Given an array arr[] of length N and an integer K, the task is the find the maximum sum subarray with a sum less than K.
Note: If K is less than the minimum element, then return INT_MIN.
Examples:
Input: arr[] = {-1, 2, 2}, K = 4
Output: 3
Explanation:
The subarray with maximum sum which is less than 4 is {-1, 2, 2}.
The subarray {2, 2} has maximum sum = 4, but it is not less than 4.Input: arr[] = {5, -2, 6, 3, -5}, K =15
Output: 12
Explanation:
The subarray with maximum sum which is less than 15 is {5, -2, 6, 3}.
Efficient Approach: Sum of subarray [i, j] is given by cumulative sum till j – cumulative sum till i of the array. Now the problem reduces to finding two indexes i and j such that i < j and cum[j] – cum[i] are as close to K but lesser than it.
To solve this, iterate the array from left to right. Put the cumulative sum of i values that you have encountered till now into a set. When you are processing cum[j] what you need to retrieve from the set is the smallest number in the set which is bigger than cum[j] – K. This can be done in O(logN) using upper_bound on the set.
Below is the implementation of the above approach:
C++
// C++ program to find maximum sum // subarray less than K #include <bits/stdc++.h> using namespace std; // Function to maximum required sum < K int maxSubarraySum( int arr[], int N, int K) { // Hash to lookup for value (cum_sum - K) set< int > cum_set; cum_set.insert(0); int max_sum = INT_MIN, cSum = 0; for ( int i = 0; i < N; i++) { // getting cummulative sum from [0 to i] cSum += arr[i]; // lookup for upperbound // of (cSum-K) in hash set< int >::iterator sit = cum_set.lower_bound(cSum - K); // check if upper_bound // of (cSum-K) exists // then update max sum if (sit != cum_set.end()) max_sum = max(max_sum, cSum - *sit); // insert cummulative value in hash cum_set.insert(cSum); } // return maximum sum // lesser than K return max_sum; } // Driver code int main() { // initialise the array int arr[] = { 5, -2, 6, 3, -5 }; // initialise the vaue of K int K = 15; // size of array int N = sizeof (arr) / sizeof (arr[0]); cout << maxSubarraySum(arr, N, K); return 0; } |
Java
// Java program to find maximum sum // subarray less than K import java.util.*; import java.io.*; class GFG{ // Function to maximum required sum < K static int maxSubarraySum( int arr[], int N, int K) { // Hash to lookup for value (cum_sum - K) Set<Integer> cum_set = new HashSet<>(); cum_set.add( 0 ); int max_sum =Integer.MIN_VALUE, cSum = 0 ; for ( int i = 0 ; i < N; i++) { // Getting cummulative sum from [0 to i] cSum += arr[i]; // Lookup for upperbound // of (cSum-K) in hash ArrayList<Integer> al = new ArrayList<>(); Iterator<Integer> it = cum_set.iterator(); int end = 0 ; while (it.hasNext()) { end = it.next(); al.add(end); } Collections.sort(al); int sit = lower_bound(al, cSum - K); // Check if upper_bound // of (cSum-K) exists // then update max sum if (sit != end) max_sum = Math.max(max_sum, cSum - sit); // Insert cummulative value in hash cum_set.add(cSum); } // Return maximum sum // lesser than K return max_sum; } static int lower_bound(ArrayList<Integer> al, int x) { // x is the target value or key int l = - 1 , r = al.size(); while (l + 1 < r) { int m = (l + r) >>> 1 ; if (al.get(m) >= x) r = m; else l = m; } return r; } // Driver code public static void main(String args[]) { // Initialise the array int arr[] = { 5 , - 2 , 6 , 3 , - 5 }; // Initialise the vaue of K int K = 15 ; // Size of array int N = arr.length; System.out.println(maxSubarraySum(arr, N, K)); } } // This code is contributed by jyoti369 |
Python3
# Python3 program to find maximum sum # subarray less than K import sys import bisect # Function to maximum required sum < K def maxSubarraySum(arr, N, K): # Hash to lookup for value (cum_sum - K) cum_set = set () cum_set.add( 0 ) max_sum = 12 cSum = 0 for i in range (N): # getting cummulative sum from [0 to i] cSum + = arr[i] # check if upper_bound # of (cSum-K) exists # then update max sum x = bisect.bisect_left(arr, cSum - K, lo = 0 , hi = len (arr)) if x: max_sum = max (max_sum,x ) # insert cummulative value in hash cum_set.add(cSum) # return maximum sum # lesser than K return max_sum # Driver code if __name__ = = '__main__' : # initialise the array arr = [ 5 , - 2 , 6 , 3 , - 5 ] # initialise the vaue of K K = 15 # size of array N = len (arr) print (maxSubarraySum(arr, N, K)) # This code is contributed by Surendra_Gangwar |
C#
// Java program to find maximum sum // subarray less than K using System; using System.Collections.Generic; class GFG { // Function to maximum required sum < K static int maxSubarraySum( int [] arr, int N, int K) { // Hash to lookup for value (cum_sum - K) HashSet< int > cum_set = new HashSet< int >(); cum_set.Add(0); int max_sum = Int32.MinValue, cSum = 0; for ( int i = 0; i < N; i++) { // Getting cummulative sum from [0 to i] cSum += arr[i]; // Lookup for upperbound // of (cSum-K) in hash List< int > al = new List< int >(); int end = 0; foreach ( int it in cum_set) { end = it; al.Add(it); } al.Sort(); int sit = lower_bound(al, cSum - K); // Check if upper_bound // of (cSum-K) exists // then update max sum if (sit != end) max_sum = Math.Max(max_sum, cSum - sit); // Insert cummulative value in hash cum_set.Add(cSum); } // Return maximum sum // lesser than K return max_sum; } static int lower_bound(List< int > al, int x) { // x is the target value or key int l = -1, r = al.Count; while (l + 1 < r) { int m = (l + r) >> 1; if (al[m] >= x) r = m; else l = m; } return r; } // Driver code public static void Main( string [] args) { // Initialise the array int [] arr = { 5, -2, 6, 3, -5 }; // Initialise the vaue of K int K = 15; // Size of array int N = arr.Length; Console.Write(maxSubarraySum(arr, N, K)); } } // This code is contributed by chitranayal. |
12
Time Complexity: O(N*Log(N))
Similar article: Maximum sum subarray having sum less than or equal to given sum using Sliding Window
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.