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[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