Maximum sum subarray having sum less than given sum using Set
Last Updated :
31 Jul, 2023
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 or equal to 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 cumulative sum from [0 to i]
cSum += arr[i];
// lookup for upperbound
// of (cSum-K) in hash
set<int>::iterator sit
= cum_set.upper_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 cumulative 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 value of K
int K = 15;
// size of array
int N = sizeof(arr) / sizeof(arr[0]);
cout << maxSubarraySum(arr, N, K);
return 0;
}
// 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 cumulative 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 cumulative 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 value 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
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 = float('-inf')
cSum = 0
for i in range(N):
# getting cumulative sum from [0 to i]
cSum += arr[i]
# lookup for upperbound of (cSum-K) in hash
al = [x for x in cum_set]
al.sort()
lower_bound_index = bisect.bisect_left(al, cSum - K)
# check if upper_bound of (cSum-K) exists then update max sum
if lower_bound_index != len(al):
max_sum = max(max_sum, cSum - al[lower_bound_index])
# // insert cumulative value in hash
cum_set.add(cSum)
# return maximum sum lesser than K
return max_sum
arr = [5, -2, 6, 3, -5]
K = 15
N = len(arr)
print(maxSubarraySum(arr, N, K))
// 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 cumulative 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 - al.ElementAt(sit));
// Insert cumulative 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 value of K
int K = 15;
// Size of array
int N = arr.Length;
Console.Write(maxSubarraySum(arr, N, K));
}
}
// This code is contributed by chitranayal.
<script>
// JavaScript program to find maximum sum
// subarray less than K
// Function to maximum required sum < K
function maxSubarraySum(arr, N, K)
{
// Hash to lookup for value (cum_sum - K)
let cum_set = new Set();
cum_set.add(0);
let max_sum = Number.MIN_SAFE_INTEGER;
let cSum = 0;
for(let i = 0; i < N; i++){
// Getting cumulative sum from [0 to i]
cSum += arr[i];
// Lookup for upperbound
// of (cSum-K) in hash
let al = [];
let end = 0;
for(let it of cum_set)
{
end = it;
al.push(it);
}
al.sort((a, b) => a - b);
let 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 cumulative value in hash
cum_set.add(cSum);
}
// Return maximum sum
// lesser than K
return max_sum;
}
let lower_bound =
(al, x) => al.filter((item) => item > x )[0]
// Driver code
// Initialise the array
let arr = [ 5, -2, 6, 3, -5 ];
// Initialise the value of K
let K = 15;
// Size of array
let N = arr.length;
document.write(maxSubarraySum(arr, N, K));
// This code is contributed by _saurabh_jaiswal
</script>
Time Complexity: O(N*Log(N)), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.
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