927. Three Equal Parts

Description

You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value.

If it is possible, return any [i, j] with i + 1 < j, such that:

• arr[0], arr[1], ..., arr[i] is the first part,
• arr[i + 1], arr[i + 2], ..., arr[j - 1] is the second part, and
• arr[j], arr[j + 1], ..., arr[arr.length - 1] is the third part.
• All three parts have equal binary values.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

 

Example 1:

Input: arr = [1,0,1,0,1]
Output: [0,3]

Example 2:

Input: arr = [1,1,0,1,1]
Output: [-1,-1]

Example 3:

Input: arr = [1,1,0,0,1]
Output: [0,2]

 

Constraints:

• 3 <= arr.length <= 3 * 104
• arr[i] is 0 or 1

 

Solutions

Solution: Simulation

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} arr
 * @return {number[]}
 */
const threeEqualParts = function (arr) {
  const n = arr.length;
  const ones = arr.reduce((total, value) => total + value);

  if (!ones) return [0, n - 1];
  if (ones % 3) return [-1, -1];

  const preOnes = ones / 3;
  let start = 0;
  let mid = 0;
  let end = 0;
  let currentOnes = 0;

  for (let index = 0; index < n; index++) {
    const value = arr[index];

    if (!value) continue;
    if (!currentOnes) start = index;
    currentOnes += 1;
    if (currentOnes === preOnes + 1) {
      mid = index;
    }
    if (currentOnes === preOnes * 2 + 1) {
      end = index;
    }
  }

  while (end < n && arr[start] === arr[mid] && arr[mid] === arr[end]) {
    start += 1;
    mid += 1;
    end += 1;
  }
  if (end === n) return [start - 1, mid];

  return [-1, -1];
};