Maximum sum subarray of size K with sum less than X

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 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



Output:

29

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

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