Maximum sum subarray of size K with sum less than X (original) (raw)

Last Updated : 19 Mar, 2021

Given an array arr[] and two integers K and X, the task is to find the maximum sum among all subarrays of size K with the sum less than X.

Examples:

Input: arr[] = {20, 2, 3, 10, 5}, K = 3, X = 20
Output: 18
Explanation: Subarray of size 3 having maximum sum less than 20 is {3, 10, 5}. Therefore, required output is 18.

Input: arr[] = {-5, 8, 7, 2, 10, 1, 20, -4, 6, 9}, K = 5, X = 30
Output: 29
Explanation: Subarray of size 5having maximum sum less than 30 is {2, 10, 1, 20, -4}. Therefore, required output is 29.

Naive Approach: The simplest approach to solve the problem is to generate all subarrays of size K and check if its sum is less than X or not. Print the maximum sum obtained among all such subarrays.

Time Complexity: O(N * K)
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to solve the problem using Sliding Window technique:

  1. Initialize a variable sum_K to store the sum of first K array elements.
  2. If sum_K is less than X, then initialize Max_Sum with sum_K.
  3. Traverse the array from (K + 1)th index and perform the following:
    1. In each iteration, subtract the first element of the previous K length subarray and add the current element to sum_K.
    2. If sum_K is less than X, then compare sum_K with Max_Sum and update Max_Sum accordingly.
  4. Finally, print Max_Sum.

Below is the implementation of the above approach:

C++ `

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;

// Function to calculate maximum sum // among all subarrays of size K // with the sum less than X void maxSumSubarr(int A[], int N, int K, int X) {

// Initialize sum_K to 0
int sum_K = 0;

// Calculate sum of first K elements
for (int i = 0; i < K; i++) {

    sum_K += A[i];
}

int Max_Sum = 0;

// If sum_K is less than X
if (sum_K < X) {

    // Initialize MaxSum with sum_K
    Max_Sum = sum_K;
}

// Iterate over the array from
// (K + 1)-th index
for (int i = K; i < N; i++) {

    // Subtract the first element
    // from the previous K elements
    // and add the next element
    sum_K -= (A[i - K] - A[i]);

    // If sum_K is less than X
    if (sum_K < X) {

        // Update the Max_Sum
        Max_Sum = max(Max_Sum, sum_K);
    }
}

cout << Max_Sum << endl;

}

// Driver Code int main() { int arr[] = { -5, 8, 7, 2, 10, 1, 20, -4, 6, 9 }; int K = 5; int X = 30;

// Size of Array
int N = sizeof(arr)
        / sizeof(arr[0]);

// Function Call
maxSumSubarr(arr, N, K, X);

return 0;

}

Java

// Java program for the above approach import java.io.*;

class GFG{

// Function to calculate maximum sum // among all subarrays of size K // with the sum less than X private static void maxSumSubarr(int A[], int N, int K, int X) {

// Initialize sum_K to 0
int sum_K = 0;

// Calculate sum of first K elements
for(int i = 0; i < K; i++)
{
    sum_K += A[i];
}

int Max_Sum = 0;

// If sum_K is less than X
if (sum_K < X)
{
    
    // Initialize MaxSum with sum_K
    Max_Sum = sum_K;
}

// Iterate over the array from
// (K + 1)-th index
for(int i = K; i < N; i++) 
{
    
    // Subtract the first element
    // from the previous K elements
    // and add the next element
    sum_K -= (A[i - K] - A[i]);
    
    // If sum_K is less than X
    if (sum_K < X)
    {
        
        // Update the Max_Sum
        Max_Sum = Math.max(Max_Sum, sum_K);
    }
}

System.out.println(Max_Sum);

}

// Driver Code public static void main (String[] args) { int arr[] = { -5, 8, 7, 2, 10, 1, 20, -4, 6, 9 }; int K = 5; int X = 30;

// Size of Array
int N = arr.length;

// Function Call
maxSumSubarr(arr, N, K, X);

} }

// This code is contributed by jithin

Python3

Python3 program for the above approach

Function to calculate maximum sum

among all subarrays of size K

with the sum less than X

def maxSumSubarr(A, N, K, X):

# Initialize sum_K to 0
sum_K = 0

# Calculate sum of first K elements
for i in range(0, K):
    sum_K += A[i]

Max_Sum = 0

# If sum_K is less than X
if (sum_K < X):

    # Initialize MaxSum with sum_K
    Max_Sum = sum_K

# Iterate over the array from
# (K + 1)-th index
for i in range(K, N):

    # Subtract the first element
    # from the previous K elements
    # and add the next element
    sum_K -= (A[i - K] - A[i])

    # If sum_K is less than X
    if (sum_K < X):
        
        # Update the Max_Sum
        Max_Sum = max(Max_Sum, sum_K)
    
print(Max_Sum)

Driver Code

arr = [ -5, 8, 7, 2, 10, 1, 20, -4, 6, 9 ] K = 5 X = 30

Size of Array

N = len(arr)

Function Call

maxSumSubarr(arr, N, K, X)

This code is contributed by sanjoy_62

C#

// C# program for the above approach using System;

class GFG{

// Function to calculate maximum sum // among all subarrays of size K // with the sum less than X private static void maxSumSubarr(int []A, int N, int K, int X) {

// Initialize sum_K to 0
int sum_K = 0;

// Calculate sum of first K elements
for(int i = 0; i < K; i++)
{
    sum_K += A[i];
}

int Max_Sum = 0;

// If sum_K is less than X
if (sum_K < X)
{
    
    // Initialize MaxSum with sum_K
    Max_Sum = sum_K;
}

// Iterate over the array from
// (K + 1)-th index
for(int i = K; i < N; i++) 
{
    
    // Subtract the first element
    // from the previous K elements
    // and add the next element
    sum_K -= (A[i - K] - A[i]);
    
    // If sum_K is less than X
    if (sum_K < X)
    {
        
        // Update the Max_Sum
        Max_Sum = Math.Max(Max_Sum, sum_K);
    }
}
Console.WriteLine(Max_Sum);

}

// Driver Code public static void Main(String[] args) { int []arr = { -5, 8, 7, 2, 10, 1, 20, -4, 6, 9 }; int K = 5; int X = 30;

// Size of Array
int N = arr.Length;

// Function Call
maxSumSubarr(arr, N, K, X);

} }

// This code is contributed by Amit Katiyar

JavaScript

`

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

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