Tape Equilibrium

12 Apr, 2021

Problem

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000];
each element of array A is an integer within the range [−1,000..1,000].

My Solution

failed

function solution(A) {
	let sumAll = A.reduce((sum, current) => sum + current, 0);
	let temp = 0;
	let diff = 0;

	for (let i in A) {
		temp += A[i];
		if (Math.abs(sumAll - temp * 2) < diff || diff === 0) {
			diff = Math.abs(sumAll - temp * 2);
		}
	}

	return diff;
}

failed

function solution(A) {
	let right = A.reduce((sum, current) => sum + current, 0);
	let left = 0;
	let diff = 0;

	for (let i = 0; i < A.length; i++) {
		left += A[i];
		right -= A[i];
		if (Math.abs(right - left) < diff || diff === 0) {
			diff = Math.abs(right - left);
		}
	}

	return diff;
}

failed

function solution(A) {
	let right = A.reduce((sum, current) => sum + current, 0);
	let left = 0;
	let diff = [];

	for (let i = 0; i < A.length; i++) {
		left += A[i];
		right -= A[i];
		diff.push(Math.abs(right - left));
	}

	return Math.min(...diff);
}

success

function solution(A) {
	let right = A.reduce((sum, current) => sum + current);
	let left = 0;
	let answer = null;

	for (let i = 0; i < A.length - 1; i++) {
		left += A[i];
		right -= A[i];
		let diff = Math.abs(left - right);
		if (answer === null || answer > diff) {
			answer = diff;
		}
	}

	return answer;
}