Posted on Jul 7, 2020

Daily Challenge #265 - Equal Sides

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!

This challenge comes from Shivo on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (9)

SavagePixie • Jul 7 '20 • Edited

JavaScript

const sum = (a, b) => a + b const doStuff = xs => xs.findIndex( (x, i, arr) => arr.slice(0, i).reduce(sum) === arr.slice(i + 1).reduce(sum) )

Miguel Manjarres • Jul 7 '20

Python

from functools import reduce def findIndex(arr): sum = lambda a, b : a + b for i in range(len(arr)): leftSum = reduce(sum, arr[0:i+1]) rightSum = reduce(sum, arr[i:len(arr)]) if leftSum == rightSum: return i return -1 print(findIndex([1,2,3,4,3,2,1])) # 3 print(findIndex([1,100,50,-51,1,1])) # 1 print(findIndex([20,10,30,10,10,15,3])) # -1 print(findIndex([-8505, -5130, 1926, -9026])) # -1

Mofid Jobakji • Jul 8 '20

javascript

 const fn = (arr) => { let sum = 0; const total = arr.reduce((a,b)=> a+b , 0); return arr.findIndex((x, i, arr) => (total - (sum += x)) * 2 + x === total) ; } console.log(fn([1, 2, 3, 4, 3, 2, 1])); // 3 console.log(fn([1, 100, 50, -51, 1, 1])); // 1 console.log(fn([20, 10, 30, 10, 10, 15, 35])); // 3 console.log(fn([-8505, -5130, 1926, -9026]));// -1 console.log(fn([11,4,30,5,10,20,15]));// 3

Hayden Mankin • Jul 7 '20

Javascript

const equalSides = (arr) => { let right = arr.reduce((a, b) => a + b); let left = 0; return arr.findIndex(i => { right -= i; if (left == right) { return true; } left += i; return false; }); }; console.log(equalSides([1, 2, 3, 4, 3, 2, 1])); // 3 console.log(equalSides([1, 100 , 50, -51 , 1 , 1])); // 1 console.log(equalSides([20, 10, 30, 10, 10, 15, 35])); // 3 console.log(equalSides([-8505, -5130, 1926, -9026])); // -1

Amin • Jul 8 '20

Haskell

halve :: [a] -> ([a], [a]) halve list = (take half list, drop (half + rest) list) where size = length list half = div size 2 rest = mod size 2 isEqualSides :: [Int] -> Int isEqualSides [] = -1 isEqualSides integers = if (sum firstHalf) == (sum secondHalf) then length firstHalf else isEqualSides (init firstHalf ++ tail secondHalf) where (firstHalf, secondHalf) = halve integers

Dimitri Lahaye • Jul 8 '20

My humble solution:

function findN(array) { reducer = (s, o) => s + o; return array.findIndex((a, i, arr) => arr.slice(0, i).reduce(reducer, 0) === arr.slice(i + 1 - array.length).reduce(reducer, 0)); }

heykieran • Jul 8 '20

Clojure

(defn equal-sides [candidate-array] (loop [c candidate-array idx 1 lt 0 rt 0] (cond (empty? c) -1 (and (= lt rt) (= 1 (count c))) idx :else (let [inc-l (> (+ rt (last c)) (+ lt (first c)))] (recur (if inc-l (rest c) (butlast c)) (if inc-l (inc idx) idx) (+ lt (if inc-l (first c) 0)) (+ rt (if-not inc-l (last c) 0)))))))

E. Choroba • Jul 8 '20

Perl

#!/usr/bin/perl use strict; use warnings; use experimental qw{ signatures }; use List::Util qw{ sum }; sub equal_sides ($arr) { my ($left, $current, $right) = (0, 0, sum(@$arr[1 .. $#$arr])); until ($current > $#$arr || $left == $right) { $left += $arr->[$current]; $right -= $arr->[++$current] // 0; } return $current > $#$arr ? -1 : $current } use Test::More tests => 4; my @tests = ([1, 2, 3, 4, 3, 2, 1], 3, [1, 100, 50, -51, 1, 1], 1, [20, 10, 30, 10, 10, 15, 35], 3, [-8505, -5130, 1926, -9026], -1); while (my ($arr, $result) = splice @tests, 0, 2) { is equal_sides($arr), $result; }

Mehmed Duhovic • Jul 7 '20