## The challenge#

You are going to be given an array of integers. Your job is to take that array and find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1.

### For example:#

Let’s say you are given the array {1,2,3,4,3,2,1}:
Your function will return the index 3, because at the 3rd position of the array, the sum of left side of the index ({1,2,3}) and the sum of the right side of the index ({3,2,1}) both equal 6.

### Let’s look at another one.#

You are given the array {1,100,50,-51,1,1}:
Your function will return the index 1, because at the 1st position of the array, the sum of left side of the index ({1}) and the sum of the right side of the index ({50,-51,1,1}) both equal 1.

### Last one:#

You are given the array {20,10,-80,10,10,15,35}
At index 0 the left side is {}
The right side is {10,-80,10,10,15,35}
They both are equal to `` when added. (Empty arrays are equal to 0 in this problem)
Index 0 is the place where the left side and right side are equal.

Note: Please remember that in most programming/scripting languages the index of an array starts at 0.

### Input:#

An integer array of length 0 < arr < 1000. The numbers in the array can be any integer positive or negative.

### Output:#

The lowest index N where the side to the left of N is equal to the side to the right of N. If you do not find an index that fits these rules, then you will return -1.

### Note:#

If you are given an array with multiple answers, return the lowest correct index.

## The solution in Java code#

Option 1:

public class Solution {
public static int findEvenIndex(int[] arr) {
int sum = 0, cSum = 0;
for (int i = 0; i < arr.length; i++) sum += arr[i];
for(int i = 0; i < arr.length; i++){
if (cSum == sum - arr[i]) return i;
cSum += arr[i];
sum -= arr[i];
}
return -1;
}
}

Option 2:

import java.util.Arrays;

public class Solution {
public static int findEvenIndex(int[] arr) {
int left = 0;
int right = Arrays.stream(arr).sum();
for (int i=0; i<arr.length; i++){
right -= arr[i];
if (left == right) return i;
left += arr[i];
}
return -1;
}
}

Option 3:

import java.util.stream.IntStream;

public class Solution {
public static int findEvenIndex(int[] arr) {
return IntStream.range(0, arr.length)
.filter(n -> IntStream.of(arr).limit(n).sum() == IntStream.of(arr).skip(n + 1).sum())
.findFirst().orElse(-1);
}
}

## Test cases to validate our solution#

import static org.junit.Assert.*;
import java.util.ArrayList;
import org.junit.Test;

public class SolutionTest {
@Test
public void test() {
assertEquals(3,Solution.findEvenIndex(new int[] {1,2,3,4,3,2,1}));
assertEquals(1,Solution.findEvenIndex(new int[] {1,100,50,-51,1,1}));
assertEquals(-1,Solution.findEvenIndex(new int[] {1,2,3,4,5,6}));
assertEquals(3,Solution.findEvenIndex(new int[] {20,10,30,10,10,15,35}));
assertEquals(-1,Solution.findEvenIndex(new int[] {-8505, -5130, 1926, -9026}));
assertEquals(1,Solution.findEvenIndex(new int[] {2824, 1774, -1490, -9084, -9696, 23094}));
assertEquals(6,Solution.findEvenIndex(new int[] {4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4}));
}
}