Count array elements exceeding sum of preceding K elements

Given an array arr[] of size N, and an integer K, the task is to find the count of array elements arr[i] which are greater than (arr[i – 1] + arr[i – 2] + … + arr[ i – K]).

Examples:

Input: arr[] = {2, 3, 8, 10, -2, 7, 5, 5, 9, 15}, K = 2 
Output: 2 
Explanation:
arr[2](> arr[1] + arr[0]) and arr[9](> arr[8] + arr[7]) are the two array elements exceeding sum of preceding K(= 2) elements.

Input: arr[] = {17, -2, 16, -8, 19, 11, 5, 15, -9, 24}, K = 3 
Output: 2 

Approach: Follow the steps below to solve the problem: 

1. Calculate and store the prefix sum of the given array.
2. Traverse the array elements present in the indices [K, N – 1].
3. For every array element, check if arr[i] exceeds (arr[i – 1] + arr[i – 2] + … + arr[i – K]) or not. Therefore, the task is to check if arr[i] exceeds prefix[i – 1] – prefix[i – (K + 1)] for K > i > N or if arr[i] exceeds prefix[i – 1] for i = K.
4. Increment count for array element satisfying the above conditions. Finally, print the required count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count array elements
// exceeding sum of preceding K elements
int countPrecedingK(int a[], int n, int K)
{
    int prefix[n];
    prefix[0] = a[0];

    // Iterate over the array
    for (int i = 1; i < n; i++) {

        // Update prefix sum
        prefix[i]
            = prefix[i - 1] + a[i];
    }

    int ctr = 0;

    // Check if arr[K] >
    // arr[0] + .. + arr[K - 1]
    if (prefix[K - 1] < a[K])

        // Increment count
        ctr++;

    for (int i = K + 1; i < n; i++) {
        // Check if arr[i] >
        // arr[i - K - 1] + .. + arr[i - 1]
        if (prefix[i - 1] - prefix[i - K - 1]
            < a[i])

            // Increment count
            ctr++;
    }

    return ctr;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 3, 8, 10, -2,
                  7, 5, 5, 9, 15 };

    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;

    cout << countPrecedingK(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to count array elements
// exceeding sum of preceding K elements
static int countPrecedingK(int a[], int n,
                           int K)
{
    int []prefix = new int[n];
    prefix[0] = a[0];

    // Iterate over the array
    for(int i = 1; i < n; i++)
    {

        // Update prefix sum
        prefix[i] = prefix[i - 1] + a[i];
    }

    int ctr = 0;

    // Check if arr[K] >
    // arr[0] + .. + arr[K - 1]
    if (prefix[K - 1] < a[K])

        // Increment count
        ctr++;

    for(int i = K + 1; i < n; i++)
    {

        // Check if arr[i] >
        // arr[i - K - 1] + .. + arr[i - 1]
        if (prefix[i - 1] -
            prefix[i - K - 1] < a[i])

            // Increment count
            ctr++;
    }
    return ctr;
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 2, 3, 8, 10, -2,
                  7, 5, 5, 9, 15 };

    int N = arr.length;
    int K = 2;

    System.out.print(countPrecedingK(arr, N, K));
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach

# Function to count array elements
# exceeding sum of preceding K elements
def countPrecedingK(a, n, K):

    prefix = [0] * n
    prefix[0] = a[0]

    # Iterate over the array
    for i in range(1, n):

        # Update prefix sum
        prefix[i] = prefix[i - 1] + a[i]

    ctr = 0

    # Check if arr[K] >
    # arr[0] + .. + arr[K - 1]
    if (prefix[K - 1] < a[K]):

        # Increment count
        ctr += 1

    for i in range(K + 1, n):

        # Check if arr[i] >
        # arr[i - K - 1] + .. + arr[i - 1]
        if (prefix[i - 1] -
            prefix[i - K - 1] < a[i]):

            # Increment count
            ctr += 1

    return ctr

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [ 2, 3, 8, 10, -2,
            7, 5, 5, 9, 15 ]

    N = len(arr)
    K = 2

    print(countPrecedingK(arr, N, K))

# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
class GFG{

// Function to count array elements
// exceeding sum of preceding K elements
static int countPrecedingK(int []a,
                           int n, int K)
{
  int []prefix = new int[n];
  prefix[0] = a[0];

  // Iterate over the array
  for(int i = 1; i < n; i++)
  {
    // Update prefix sum
    prefix[i] = prefix[i - 1] + a[i];
  }

  int ctr = 0;

  // Check if arr[K] >
  // arr[0] + .. + arr[K - 1]
  if (prefix[K - 1] < a[K])

    // Increment count
    ctr++;

  for(int i = K + 1; i < n; i++)
  {
    // Check if arr[i] >
    // arr[i - K - 1] + .. + arr[i - 1]
    if (prefix[i - 1] -
        prefix[i - K - 1] < a[i])

      // Increment count
      ctr++;
  }
  return ctr;
}

// Driver Code
public static void Main(String[] args)
{
  // Given array
  int []arr = {2, 3, 8, 10, -2,
               7, 5, 5, 9, 15};

  int N = arr.Length;
  int K = 2;

  Console.Write(countPrecedingK(arr, N, K));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count array elements
// exceeding sum of preceding K elements
function countPrecedingK(a, n, K)
{
    let prefix = new Array(n).fill(0);
    prefix[0] = a[0];

    // Iterate over the array
    for(let i = 1; i < n; i++)
    {

        // Update prefix sum
        prefix[i] = prefix[i - 1] + a[i];
    }

    let ctr = 0;

    // Check if arr[K] >
    // arr[0] + .. + arr[K - 1]
    if (prefix[K - 1] < a[K])

        // Increment count
        ctr++;

    for(let i = K + 1; i < n; i++)
    {

        // Check if arr[i] >
        // arr[i - K - 1] + .. + arr[i - 1]
        if (prefix[i - 1] -
            prefix[i - K - 1] < a[i])

            // Increment count
            ctr++;
    }
    return ctr;
}

// Driver Code

     // Given array
    let arr = [ 2, 3, 8, 10, -2,
                  7, 5, 5, 9, 15 ];

    let N = arr.length;
    let K = 2;

    document.write(countPrecedingK(arr, N, K));

</script>

Output:

2

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