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1422. Maximum Score After Splitting a String

Description

Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

 

Example 1:

Input: s = "011101"
Output: 5 
Explanation: 
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5 
left = "01" and right = "1101", score = 1 + 3 = 4 
left = "011" and right = "101", score = 1 + 2 = 3 
left = "0111" and right = "01", score = 1 + 1 = 2 
left = "01110" and right = "1", score = 2 + 1 = 3

Example 2:

Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5

Example 3:

Input: s = "1111"
Output: 3

 

Constraints:

  • 2 <= s.length <= 500
  • The string s consists of characters '0' and '1' only.

Solutions

  • class Solution {
    public int maxScore(String s) {
        int ans = 0;
        for (int i = 1; i < s.length(); ++i) {
            int t = 0;
            for (int j = 0; j < i; ++j) {
                if (s.charAt(j) == '0') {
                    ++t;
                }
            }
            for (int j = i; j < s.length(); ++j) {
                if (s.charAt(j) == '1') {
                    ++t;
                }
            }
            ans = Math.max(ans, t);
        }
        return ans;
    }
}

  • class Solution {
public:
    int maxScore(string s) {
        int ans = 0;
        for (int i = 1, n = s.size(); i < n; ++i) {
            int t = 0;
            for (int j = 0; j < i; ++j) {
                t += s[j] == '0';
            }
            for (int j = i; j < n; ++j) {
                t += s[j] == '1';
            }
            ans = max(ans, t);
        }
        return ans;
    }
};

  • class Solution:
    def maxScore(self, s: str) -> int:
        return max(s[:i].count('0') + s[i:].count('1') for i in range(1, len(s)))


  • func maxScore(s string) int {
	ans := 0
	for i, n := 1, len(s); i < n; i++ {
		t := 0
		for j := 0; j < i; j++ {
			if s[j] == '0' {
				t++
			}
		}
		for j := i; j < n; j++ {
			if s[j] == '1' {
				t++
			}
		}
		ans = max(ans, t)
	}
	return ans
}

  • function maxScore(s: string): number {
    const n = s.length;
    let res = 0;
    let score = 0;
    if (s[0] === '0') {
        score++;
    }
    for (let i = 1; i < n; i++) {
        if (s[i] === '1') {
            score++;
        }
    }
    res = Math.max(res, score);
    for (let i = 1; i < n - 1; i++) {
        if (s[i] === '0') {
            score++;
        } else if (s[i] === '1') {
            score--;
        }
        res = Math.max(res, score);
    }
    return res;
}


  • impl Solution {
    pub fn max_score(s: String) -> i32 {
        let n = s.len();
        let mut res = 0;
        let mut score = 0;
        let bs = s.as_bytes();
        if bs[0] == b'0' {
            score += 1;
        }
        for i in 1..n {
            if bs[i] == b'1' {
                score += 1;
            }
        }
        res = res.max(score);
        for i in 1..n - 1 {
            if bs[i] == b'0' {
                score += 1;
            } else if bs[i] == b'1' {
                score -= 1;
            }
            res = res.max(score);
        }
        res
    }
}

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