Maximum number of operations to move ones to the end

String Greedy Counting

67.1% acceptance

Feb 25, 2026

530

You are given a binary string s. You can perform the following operation on the string any number of times: Choose any index i from the string where i + 1 < s.length such that s[i] == '1' and s[i + 1] == '0'. Move the character s[i] to the right until it reaches the end of the string or another '1'. Return the maximum number of operations that you can perform.

Solution

Rust

Time O(n)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn max_operations(s: String) -> i32 {
    // Each '1' contributes 1 op for each '0'-group to its right.
    // Scan right to left: track # of '0'-groups seen so far.
    let bytes = s.as_bytes();
    let n = bytes.len();
    let mut ops = 0i32;
    let mut groups_right = 0i32;
    let mut in_zero = false;
    for i in (0..n).rev() {
      if bytes[i] == b'0' {
        in_zero = true;
      } else {
        if in_zero {
          groups_right += 1;
          in_zero = false;
        }
        ops += groups_right;
      }
    }
    ops
  }
}