as they will also form a subarray with the current one satisfying all conditions.
- After finding it for an index, add the current value to a multiset.
Return the final answer.

Note: For efficiently calculating the number of elements with a value less than or equal to Y, use policy-based data structures.

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include <iostream>
using namespace __gnu_pbds;
using namespace std;

// A new data structure defined.
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;

// Function to find the Number of
// subarrays  with median greater than
// or equal to X.
long long findNumberOfSubarray(int arr[],
                               int n, int X)
{
    // Build new array by comparing it with X
    int new_array[n];

    for (int i = 0; i < n; i++) {
        if (arr[i] >= X) {
            new_array[i] = 1;
        }
        else {
            new_array[i] = -1;
        }
    }

    // Build new array in which
    // at i-th index, Sum of first i elements
    // are stored
    int pref_sum[n];
    pref_sum[0] = new_array[0];
    for (int i = 1; i < n; i++) {
        pref_sum[i] = pref_sum[i - 1]
                      + new_array[i];
    }

    // Store the answer
    // Using long long because
    // it can exceed the storage limit of int
    long long ans = 0;

    // For storing already traversed values
    ordered_set s;
    s.insert(0);

    // Iterating forwards from 0 to n-1.
    for (int i = 0; i < n; i++) {
        int less_than
            = s.order_of_key(pref_sum[i] + 1);
        ans += less_than;
        s.insert(pref_sum[i]);
    }

    return ans;
}

// Driver Code
int main()
{
    int N = 4, X = 4;
    int arr[] = { 5, 2, 4, 1 };

    // Function call
    long long ans
        = findNumberOfSubarray(arr, N, X);
    cout << ans;
    return 0;
}

Java

// Java code to implement the above approach

import java.util.*;

class GFG {
    // Function to find the Number of
   // subarrays  with median greater than
  // or equal to X.
    public static int findNumberOfSubarray(int[] arr, int n, int X) {
    int[] newArray = new int[n];
    for (int i = 0; i < n; i++) {
        if (arr[i] >= X) {
            newArray[i] = 1;
        } else {
            newArray[i] = -1;
        }
    }
    // Build new array in which
    // at i-th index, Sum of first i elements
    // are stored
    int[] prefSum = new int[n];
    prefSum[0] = newArray[0];
    for (int i = 1; i < n; i++) {
        prefSum[i] = prefSum[i - 1] + newArray[i];
    }

    int ans = 0;
    HashSet<Integer> set = new HashSet<>();
    set.add(0);
    // Iterating forwards from 0 to n-1.
    for (int i = 0; i < n; i++) {
        int lessThan = Collections.binarySearch(new ArrayList<>(set), prefSum[i] + 1);
        if (lessThan < 0) {
            lessThan = -(lessThan + 1);
        }
        if (set.contains(prefSum[i] + 1)) {
            lessThan++;
        }
        ans += lessThan;
        set.add(prefSum[i]);
    }

    return ans;
}
// Driver Code
public static void main(String[] args) {
    int N = 4, X = 4;
    int[] arr = {5, 2, 4, 1};

    int ans = findNumberOfSubarray(arr, N, X);
    System.out.println(ans);
}
}

Python3

# python3 code to implement the above approach
import bisect

# Function to find the Number of
# subarrays with median greater than
# or equal to X.
def findNumberOfSubarray(arr, n, X):

    # Build new array by comparing it with X
    new_array = [0 for _ in range(n)]

    for i in range(0, n):
        if (arr[i] >= X):
            new_array[i] = 1

        else:
            new_array[i] = -1

    # Build new array in which
    # at i-th index, Sum of first i elements
    # are stored
    pref_sum = [0 for _ in range(n)]
    pref_sum[0] = new_array[0]
    for i in range(1, n):
        pref_sum[i] = pref_sum[i - 1] + new_array[i]

    # Store the answer
    # Using long long because
    # it can exceed the storage limit of int
    ans = 0

    # For storing already traversed values
    s = set()
    s.add(0)

    # Iterating forwards from 0 to n-1.
    for i in range(0, n):

        less_than = bisect.bisect_left(
            sorted(s), pref_sum[i]+1, lo=0, hi=len(s))
        if pref_sum[i] + 1 in s:
            less_than += 1
        ans += less_than
        s.add(pref_sum[i])

    return ans

# Driver Code
if __name__ == "__main__":

    N, X = 4, 4
    arr = [5, 2, 4, 1]

    # Function call
    ans = findNumberOfSubarray(arr, N, X)
    print(ans)

    # This code is contributed by rakeshsahni

// C# code to implement the equivalent of the above Java code

using System;
using System.Collections.Generic;

class GFG
{
  // Function to find the number of subarrays with 
  // median greater than or equal to X
  public static int findNumberOfSubarray(int[] arr, int n, int X)
  {
    // Create a new array where elements >= X are marked 
    // as 1 and elements < X are marked as -1
    int[] newArray = new int[n];
    for (int i = 0; i < n; i++)
    {
      if (arr[i] >= X)
      {
        newArray[i] = 1;
      }
      else
      {
        newArray[i] = -1;
      }
    }

    // Build a new array where the sum of the first
    // i elements is stored at the i-th index
    int[] prefSum = new int[n];
    prefSum[0] = newArray[0];
    for (int i = 1; i < n; i++)
    {
      prefSum[i] = prefSum[i - 1] + newArray[i];
    }

    int ans = 0;

    // Store the prefix sum values in a HashSet
    HashSet<int> set = new HashSet<int>();
    set.Add(0);

    // Iterate forwards from 0 to n-1
    for (int i = 0; i < n; i++)
    {
      // Find the number of elements less than (prefSum[i] + 1) using binary search
      int lessThan = Array.BinarySearch(new List<int>(set).ToArray(), prefSum[i] + 1);
      if (lessThan < 0)
      {
        lessThan = -(lessThan + 1);
      }

      // If (prefSum[i] + 1) is in the set, increment lessThan
      if (set.Contains(prefSum[i] + 1))
      {
        lessThan++;
      }

      // Add the number of subarrays ending at i with median >= X
      ans += lessThan;

      // Add the current prefix sum value to the set
      set.Add(prefSum[i]);
    }

    return ans;
  }

  // Driver Code
  static void Main(string[] args)
  {
    int N = 4, X = 4;
    int[] arr = { 5, 2, 4, 1 };

    int ans = findNumberOfSubarray(arr, N, X);
    Console.WriteLine(ans);
  }
}

JavaScript

// javascript code to implement the above approach

// Function to find the Number of
// subarrays with median greater than
// or equal to X.
function findNumberOfSubarray(arr, n, X) {


    // Build new array by comparing it with X
    let new_array = Array(n).fill(0);

    for (let i = 0; i < n; i++) {
        if (arr[i] >= X) {
            new_array[i] = 1;
        } else {
            new_array[i] = -1;
        }
    }

    // Build new array in which
    // at i-th index, Sum of first i elements
    // are stored
    let pref_sum = Array(n).fill(0);
    pref_sum[0] = new_array[0];
    for (let i = 1; i < n; i++) {
        pref_sum[i] = pref_sum[i - 1] + new_array[i];
    }

    // Store the answer
    // Using BigInt because it can exceed the storage limit of int
    let ans = BigInt(0);

    // For storing already traversed values
    let s = new Set();
    s.add(0);

    // Iterating forwards from 0 to n-1.
    for (let i = 0; i < n; i++) {

        // less_than = bisect.bisect_left(sorted(s), pref_sum[i]+1, lo=0, hi=len(s))
        let sortedS = Array.from(s).sort(function(a, b) {
            return a - b
        });
        let less_than = bisect_left(sortedS, pref_sum[i] + 1, 0, sortedS.length);
        if (s.has(pref_sum[i] + 1)) {
            less_than += 1;
        }
        ans += BigInt(less_than);
        s.add(pref_sum[i]);
    }

    return ans;

}

// binary search equivalent to Python's bisect.bisect_left
function bisect_left(sortedArr, x, lo, hi) {
    lo = lo || 0;
    hi = hi || sortedArr.length;
    while (lo < hi) {
        let mid = (lo + hi) >>> 1;
        if (sortedArr[mid] < x) {
            lo = mid + 1;
        } else {
            hi = mid;
        }
    }
    return lo;
}

// Driver Code
let N = 4,
    X = 4;
let arr = [5, 2, 4, 1];

// Function call
let ans = findNumberOfSubarray(arr, N, X);
console.log(ans);

// This code is contributed by phasing17

Output

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

Analysis of Algorithms

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