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Finding subarray with given sum

In the article, we are going to find a subarray which sums to an input sum. This problem came in coding round of Visa, Amazon. Submitted by Radib Kar, on November 22, 2018

Problem Statement: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to the given number.

Algorithm

1. A in the input array, n is the length of the array & s in the given sum.
2. Initialize vector b. (for storing indexes of subarray)
3. Initialize a variable cur_sum=0
4. for i=0:n-1

	set cur_sum=A[i]
	for j=i+1:n-1
		add A[j] to cur_sum
		if(cur_sum==s)
			i is the starting index of subarray found
			j is the end index of subarray found
			print elements between indexes i & j
            (including i & j) & return to main
		end if
		if cur_sum become greater than s then break loop
	end for
end for

5. If control comes out of the loop that means so such subarray exists, thus print "no subarray found".

C++ implementation to find subarray with given sum

#include <bits/stdc++.h>
using namespace std;

vector<int> find(vector<int> a,int n,int s){
	vector<int> b; //output vector
	int cur_sum=0;
	for(int i=0;i<n;i++){
		cur_sum=a[i];//cur_suming element
		for(int j=i+1;j<n;j++){
			//add next element contigeously 
			cur_sum=cur_sum+a[j]; 
			// if current sum  is eaqual to the sum value
			if(cur_sum==s){ 
				//insert array index of the cur_sum element in sub array
				b.push_back(i); 
				//insert array index of the last element in sub array
				b.push_back(j); 
				return b;
			}
			if(cur_sum>s)
				break;
		}
	}
	// if no such sub array exists
	b.push_back(-1); 
	return b;
} 

int main() {
	//code
	int s,n,no;
	cout<<"enter array length & Sum respectively"<<endl;
	scanf("%d%d",&n,&s);

	vector<int> a; //input array

	cout<<"enter array elements........"<<endl;
	for(int j=0;j<n;j++){
		scanf("%d",&no);
		a.push_back(no); //inserting array elements
	}

	vector<int> b=find(a,n,s);
	if(b[0]==-1)
		printf("subarray not found");
	else{
		cout<<"Subarray is: ";
		for(int i=b[0];i<=b[1];i++)
			cout<<a[i]<<" ";
		cout<<endl;
	}	    
	return 0;
}

Output

First run:
enter array length & Sum respectively
8
20 
enter array elements........ 
4
2
10 
3
-3 
10 
5
5
Subarray is: 10 3 -3 10 


Second run:
enter array length & Sum respectively
6
15 
enter array elements........ 
5
1
-6 
7
7
3
subarray not found