Subarray whose sum is closest to K
Last Updated :
14 Feb, 2025
Given an array of positive and negative integers and an integer K. The task is to find the subarray with its sum closest to k. In case of multiple answers, print anyone.
Note: Closest here means abs(sum-k) should be minimal.
Examples:
Input: arr[] = [ -5, 12, -3, 4, -15, 6, 1 ], k = 2
Output: 1
Explanation: The subarray {-3, 4} or {1} has sum = 1 which is the closest to k.
Input: arr[] = [ 2, 2, -1, 5, -3, -2 ], k = 7
Output: 6
Explanation: Here the output can be 6 or 8. The subarray {2, 2, -1, 5} gives sum as 8 which has abs(8-7) = 1 which is same as that of the subarray {2, -1, 5} which has abs(6-7) = 1.
[Naive Approach] - Using Nested Loops - O(n ^ 2) Time and O(1) Space
The idea is to use nested loops to generate all possible subarrays and find the subarray with sum of its elements closest to k. The outer loop marks the starting of subarray and the inner loops marks the end.
Below is given the implementation:
C++
// Function to find the sum of subarray
// whose sum is closest to K
#include <bits/stdc++.h>
using namespace std;
int closestSubarraySumToK(vector<int>& arr, int k) {
// to store the minimum difference
int diff = INT_MAX;
// to store the answer
int ans = -1;
// generate sum of all possible subarrays
for (int i = 0; i < arr.size(); i++) {
// to store the sum of subarray
int sum = 0;
for (int j = i; j < arr.size(); j++) {
sum += arr[j];
if (abs(k - sum) < diff) {
diff = abs(k - sum);
ans = sum;
}
}
}
return ans;
}
// Driver Code
int main() {
vector<int> arr = {-5, 12, -3, 4, -15, 6, 1};
int k = 2;
cout << closestSubarraySumToK(arr, k) << endl;
return 0;
}
// Java program to find the sum of subarray
// whose sum is closest to K
import java.util.*;
class GFG {
// Function to find the sum of subarray
// whose sum is closest to K
static int closestSubarraySumToK(List<Integer> arr, int k) {
// to store the minimum difference
int diff = Integer.MAX_VALUE;
// to store the answer
int ans = -1;
// generate sum of all possible subarrays
for (int i = 0; i < arr.size(); i++) {
// to store the sum of subarray
int sum = 0;
for (int j = i; j < arr.size(); j++) {
sum += arr.get(j);
if (Math.abs(k - sum) < diff) {
diff = Math.abs(k - sum);
ans = sum;
}
}
}
return ans;
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(-5, 12, -3, 4, -15, 6, 1);
int k = 2;
System.out.println(closestSubarraySumToK(arr, k));
}
}
# Python program to find the sum of subarray
# whose sum is closest to K
# Function to find the sum of subarray
# whose sum is closest to K
def closestSubarraySumToK(arr, k):
# to store the minimum difference
diff = float('inf')
# to store the answer
ans = -1
# generate sum of all possible subarrays
for i in range(len(arr)):
# to store the sum of subarray
sum = 0
for j in range(i, len(arr)):
sum += arr[j]
if abs(k - sum) < diff:
diff = abs(k - sum)
ans = sum
return ans
# Driver Code
arr = [-5, 12, -3, 4, -15, 6, 1]
k = 2
print(closestSubarraySumToK(arr, k))
// C# program to find the sum of subarray
// whose sum is closest to K
using System;
using System.Collections.Generic;
class GFG {
// Function to find the sum of subarray
// whose sum is closest to K
static int ClosestSubarraySumToK(List<int> arr, int k) {
// to store the minimum difference
int diff = int.MaxValue;
// to store the answer
int ans = -1;
// generate sum of all possible subarrays
for (int i = 0; i < arr.Count; i++) {
// to store the sum of subarray
int sum = 0;
for (int j = i; j < arr.Count; j++) {
sum += arr[j];
if (Math.Abs(k - sum) < diff) {
diff = Math.Abs(k - sum);
ans = sum;
}
}
}
return ans;
}
// Driver Code
static void Main() {
List<int> arr = new List<int> { -5, 12, -3, 4, -15, 6, 1 };
int k = 2;
Console.WriteLine(ClosestSubarraySumToK(arr, k));
}
}
// JavaScript program to find the sum of subarray
// whose sum is closest to K
// Function to find the sum of subarray
// whose sum is closest to K
function closestSubarraySumToK(arr, k) {
// to store the minimum difference
let diff = Number.MAX_VALUE;
// to store the answer
let ans = -1;
// generate sum of all possible subarrays
for (let i = 0; i < arr.length; i++) {
// to store the sum of subarray
let sum = 0;
for (let j = i; j < arr.length; j++) {
sum += arr[j];
if (Math.abs(k - sum) < diff) {
diff = Math.abs(k - sum);
ans = sum;
}
}
}
return ans;
}
// Driver Code
let arr = [-5, 12, -3, 4, -15, 6, 1];
let k = 2;
console.log(closestSubarraySumToK(arr, k));
Time Complexity: O(n^2), as we are using nested loops two generate all possible subarrays.
Space Complexity: O(1)
[Expected Approach] - Using Tree Set - O(n * log n) Time and O(n) Space
- The idea is to use Tree Set (or Sorted Set) to store prefix sums in sorted order and use binary search to find the closest sum.
- Initialize the set by inserting the first element of array arr[], and iterate for all the elements and keep adding the sum to set.
- Now use binary search to find the prefix sum which is closest to (sum - k).
Below is the implementation of the above approach.
C++
// C++ program to find the
// sum of subarray whose sum is
// closest to K
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of subarray
// whose sum is closest to K
int closestSubarraySumToK(int a[], int n, int k)
{
// Declare a set
set<int> s;
// initially consider the
// first subarray as the first
// element in the array
int presum = a[0];
// insert
s.insert(a[0]);
// Initially let this difference
// be the minimum
int mini = abs(a[0] - k);
// let this be the sum
// of the subarray
// to be searched initially
int sum = presum;
// iterate for all the array elements
for (int i = 1; i < n; i++) {
// calculate the prefix sum
presum += a[i];
// insert the current prefix sum
s.insert(presum);
// find the closest subarray
// sum to by using lower_bound
auto it = s.lower_bound(presum - k);
// if k present in set
// return it
if (s.find(k) != s.end())
return k;
// if it is the first element
// in the set
if (it == s.begin()) {
// get the prefix sum till start
// of the subarray
int diff = *it;
// if the subarray sum is closest to K
// than the previous one
if (abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
if (abs(presum - k) < mini) {
// update the minimal difference
mini = abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
}
// if the difference is
// present in between
else if (it != s.end()) {
// get the prefix sum till start
// of the subarray
int diff = *it;
// if the subarray sum is closest to K
// than the previous one
if (abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
// also check for the one before that
// since the sum can be greater than
// or less than K also
it--;
// get the prefix sum till start
// of the subarray
diff = *it;
// if the subarray sum is closest to K
// than the previous one
if (abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
}
// if there exists no such prefix sum
// then the current prefix sum is
// checked and updated
else {
// if the subarray sum is closest to K
// than the previous one
if (abs(presum - k) < mini) {
// update the minimal difference
mini = abs(presum - k);
// update the sum
sum = presum;
}
}
}
return sum;
}
// Driver Code
int main()
{
int a[] = { -5, 12, -3, 4, -15, 6, 1 };
int n = sizeof(a) / sizeof(a[0]);
int k = 2;
cout << closestSubarraySumToK(a, n, k);
return 0;
}
// This code is modified by Susobhan Akhuli
// Java program to find the sum of subarray
// whose sum is closest to K
import java.util.*;
class GFG {
// Function to find the sum of subarray
// whose sum is closest to K
static int closestSubarraySumToK(List<Integer> arr, int k) {
int n = arr.size();
// Declare a set
TreeSet<Integer> s = new TreeSet<>();
// initially consider the
// first subarray as the first
// element in the array
int presum = arr.get(0);
// insert
s.add(arr.get(0));
// Initially let this difference
// be the minimum
int mini = Math.abs(arr.get(0) - k);
// let this be the sum
// of the subarray
// to be searched initially
int sum = presum;
// iterate for all the array elements
for (int i = 1; i < n; i++) {
// calculate the prefix sum
presum += arr.get(i);
// insert the current prefix sum
s.add(presum);
// find the closest subarray
// sum to by using ceiling (lower_bound in C++)
Integer it = s.ceiling(presum - k);
// if k present in set
// return it
if (s.contains(k))
return k;
// if it is the first element
// in the set
if (it == null) {
// if the subarray sum is closest to K
// than the previous one
if (Math.abs(presum - k) < mini) {
// update the minimal difference
mini = Math.abs(presum - k);
// update the sum
sum = presum;
}
} else {
// get the prefix sum till start
// of the subarray
int diff = it;
// if the subarray sum is closest to K
// than the previous one
if (Math.abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = Math.abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
// also check for the one before that
Integer itLower = s.lower(it);
if (itLower != null) {
diff = itLower;
if (Math.abs((presum - diff) - k) < mini) {
mini = Math.abs((presum - diff) - k);
sum = presum - diff;
}
}
}
}
return sum;
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(-5, 12, -3, 4, -15, 6, 1);
int k = 2;
System.out.println(closestSubarraySumToK(arr, k));
}
}
# Python program to find the sum of subarray
# whose sum is closest to K
import bisect
# Function to find the sum of subarray
# whose sum is closest to K
def closestSubarraySumToK(arr, k):
n = len(arr)
# Declare a sorted list
s = []
# initially consider the
# first subarray as the first
# element in the array
presum = arr[0]
# insert
s.append(arr[0])
# Initially let this difference
# be the minimum
mini = abs(arr[0] - k)
# let this be the sum
# of the subarray
# to be searched initially
sum = presum
# iterate for all the array elements
for i in range(1, n):
# calculate the prefix sum
presum += arr[i]
# insert the current prefix sum
bisect.insort(s, presum)
# find the closest subarray
# sum to by using bisect_right (lower_bound in C++)
pos = bisect.bisect_left(s, presum - k)
# if k present in set
# return it
if k in s:
return k
if pos < len(s):
diff = s[pos]
if abs((presum - diff) - k) < mini:
mini = abs((presum - diff) - k)
sum = presum - diff
if pos > 0:
diff = s[pos - 1]
if abs((presum - diff) - k) < mini:
mini = abs((presum - diff) - k)
sum = presum - diff
return sum
# Driver Code
arr = [-5, 12, -3, 4, -15, 6, 1]
k = 2
print(closestSubarraySumToK(arr, k))
// C# program to find the sum of subarray
// whose sum is closest to K
using System;
using System.Collections.Generic;
class GFG {
// Function to find the sum of subarray
// whose sum is closest to K
static int ClosestSubarraySumToK(List<int> arr, int k) {
int n = arr.Count;
// Declare a sorted set
SortedSet<int> s = new SortedSet<int>();
// initially consider the
// first subarray as the first
// element in the array
int presum = arr[0];
// insert
s.Add(arr[0]);
// Initially let this difference
// be the minimum
int mini = Math.Abs(arr[0] - k);
// let this be the sum
// of the subarray
// to be searched initially
int sum = presum;
// iterate for all the array elements
for (int i = 1; i < n; i++) {
// calculate the prefix sum
presum += arr[i];
// insert the current prefix sum
s.Add(presum);
// find the closest subarray
// sum to by using GetViewBetween (lower_bound in C++)
int? it = s.GetViewBetween(presum - k, int.MaxValue).Min;
// if k present in set
// return it
if (s.Contains(k))
return k;
if (it.HasValue) {
int diff = it.Value;
if (Math.Abs((presum - diff) - k) < mini) {
mini = Math.Abs((presum - diff) - k);
sum = presum - diff;
}
}
// also check for the one before that
int? lower = s.GetViewBetween(int.MinValue, presum - k).Max;
if (lower.HasValue) {
int diff = lower.Value;
if (Math.Abs((presum - diff) - k) < mini) {
mini = Math.Abs((presum - diff) - k);
sum = presum - diff;
}
}
}
return sum;
}
static void Main() {
List<int> arr = new List<int> { -5, 12, -3, 4, -15, 6, 1 };
int k = 2;
Console.WriteLine(ClosestSubarraySumToK(arr, k));
}
}
// JavaScript program to find the sum of subarray
// whose sum is closest to K
class GFG {
// Function to find the sum of subarray
// whose sum is closest to K
static closestSubarraySumToK(arr, k) {
let n = arr.length;
// Declare a sorted set (simulated using an array)
let s = [];
// initially consider the
// first subarray as the first
// element in the array
let presum = arr[0];
// insert
s.push(arr[0]);
// Initially let this difference
// be the minimum
let mini = Math.abs(arr[0] - k);
// let this be the sum
// of the subarray
// to be searched initially
let sum = presum;
// iterate for all the array elements
for (let i = 1; i < n; i++) {
// calculate the prefix sum
presum += arr[i];
// insert the current prefix sum (keeping it sorted)
let index = this.lowerBound(s, presum);
s.splice(index, 0, presum);
// find the closest subarray
// sum to by using lower_bound
let itIndex = this.lowerBound(s, presum - k);
// if k present in set
// return it
if (s.includes(k))
return k;
// if it is the first element
// in the set
if (itIndex === 0) {
// get the prefix sum till start
// of the subarray
let diff = s[itIndex];
// if the subarray sum is closest to K
// than the previous one
if (Math.abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = Math.abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
if (Math.abs(presum - k) < mini) {
mini = Math.abs(presum - k);
sum = presum;
}
}
// if the difference is
// present in between
else if (itIndex < s.length) {
// get the prefix sum till start
// of the subarray
let diff = s[itIndex];
// if the subarray sum is closest to K
// than the previous one
if (Math.abs((presum - diff) - k) < mini) {
// update the minimal difference
mini = Math.abs((presum - diff) - k);
// update the sum
sum = presum - diff;
}
// also check for the one before that
// since the sum can be greater than
// or less than K also
if (itIndex > 0) {
diff = s[itIndex - 1];
if (Math.abs((presum - diff) - k) < mini) {
mini = Math.abs((presum - diff) - k);
sum = presum - diff;
}
}
}
// if there exists no such prefix sum
// then the current prefix sum is
// checked and updated
else {
if (Math.abs(presum - k) < mini) {
mini = Math.abs(presum - k);
sum = presum;
}
}
}
return sum;
}
// Function to find the lower_bound (binary search)
static lowerBound(arr, target) {
let left = 0, right = arr.length;
while (left < right) {
let mid = Math.floor((left + right) / 2);
if (arr[mid] < target) left = mid + 1;
else right = mid;
}
return left;
}
}
// Driver Code
let arr = [-5, 12, -3, 4, -15, 6, 1];
let k = 2;
console.log(GFG.closestSubarraySumToK(arr, k));
Time Complexity:O(n log n). Binary search operating takes O(log n) Time, and we are performing binary search at each index, thus the overall time complexity will be O(n * log(n)).
Auxiliary Space: O(n), to store the prefix sum in the set.
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