Smallest Subarray with Sum K from an Array
Last Updated :
20 Jan, 2023
Given an array arr[] consisting of N integers, the task is to find the length of the Smallest subarray with a sum equal to K.
Examples:
Input: arr[] = {2, 4, 6, 10, 2, 1}, K = 12
Output: 2
Explanation: All possible subarrays with sum 12 are {2, 4, 6} and {10, 2}.
Input: arr[] = {-8, -8, -3, 8}, K = 5
Output: 2
Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays and for each subarray, check if its sum is equal to K or not. Print the minimum length of all such subarrays.
C++
// c++ code to implement the above idea
#include <bits/stdc++.h>
using namespace std;
// Function to find the Smallest Subarray with
// Sum K from an Array
int smallestSubarraySumK(vector<int>& arr, int k)
{
int result = INT_MAX;
for (int i = 0; i < arr.size(); ++i) {
int sum = 0;
for (int j = i; j < arr.size(); j++) {
sum += arr[j];
if (sum == k) {
result = min(result, (j - i + 1));
}
}
}
// Return result
return result;
}
// Driver code
int main()
{
vector<int> arr = { -8, -8, -3, 8 };
int k = 5;
int result = smallestSubarraySumK(arr, k);
if (result == INT_MAX)
cout << -1;
else
cout << result;
return 0;
}
/*package whatever //do not write package name here */
import java.util.*;
class GFG {
// Function to find the Smallest Subarray with
// Sum K from an Array
static int smallestSubarraySumK(int[] arr, int k)
{
int result = Integer.MAX_VALUE;
for (int i = 0; i < arr.length; ++i) {
int sum = 0;
for (int j = i; j < arr.length; j++) {
sum += arr[j];
if (sum == k) {
result = Math.min(result, (j - i + 1));
}
}
}
// Return result
return result;
}
public static void main (String[] args) {
int[] arr = { -8, -8, -3, 8 };
int k = 5;
int result = smallestSubarraySumK(arr, k);
if (result == Integer.MAX_VALUE)
System.out.println(-1);
else
System.out.println(result);
}
}
// This code is contributed by aadityaburujwale.
# python code to implement the above idea
import math
# Function to find the Smallest Subarray with
# Sum K from an Array
def smallestSubarraySumK(arr, k):
result = 2147483647;
for i in range(0, len(arr)):
ans = 0;
for j in range(i, len(arr)):
ans += arr[j];
if (ans == k):
result = min(result, (j - i + 1));
# Return result
return result;
# Driver code
arr = [ -8, -8, -3, 8 ];
k = 5;
result = smallestSubarraySumK(arr, k);
if (result == 2147483647):
print(-1);
else:
print(result);
# This code is contributed by ritaagarwal.
// Method for checking if we can
// make adjacents different or not
// c# code to implement the above idea
using System;
using System.Collections.Generic;
class GFG {
// Function to find the Smallest Subarray with
// Sum K from an Array
static int smallestSubarraySumK(List<int> arr, int k)
{
int result = Int32.MaxValue;
for (int i = 0; i < arr.Count; ++i) {
int sum = 0;
for (int j = i; j < arr.Count; j++) {
sum += arr[j];
if (sum == k) {
result = Math.Min(result, (j - i + 1));
}
}
}
// Return result
return result;
}
// Driver code
public static void Main()
{
List<int> arr = new List<int>{ -8, -8, -3, 8 };
int k = 5;
int result = smallestSubarraySumK(arr, k);
if (result == Int32.MaxValue)
Console.Write(-1);
else
Console.Write(result);
}
}
// This code is contributed by poojaagarwal2.
// Javascript code to implement the above idea
// Function to find the Smallest Subarray with
// Sum K from an Array
function smallestSubarraySumK(arr, k)
{
let result = Number.MAX_SAFE_INTEGER;
for (let i = 0; i < arr.length; ++i) {
let sum = 0;
for (let j = i; j < arr.length; j++) {
sum += arr[j];
if (sum == k) {
result = Math.min(result, (j - i + 1));
}
}
}
// Return result
return result;
}
// Driver code
let arr = [ -8, -8, -3, 8 ];
let k = 5;
let result = smallestSubarraySumK(arr, k);
if (result == Number.MAX_SAFE_INTEGER)
document.write(-1);
else
document.write(result);
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: The above approach can be further optimized using the Prefix Sum technique and Map.
Follow the steps below to solve the problem:
- currPrefixSum will store the prefix sum ending at ith index.
- Iterate over the array and keep calculating currPrefixSum.
- Check if, currPrefixSum is equal to K.
- If yes, then use this length of subarray (currPrefixSum) to minimize the result.
- Also, check if the required prefix sum (currPrefixSum - K) has been calculated previously or not.
- If the required prefix sum is calculated previously then find its last occurrence of that prefix sum and use it to calculate the length of the current prefix sum equal to K by (current index - last occurrence of required prefix sum) and use it to minimize the result.
- Store the currPrefixSum which is ending at ith index into the map.
- Finally, return the result.
Below is the implementation of the above approach:
C++
// c++ code to implement the above idea
#include <bits/stdc++.h>
using namespace std;
// Function to find the Smallest Subarray with
// Sum K from an Array
int smallestSubarraySumK(vector<int> & arr, int k ){
// Use map here to store the prefixSum ending
// at ith index.
unordered_map<long long, int> unmap;
int n = arr.size();
// Store the current Prefix sum till ith index;
long long currPrefixSum = 0;
// Store the minimum size subarray whose sum is K
long long result = INT_MAX;
for(int i = 0; i < n; i++){
currPrefixSum += arr[i];
// Check if the current prefix sum is equals to K
if(currPrefixSum == k){
long long currLen = i + 1;
result = min(result, currLen);
}
// Required PrefixSum
long long requirePrefixSum
= currPrefixSum - k;
// Check if there exist any required
// Prefix Sum or not
if(unmap.count(requirePrefixSum)){
long long foundIdx =
unmap[requirePrefixSum];
long long currIdx = i;
result = min(result,
currIdx - foundIdx);
}
// Store the current prefix sum ending
// at i
unmap[currPrefixSum] = i;
}
if(result >= INT_MAX) return -1;
// return the result
return result;
}
// Driver code
int main(){
vector<int> arr = {-8, -8, -3, 8};
int k = 5;
cout << smallestSubarraySumK(arr, k);
return 0;
}
// Java Program to implement
// the above approach
import java.util.*;
public class Main {
// Function to find the Smallest Subarray with Sum
// K from an Array
static int smallestSubarraySumK(int arr[], int n, int K)
{
// Use map here to store the prefixSum ending
// at ith index.
HashMap<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// Store the current Prefix sum till ith
// index;
int currPrefixSum = 0;
// Store the minimum size subarray whose
// sum is K
int result = Integer.MAX_VALUE;
for(int i = 0; i < n; i++){
currPrefixSum += arr[i];
// Check if the current prefix sum is
// equals to K
if(currPrefixSum == K){
int currLen = i + 1;
result = Math.min(result, currLen);
}
// Required PrefixSum
int requirePrefixSum = currPrefixSum - K;
// Check if there exist any required Prefix Sum or not
if(mp.containsKey(requirePrefixSum)){
int foundIdx = mp.get(requirePrefixSum);
int currIdx = i;
result = Math.min(result, currIdx
- foundIdx);
}
// Store the current prefix sum ending at i
mp.put(currPrefixSum, i);
}
if(result >= Integer.MAX_VALUE) return -1;
// return the result
return result;
}
// Driver Code
public static void main(String[] args){
int arr[] = {-8, -8, -3, 8};
int n = arr.length;
int K = 5;
System.out.println(smallestSubarraySumK(arr,
n, K));
}
}
# Python3 program to implement
# the above approach
from collections import defaultdict
import sys
# Function to find the length of the
# smallest subarray with sum K
def subArraylen(arr, n, K):
mp = defaultdict(lambda : 0)
currPrefixSum = 0
result = sys.maxsize
for i in range(n):
currPrefixSum += arr[i]
if(currPrefixSum == K):
currLen = i + 1
result = min(result, currLen)
requirePrefixSum = currPrefixSum - K
if(requirePrefixSum in mp.keys()):
foundIdx = mp[requirePrefixSum]
currIdx = i
result = min(result, currIdx - foundIdx)
mp[currPrefixSum] = i
return result
# Driver Code
if __name__ == "__main__":
arr = [-8, -8, -3, 8]
n = len(arr)
K = 5
ln = subArraylen(arr, n, K)
# Function call
if(ln == sys.maxsize):
print("-1")
else:
print(ln)
# This code is contributed by Shivam Singh
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the Smallest Subarray with Sum
// K from an Array
static int smallestSubarraySumK(int[] arr, int n, int K)
{
// Use map here to store the prefixSum ending
// at ith index.
Dictionary<int, int> mp = new Dictionary<int, int>();
// Store the current Prefix sum till ith
// index;
int currPrefixSum = 0;
// Store the minimum size subarray whose
// sum is K
int result = int.MaxValue;
for (int i = 0; i < n; i++)
{
currPrefixSum += arr[i];
// Check if the current prefix sum is
// equals to K
if (currPrefixSum == K)
{
int currLen = i + 1;
result = Math.Min(result, currLen);
}
// Required PrefixSum
int requirePrefixSum = currPrefixSum - K;
// Check if there exist any required Prefix Sum or not
if (mp.ContainsKey(requirePrefixSum))
{
int foundIdx = mp[requirePrefixSum];
int currIdx = i;
result = Math.Min(result, currIdx
- foundIdx);
}
// Store the current prefix sum ending at i
mp[currPrefixSum] = i;
}
if (result >= int.MaxValue) return -1;
// return the result
return result;
}
// Driver Code
public static void Main()
{
int[] arr = { -8, -8, -3, 8 };
int n = arr.Length;
int K = 5;
Console.Write(smallestSubarraySumK(arr,
n, K));
}
}
// This code is contributed by saurabh_jaiswal.
<script>
// JavaScript code to implement the above idea
// Function to find the Smallest Subarray with
// Sum K from an Array
function smallestSubarraySumK(arr,k)
{
// Use map here to store the prefixSum ending
// at ith index.
let unmap = new Map();
let n = arr.length
// Store the current Prefix sum till ith index;
let currPrefixSum = 0;
// Store the minimum size subarray whose sum is K
let result = Number.MAX_VALUE;
for(let i = 0; i < n; i++)
{
currPrefixSum += arr[i];
// Check if the current prefix sum is equals to K
if(currPrefixSum == k){
let currLen = i + 1;
result = Math.min(result, currLen);
}
// Required PrefixSum
let requirePrefixSum
= currPrefixSum - k;
// Check if there exist any required
// Prefix Sum or not
if(unmap.has(requirePrefixSum)){
let foundIdx =
unmap.get(requirePrefixSum);
let currIdx = i;
result = Math.min(result,
currIdx - foundIdx);
}
// Store the current prefix sum ending
// at i
unmap.set(currPrefixSum, i);
}
if(result >= Number.MAX_VALUE) return -1;
// return the result
return result;
}
// Driver code
let arr = [-8, -8, -3, 8];
let k = 5;
document.write(smallestSubarraySumK(arr, k));
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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