Find subarray with given sum | Set 2 (Handles Negative Numbers)
Given an unsorted array of integers, find a subarray that adds to a given number. If there is more than one subarray with the sum of the given number, print any of them.
Examples:
Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33 Output: Sum found between indexes 2 and 4 Explanation: Sum of elements between indices 2 and 4 is 20 + 3 + 10 = 33 Input: arr[] = {10, 2, -2, -20, 10}, sum = -10 Output: Sum found between indexes 0 to 3 Explanation: Sum of elements between indices 0 and 3 is 10 + 2 - 2 - 20 = -10 Input: arr[] = {-10, 0, 2, -2, -20, 10}, sum = 20 Output: No subarray with given sum exists Explanation: There is no subarray with the given sum
Note: We have discussed a solution that does not handle negative integers here. In this post, negative integers are also handled.
Simple Approach: A simple solution is to consider all subarrays one by one and check the sum of every subarray. Following program implements the simple solution. Run two loops: the outer loop picks a starting point I and the inner loop tries all subarrays starting from i.
Algorithm:
- Traverse the array from start to end.
- From every index start another loop from i to the end of the array to get all subarray starting from i, keep a variable sum to calculate the sum. For every index in the inner loop update sum = sum + array[j]If the sum is equal to the given sum then print the subarray.
- For every index in the inner loop update sum = sum + array[j]
- If the sum is equal to the given sum then print the subarray.
C++
/* A simple program to print subarray with sum as given sum */ #include <bits/stdc++.h> using namespace std; /* Returns true if the there is a subarray of arr[] with sum equal to 'sum' otherwise returns false. Also, prints the result */ int subArraySum( int arr[], int n, int sum) { int curr_sum, i, j; // Pick a starting point for (i = 0; i < n; i++) { curr_sum = 0; // try all subarrays starting with 'i' for (j = i ; j < n; j++) { curr_sum = curr_sum + arr[j]; if (curr_sum == sum) { cout << "Sum found between indexes " << i << " and " << j ; return 1; } } } cout << "No subarray found" ; return 0; } // Driver Code int main() { int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 }; int n = sizeof (arr) / sizeof (arr[0]); int sum = 23; subArraySum(arr, n, sum); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { /* Returns true if the there is a subarray of arr[] with sum equal to 'sum' otherwise returns false. Also, prints the result */ static int subArraySum( int arr[], int n, int sum) { int curr_sum, i, j; // Pick a starting point for (i = 0 ; i < n; i++) { curr_sum = 0 ; // try all subarrays starting with 'i' for (j = i ; j < n; j++) { curr_sum = curr_sum + arr[j]; if (curr_sum == sum) { System.out.print( "Sum found between indexes " + i + " and " + j); return 1 ; } } } System.out.print( "No subarray found" ); return 0 ; } // Driver Code public static void main (String[] args) { int arr[] = { 15 , 2 , 4 , 8 , 9 , 5 , 10 , 23 }; int n = arr.length; int sum = 23 ; subArraySum(arr, n, sum); } } // This code is contributed by code_hunt. |
Sum found between indexes 1 and 4
Approach: The idea is to store the sum of elements of every prefix of the array in a hashmap, i.e, every index stores the sum of elements up to that index hashmap. So to check if there is a subarray with a sum equal to s, check for every index i, and sum up to that index as x. If there is a prefix with a sum equal to x – s, then the subarray with the given sum is found.
Algorithm:
- Create a Hashmap (hm) to store a key-value pair, i.e, key = prefix sum and value = its index, and a variable to store the current sum (sum = 0) and the sum of the subarray as s
- Traverse through the array from start to end.
- For every element update the sum, i.e sum = sum + array[i]
- If the sum is equal to s then print that the subarray with the given sum is from 0 to i
- If there is any key in the HashMap which is equal to sum – s then print that the subarray with the given sum is from hm[sum – s] to i
- Put the sum and index in the hashmap as a key-value pair.
Dry-run of the above approach:
Implementation:
C++
// C++ program to print subarray with sum as given sum #include<bits/stdc++.h> using namespace std; // Function to print subarray with sum as given sum void subArraySum( int arr[], int n, int sum) { // create an empty map unordered_map< int , int > map; // Maintains sum of elements so far int curr_sum = 0; for ( int i = 0; i < n; i++) { // add current element to curr_sum curr_sum = curr_sum + arr[i]; // if curr_sum is equal to target sum // we found a subarray starting from index 0 // and ending at index i if (curr_sum == sum) { cout << "Sum found between indexes " << 0 << " to " << i << endl; return ; } // If curr_sum - sum already exists in map // we have found a subarray with target sum if (map.find(curr_sum - sum) != map.end()) { cout << "Sum found between indexes " << map[curr_sum - sum] + 1 << " to " << i << endl; return ; } map[curr_sum] = i; } // If we reach here, then no subarray exists cout << "No subarray with given sum exists" ; } // Driver program to test above function int main() { int arr[] = {10, 2, -2, -20, 10}; int n = sizeof (arr)/ sizeof (arr[0]); int sum = -10; subArraySum(arr, n, sum); return 0; } |
Java
// Java program to print subarray with sum as given sum import java.util.*; class GFG { public static void subArraySum( int [] arr, int n, int sum) { //cur_sum to keep track of cumulative sum till that point int cur_sum = 0 ; int start = 0 ; int end = - 1 ; HashMap<Integer, Integer> hashMap = new HashMap<>(); for ( int i = 0 ; i < n; i++) { cur_sum = cur_sum + arr[i]; //check whether cur_sum - sum = 0, if 0 it means //the sub array is starting from index 0- so stop if (cur_sum - sum == 0 ) { start = 0 ; end = i; break ; } //if hashMap already has the value, means we already // have subarray with the sum - so stop if (hashMap.containsKey(cur_sum - sum)) { start = hashMap.get(cur_sum - sum) + 1 ; end = i; break ; } //if value is not present then add to hashmap hashMap.put(cur_sum, i); } // if end is -1 : means we have reached end without the sum if (end == - 1 ) { System.out.println( "No subarray with given sum exists" ); } else { System.out.println( "Sum found between indexes " + start + " to " + end); } } // Driver code public static void main(String[] args) { int [] arr = { 10 , 2 , - 2 , - 20 , 10 }; int n = arr.length; int sum = - 10 ; subArraySum(arr, n, sum); } } |
Python3
# Python3 program to print subarray with sum as given sum # Function to print subarray with sum as given sum def subArraySum(arr, n, Sum ): # create an empty map Map = {} # Maintains sum of elements so far curr_sum = 0 for i in range ( 0 ,n): # add current element to curr_sum curr_sum = curr_sum + arr[i] # if curr_sum is equal to target sum # we found a subarray starting from index 0 # and ending at index i if curr_sum = = Sum : print ( "Sum found between indexes 0 to" , i) return # If curr_sum - sum already exists in map # we have found a subarray with target sum if (curr_sum - Sum ) in Map : print ( "Sum found between indexes" , \ Map [curr_sum - Sum ] + 1 , "to" , i) return Map [curr_sum] = i # If we reach here, then no subarray exists print ( "No subarray with given sum exists" ) # Driver program to test above function if __name__ = = "__main__" : arr = [ 10 , 2 , - 2 , - 20 , 10 ] n = len (arr) Sum = - 10 subArraySum(arr, n, Sum ) # This code is contributed by Rituraj Jain |
C#
using System; using System.Collections.Generic; // C# program to print subarray with sum as given sum public class GFG { public static void subArraySum( int [] arr, int n, int sum) { //cur_sum to keep track of cumulative sum till that point int cur_sum = 0; int start = 0; int end = -1; Dictionary< int , int > hashMap = new Dictionary< int , int >(); for ( int i = 0; i < n; i++) { cur_sum = cur_sum + arr[i]; //check whether cur_sum - sum = 0, if 0 it means //the sub array is starting from index 0- so stop if (cur_sum - sum == 0) { start = 0; end = i; break ; } //if hashMap already has the value, means we already // have subarray with the sum - so stop if (hashMap.ContainsKey(cur_sum - sum)) { start = hashMap[cur_sum - sum] + 1; end = i; break ; } //if value is not present then add to hashmap hashMap[cur_sum] = i; } // if end is -1 : means we have reached end without the sum if (end == -1) { Console.WriteLine( "No subarray with given sum exists" ); } else { Console.WriteLine( "Sum found between indexes " + start + " to " + end); } } // Driver code public static void Main( string [] args) { int [] arr = new int [] {10, 2, -2, -20, 10}; int n = arr.Length; int sum = -10; subArraySum(arr, n, sum); } } // This code is contributed by Shrikant13 |
Javascript
<script> // Javascript program to print subarray with sum as given sum function subArraySum(arr, n, sum) { //cur_sum to keep track of cumulative sum till that point let cur_sum = 0; let start = 0; let end = -1; let hashMap = new Map(); for (let i = 0; i < n; i++) { cur_sum = cur_sum + arr[i]; //check whether cur_sum - sum = 0, if 0 it means //the sub array is starting from index 0- so stop if (cur_sum - sum == 0) { start = 0; end = i; break ; } //if hashMap already has the value, means we already // have subarray with the sum - so stop if (hashMap.has(cur_sum - sum)) { start = hashMap.get(cur_sum - sum) + 1; end = i; break ; } //if value is not present then add to hashmap hashMap.set(cur_sum, i); } // if end is -1 : means we have reached end without the sum if (end == -1) { document.write( "No subarray with given sum exists" ); } else { document.write( "Sum found between indexes " + start + " to " + end); } } // Driver program let arr = [10, 2, -2, -20, 10]; let n = arr.length; let sum = -10; subArraySum(arr, n, sum); </script> |
Sum found between indexes 0 to 3
Complexity Analysis:
- Time complexity: O(N).
If hashing is performed with the help of an array, then this is the time complexity. In case the elements cannot be hashed in an array a hash map can also be used as shown in the above code. - Auxiliary space: O(n).
As a HashMap is needed, this takes linear space.
Find subarray with given sum with negatives allowed in constant space
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