2155

2155. All Divisions With the Highest Score of a Binary Array (Medium)

You are given a 0-indexed binary array nums of length n. nums can be divided at index i (where 0 <= i <= n) into two arrays (possibly empty) numsleft and numsright:

• numsleft has all the elements of nums between index 0 and i - 1 (inclusive), while numsright has all the elements of nums between index i and n - 1 (inclusive).
• If i == 0, numsleft is empty, while numsright has all the elements of nums.
• If i == n, numsleft has all the elements of nums, while numsright is empty.

The division score of an index i is the sum of the number of 0's in numsleft and the number of 1's in numsright.

Return all distinct indices that have the highest possible division score. You may return the answer in any order.

 

Example 1:

Input: nums = [0,0,1,0]
Output: [2,4]
Explanation: Division at index
- 0: numsleft is []. numsright is [0,0,1,0]. The score is 0 + 1 = 1.
- 1: numsleft is [0]. numsright is [0,1,0]. The score is 1 + 1 = 2.
- 2: numsleft is [0,0]. numsright is [1,0]. The score is 2 + 1 = 3.
- 3: numsleft is [0,0,1]. numsright is [0]. The score is 2 + 0 = 2.
- 4: numsleft is [0,0,1,0]. numsright is []. The score is 3 + 0 = 3.
Indices 2 and 4 both have the highest possible division score 3.
Note the answer [4,2] would also be accepted.

Example 2:

Input: nums = [0,0,0]
Output: [3]
Explanation: Division at index
- 0: numsleft is []. numsright is [0,0,0]. The score is 0 + 0 = 0.
- 1: numsleft is [0]. numsright is [0,0]. The score is 1 + 0 = 1.
- 2: numsleft is [0,0]. numsright is [0]. The score is 2 + 0 = 2.
- 3: numsleft is [0,0,0]. numsright is []. The score is 3 + 0 = 3.
Only index 3 has the highest possible division score 3.

Example 3:

Input: nums = [1,1]
Output: [0]
Explanation: Division at index
- 0: numsleft is []. numsright is [1,1]. The score is 0 + 2 = 2.
- 1: numsleft is [1]. numsright is [1]. The score is 0 + 1 = 1.
- 2: numsleft is [1,1]. numsright is []. The score is 0 + 0 = 0.
Only index 0 has the highest possible division score 2.

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• nums[i] is either 0 or 1.

Similar Questions:

• Ones and Zeroes (Medium)
• Max Consecutive Ones II (Medium)
• Count Subarrays With More Ones Than Zeros (Medium)
• Array Partition I (Easy)
• Divide Array in Sets of K Consecutive Numbers (Medium)

Solution 1. Count Left Ones and Right Zeros on the fly

Let zero/one be the number of zeros/ones in the left/right part. Initially one = SUM(A).

For each i, score = zero + one. And we can update zero and one when we move A[i] from right part to left part.

// OJ: https://leetcode.com/problems/all-divisions-with-the-highest-score-of-a-binary-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> maxScoreIndices(vector<int>& A) {
        int N = A.size(), one = accumulate(begin(A), end(A), 0), zero = 0, mx = 0;
        vector<int> ans;
        for (int i = 0; i <= N; ++i) {
            int score = one + zero;
            if (score > mx) {
                ans = {i};
                mx = score;
            } else if (score == mx) ans.push_back(i);
            if (i < N) {
                one -= A[i] == 1;
                zero += A[i] == 0;
            }
        }
        return ans;
    }
};

Note that here I marked extra space complexity as O(N) because the ans is used to store some temporary results whose length might be as long as N-1 while the answer might be just of length 1. So, the extra space we used might be linear to the input but not strictly confined by the size of the real answer.

That said, we can make it O(1) space by calculating the maximum score in one loop first then gather the ans array in another loop. In this way, the size of the ans is absolutely the same as the size of the real answer and won't be linear to the input size. But, this requires two loops.

Solution 2. Prefix Sum

Let sum[i+1] be A[0] + ... + A[i].

For index i, there are i - sum[i] zeros in the left part and sum[N] - sum[i] ones in the right part.

So the score for i is i - sum[i] + sum[N] - sum[i].

// OJ: https://leetcode.com/problems/all-divisions-with-the-highest-score-of-a-binary-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> maxScoreIndices(vector<int>& A) {
        int N = A.size(), mx = 0;
        vector<int> sum(N + 1, 0), ans;
        for (int i = 0; i < N; ++i) sum[i + 1] = sum[i] + A[i];
        for (int i = 0; i <= N; ++i) {
            int score = i - sum[i] + sum[N] - sum[i];
            if (score > mx) {
                ans = {i};
                mx = score;
            } else if (score == mx) ans.push_back(i);
        }
        return ans;
    }
};

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https://leetcode.com/problems/all-divisions-with-the-highest-score-of-a-binary-array/discuss/1730117