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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0927_three_equal_parts;

// #Hard #Array #Math #2022_03_29_Time_2_ms_(90.20%)_Space_47.4_MB_(96.08%)

/**
 * 927 - Three Equal Parts\.
 *
 * Hard
 *
 * 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`
**/
public class Solution {
    public int[] threeEqualParts(int[] arr) {
        int ones = 0;
        for (int num : arr) {
            ones += num;
        }
        if (ones == 0) {
            return new int[] {0, 2};
        } else if (ones % 3 != 0) {
            return new int[] {-1, -1};
        }
        ones /= 3;
        int index1 = -1;
        int index2 = -1;
        int index3 = -1;
        int totalOnes = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == 0) {
                continue;
            }
            totalOnes += arr[i];
            if (totalOnes == 1) {
                index1 = i;
            } else if (totalOnes == ones + 1) {
                index2 = i;
            } else if (totalOnes == 2 * ones + 1) {
                index3 = i;
            }
        }
        while (index3 < arr.length) {
            if (arr[index1] == arr[index3] && arr[index2] == arr[index3]) {
                ++index1;
                ++index2;
                ++index3;
            } else {
                return new int[] {-1, -1};
            }
        }
        return new int[] {index1 - 1, index2};
    }
}