# 알고리즘 TapeEquilibrium

taewo·2020년 6월 23일
0

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].

function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
let result = null
if (A.length == 2) {
result = Math.abs(A[0] - A[1])
}

for (let i=1; i<A.length; i++) {
const pVal = Math.abs(sum(A.slice(0,i)) - sum(A.slice(i)))
if (result === null) {
result = pVal
} else if (result > pVal){
result = pVal
}
}

return result
}

function sum (arr) {
return arr.reduce((acc, cur) => {
return acc + cur
})
}