927. Three Equal Parts

Problem 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

Solution

This video explains well

Python

class Solution:
    def threeEqualParts(self, arr: List[int]) -> List[int]:
        total = sum(arr)
        if total % 3 != 0:
            return [-1, -1]

        if total == 0:
            return [0, len(arr) - 1]

        ones = total // 3

        i = len(arr) - 1
        while i > 0 and ones > 0:
            if arr[i] == 1:
                ones -= 1
            i -= 1

        # arr[i+1:] is the number
        digit = arr[i + 1:]
        ret = []
        start = 0
        while start < len(arr) and arr[start] == 0:
            start += 1

        if not self.check(arr, start, digit):
            return [-1, -1]

        ret.append(start + len(digit) - 1)

        start = start + len(digit)
        while start < len(arr) and arr[start] == 0:
            start += 1

        if not self.check(arr, start, digit):
            return [-1, -1]

        ret.append(start + len(digit))
        return ret


    def check(self, arr, pointer, digit):
        if arr[pointer: pointer + len(digit)] == digit:
            return True
        return False

Java

class Solution {
    public int[] threeEqualParts(int[] arr) {
        int sum = 0;
        for (int i = 0; i < arr.length; i++) {
            sum += arr[i];
        }

        if (sum % 3 != 0) {
            return new int[]{-1, -1};
        }

        if (sum == 0) {
            return new int[]{0, arr.length - 1};
        }

        int[] ret = new int[2];
        int count = (int) sum / 3;

        // count from end to start
        int j = arr.length - 1;
        while (j >= 0 && count > 0) {
            if (arr[j] == 1) {
                count--;
            }
            j--;
        }

        // j point to the first `1` in the third part
        j++;


        // check for the first part
        int i = 0;
        while (i < arr.length && arr[i] == 0) {
            i++;
        }

        // i points to the first `1`

        int pointer = j;
        while (pointer < arr.length) {
            if (arr[i] == arr[pointer]) {
                i++;
                pointer++;
            } else {
                break;
            }
        }

        if (pointer != arr.length) {
            return new int[]{-1, -1};
        }

        ret[0] = i - 1;

        // check for the second part
        while (i < arr.length && arr[i] == 0) {
            i++;
        }

        pointer = j;
        while (pointer < arr.length) {
            if (arr[i] == arr[pointer]) {
                i++;
                pointer++;
            } else {
                break;
            }
        }

        if (pointer != arr.length) {
            return new int[]{-1, -1};
        }

        ret[1] = i;

        return ret;
    }
}

Both solutions can extract the “repeated” part to helper functions, I am just lazy

Complexity Analysis

• Time Complexity
  • O(n)
• Space Complexity
  • O(1)