[CodeWars] Equal Sides Of An Array

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 0 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.

An empty array should be treated like a 0 in this problem.


My Solution

#include <vector>
using namespace std;

int find_even_index(const vector <int> numbers) {
    int left = 0;
    int right = 0;

    for (int i = 0; i < numbers.size(); ++i)
    {
        for (int j = 0; j < i; ++j)
            left += numbers[j];

        for (int k = numbers.size()1; k > i; k)
            right += numbers[k];

        if (left == right)
            return i;
        else { left = 0; right = 0; }
    }

    return 1;
}
cs

I used two variables, one is the sum of left, the other is the sum of right. Then, search which index is correct for the condition of the problem. If there’s not, returns -1.

글쓴이: BakJH

Student of Daedeok SW Meister Highschool, in Korea.

답글 남기기

아래 항목을 채우거나 오른쪽 아이콘 중 하나를 클릭하여 로그 인 하세요:

WordPress.com 로고

WordPress.com의 계정을 사용하여 댓글을 남깁니다. 로그아웃 /  변경 )

Google photo

Google의 계정을 사용하여 댓글을 남깁니다. 로그아웃 /  변경 )

Twitter 사진

Twitter의 계정을 사용하여 댓글을 남깁니다. 로그아웃 /  변경 )

Facebook 사진

Facebook의 계정을 사용하여 댓글을 남깁니다. 로그아웃 /  변경 )

%s에 연결하는 중