2155. All Divisions With the Highest Score of a Binary Array 

Description

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) nums_left and nums_right:

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

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

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: nums_left is []. nums_right is [0,0,1,0]. The score is 0 + 1 = 1.
- 1: nums_left is [0]. nums_right is [0,1,0]. The score is 1 + 1 = 2.
- 2: nums_left is [0,0]. nums_right is [1,0]. The score is 2 + 1 = 3.
- 3: nums_left is [0,0,1]. nums_right is [0]. The score is 2 + 0 = 2.
- 4: nums_left is [0,0,1,0]. nums_right 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: nums_left is []. nums_right is [0,0,0]. The score is 0 + 0 = 0.
- 1: nums_left is [0]. nums_right is [0,0]. The score is 1 + 0 = 1.
- 2: nums_left is [0,0]. nums_right is [0]. The score is 2 + 0 = 2.
- 3: nums_left is [0,0,0]. nums_right 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: nums_left is []. nums_right is [1,1]. The score is 0 + 2 = 2.
- 1: nums_left is [1]. nums_right is [1]. The score is 0 + 1 = 1.
- 2: nums_left is [1,1]. nums_right 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.

Solutions

Solution 1: Prefix Sum

We start from $i = 0$, using two variables $\textit{l0}$ and $\textit{r1}$ to respectively record the number of $1$s to the left and right of $i$. Initially, $\textit{l0} = 0$, while $\textit{r1} = \sum \textit{nums}$.

We iterate through the array $\textit{nums}$. For each $i$, we update $\textit{l0}$ and $\textit{r1}$, calculate the current grouping score $t = \textit{l0} + \textit{r1}$. If $t$ equals the current maximum grouping score $\textit{mx}$, then we add $i$ to the answer array. If $t$ is greater than $\textit{mx}$, we update $\textit{mx}$ to $t$, clear the answer array, and then add $i$ to the answer array.

After the iteration ends, we return the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def maxScoreIndices(self, nums: List[int]) -> List[int]:
        l0, r1 = 0, sum(nums)
        mx = r1
        ans = [0]
        for i, x in enumerate(nums, 1):
            l0 += x ^ 1
            r1 -= x
            t = l0 + r1
            if mx == t:
                ans.append(i)
            elif mx < t:
                mx = t
                ans = [i]
        return ans

Java

class Solution {
    public List<Integer> maxScoreIndices(int[] nums) {
        int l0 = 0, r1 = Arrays.stream(nums).sum();
        int mx = r1;
        List<Integer> ans = new ArrayList<>();
        ans.add(0);
        for (int i = 1; i <= nums.length; ++i) {
            int x = nums[i - 1];
            l0 += x ^ 1;
            r1 -= x;
            int t = l0 + r1;
            if (mx == t) {
                ans.add(i);
            } else if (mx < t) {
                mx = t;
                ans.clear();
                ans.add(i);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> maxScoreIndices(vector<int>& nums) {
        int l0 = 0, r1 = accumulate(nums.begin(), nums.end(), 0);
        int mx = r1;
        vector<int> ans = {0};
        for (int i = 1; i <= nums.size(); ++i) {
            int x = nums[i - 1];
            l0 += x ^ 1;
            r1 -= x;
            int t = l0 + r1;
            if (mx == t) {
                ans.push_back(i);
            } else if (mx < t) {
                mx = t;
                ans = {i};
            }
        }
        return ans;
    }
};

Go

func maxScoreIndices(nums []int) []int {
	l0, r1 := 0, 0
	for _, x := range nums {
		r1 += x
	}
	mx := r1
	ans := []int{0}
	for i, x := range nums {
		l0 += x ^ 1
		r1 -= x
		t := l0 + r1
		if mx == t {
			ans = append(ans, i+1)
		} else if mx < t {
			mx = t
			ans = []int{i + 1}
		}
	}
	return ans
}

TypeScript

function maxScoreIndices(nums: number[]): number[] {
    const n = nums.length;
    let [l0, r1] = [0, nums.reduce((a, b) => a + b, 0)];
    let mx = r1;
    const ans: number[] = [0];
    for (let i = 1; i <= n; ++i) {
        const x = nums[i - 1];
        l0 += x ^ 1;
        r1 -= x;
        const t = l0 + r1;
        if (mx === t) {
            ans.push(i);
        } else if (mx < t) {
            mx = t;
            ans.length = 0;
            ans.push(i);
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn max_score_indices(nums: Vec<i32>) -> Vec<i32> {
        let mut l0 = 0;
        let mut r1: i32 = nums.iter().sum();
        let mut mx = r1;
        let mut ans = vec![0];

        for i in 1..=nums.len() {
            let x = nums[i - 1];
            l0 += x ^ 1;
            r1 -= x;
            let t = l0 + r1;
            if mx == t {
                ans.push(i as i32);
            } else if mx < t {
                mx = t;
                ans = vec![i as i32];
            }
        }

        ans
    }
}