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.
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.
Tests
{1,2,3,4,3,2,1}
{1,100,50,-51,1,1}
{20,10,30,10,10,15,35}
{-8505, -5130, 1926, -9026}
Good luck!
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Discussion (11)
JavaScript
This code is clever, however, when we talk about best performance with O(n) , this code is slower. See the JsPerf Performance of your code here: pancake-island-lollipop. You can try this instead:
let answer = (a, b = a => a.reduce((a, b) => a + b, 0)) => a.findIndex((_, i) => b(a.slice(0, i)) == b(a.slice(i + 1))) console.log(answer([1, 2, 3, 4, 3, 2, 1])); // prints '3' console.log(answer([1, 100, 50, -51, 1, 1])); // prints '1' console.log(answer([20, 10, 30, 10, 10, 15, 35])); // prints '3' console.log(answer([-8505, -5130, 1926, -9026])); // prints '-1'
Currently, the fastest form of loop is a standard for loop with length caching
This code also solves the problem with length caching, in my opinion, this is also the fastest in general:
function answer(arr) { var left = 0, right = arr.reduce((p, c) => p + c, 0) for (var i = 0, len=arr.length; i < len; i++) { if (i > 0) left += arr[i - 1]; right -= arr[i]; if (left == right) return i; } return -1; } console.log(answer([1, 2, 3, 4, 3, 2, 1])); // prints '3' console.log(answer([1, 100, 50, -51, 1, 1])); // prints '1' console.log(answer([20, 10, 30, 10, 10, 15, 35])); // prints '3' console.log(answer([-8505, -5130, 1926, -9026])); // prints '-1'
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My humble solution:
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