# 01.01.2025 [1422. Maximum Score After Splitting a String]

Max(left_zeros + right_ones) #easy

01.12.2025

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Problem TLDR

Max(left_zeros + right_ones) #easy

Intuition

The brute-force works: try every possible position split. The better way is two-pass: count the total ones, then decrease it at every step.

The optimal solution is a single pass: notice, how the sum = zeros + ones changes at every move, we actually computing the balance around the total ones.

Approach

• try every solution

• how short can it be?

Complexity

• Time complexity: $$O(n)$$

• Space complexity: $$O(1)$$

Code

    fun maxScore(s: String): Int {
        var ones = s.last() - '0'; var b = 0
        return s.dropLast(1).maxOf {
            if (it > '0') { ones++; --b } else ++b
        } + ones
    }

    pub fn max_score(s: String) -> i32 {
        let (mut ones, mut b) = (0, 0);
        s.bytes().enumerate().map(|(i, c)| {
            ones += (c > b'0') as i32;
            if i < s.len() - 1 { 
              b -= (c > b'0') as i32 * 2 - 1; } b
        }).max().unwrap() + ones
    }

    int maxScore(string s) {
        int o = s[s.size() - 1] == '1', b = 0, r = -1;
        for (int i = 0; i < s.size() - 1; ++i) {
            o += s[i] > '0';
            b -= (s[i] > '0') * 2 - 1;
            r = max(r, b);
        } return r + o;
    }