927

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

Companies:
Hotstar

Related Topics:
Array, Math

Solution 1.

// OJ: https://leetcode.com/problems/three-equal-parts/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> threeEqualParts(vector<int>& A) {
        int sum = accumulate(begin(A), end(A), 0);
        if (sum % 3) return {-1,-1};
        if (sum == 0) return {0, (int)A.size() - 1};
        sum /= 3; 
        int last = A.size() - 1;
        while (A[last] == 0) --last;
        int trailingZeros = A.size() - last - 1, firstBegin = 0;
        while (A[firstBegin] == 0) ++firstBegin;
        int firstEnd = firstBegin, cnt = 0;
        while (cnt < sum) cnt += A[firstEnd++];
        for (int i = 0; i < trailingZeros; ++i, ++firstEnd);
        int j = firstEnd, i = firstBegin;
        while (A[j] == 0) ++j;
        while (i < firstEnd && A[i] == A[j]) ++i, ++j;
        if (i < firstEnd) return {-1, -1};
        int secondEnd = j;
        while (A[j] == 0) ++j;
        i = firstBegin;
        while (i < firstEnd && A[i] == A[j]) ++i, ++j;
        if (i < firstEnd) return {-1, -1};
        return {firstEnd - 1, secondEnd};
    }
};