Sherlock and Array
Problem Statement :
Watson gives Sherlock an array of integers. His challenge is to find an element of the array such that the sum of all elements to the left is equal to the sum of all elements to the right. Function Description Complete the balancedSums function in the editor below. balancedSums has the following parameter(s): int arr[n]: an array of integers Returns string: either YES or NO Input Format The first line contains , the number of test cases. The next pairs of lines each represent a test case. - The first line contains , the number of elements in the array . - The second line contains space-separated integers where .
Solution :
Solution in C :
In C++ :
#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define eps 1e-11
//#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 256
using namespace std;
long n,tests,ar[200000],s[200000];
long ans;
int main(){
//freopen("dagger.in","r",stdin);
//freopen("dagger.out","w",stdout);
//freopen("C:/input.txt","r",stdin);
//freopen("C:/output.txt","w",stdout);
ios_base::sync_with_stdio(0);
//cin.tie(0);
cin>>tests;
for (;tests;--tests)
{
cin>>n;
for (int i=1;i<=n;i++)
{
cin>>ar[i];
s[i]=s[i-1]+ar[i];
}
ans=0;
for (int i=1;i<=n;i++)
if(s[i-1]==s[n]-s[i])ans=1;
if (ans)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
cin.get();cin.get();
return 0;}
In Java :
import java.io.*;
import java.util.*;
public class Solution {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
void solve() throws IOException {
int n = nextInt();
int[] a = new int[n];
int sum = 0;
for (int i = 0; i < n; i++) {
a[i] = nextInt();
sum += a[i];
}
for (int i = 0, pref = 0; i < n; i++) {
if (sum - pref - a[i] == pref) {
out.println("YES");
return;
}
pref += a[i];
}
out.println("NO");
}
Solution() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int t = nextInt();
while (t-- > 0) {
solve();
}
out.close();
}
public static void main(String[] args) throws IOException {
new Solution();
}
String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}
String nextString() {
try {
return br.readLine();
} catch (IOException e) {
eof = true;
return null;
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}
In C :
#include<stdio.h>
int main()
{
int t,n,i;
scanf("%d",&t);
while(t--)
{
int a[100010],b[100010];
int sum=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
b[i]=sum;
scanf("%d",&a[i]);
sum+=a[i];
}
for(i=1;i<=n;i++)
{
if(b[i]==sum-b[i]-a[i])
{
printf("YES\n");
goto A;
}
}
printf("NO\n");
A:;
}
return(0);
}
In Python3 :
for t in range(int(input())):
n = int(input())
a = [int(b) for b in input().split()]
s = sum(a)
count = 0
for i in range(n):
if 2*count == s-a[i]:
print("YES")
break
count += a[i]
else:
print("NO")
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