오늘의 기억

# [Algorithm] TapeEquilibrium

## Question

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].

## Solution (TypeScript)

``````function solution(n: number[]) {
const length = n.length;
if (length === 2) {
return Math.abs(n[0] - n[1]);
}

const sumArray = [];

for (let i = 0; i < length; i += 1) {
const val = n[i];
if (0 === i) {
sumArray[0] = val;
} else {
sumArray[i] = sumArray[i - 1] + val;
}
}

let result = Number.MAX_VALUE;
for (let i = 0; i < length - 1; i += 1) {
const tmp = Math.abs(sumArray[i] - (sumArray[length - 1] - sumArray[i]));
result = Math.min(result, tmp);
}
return result;
}
``````

Author: Codility, Lesson 3

20200105 Charyum.Park