Minimum operations to make a special number

Math String Greedy Enumeration

38.5% acceptance

Feb 25, 2026

373

You are given a 0-indexed string num representing a non-negative integer. In one operation, you can pick any digit of num and delete it. Note that if you delete all the digits of num, num becomes 0. Return the minimum number of operations required to make num special. An integer x is considered special if it is divisible by 25.

Solution

Rust

Time O(n³)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn minimum_operations(num: String) -> i32 {
    let b = num.as_bytes();
    let n = b.len();
    // Divisible by 25: last two digits in {00,25,50,75}
    let endings: &[(u8, u8)] = &[(b'0', b'0'), (b'2', b'5'), (b'5', b'0'), (b'7', b'5')];
    // Keep a single '0' anywhere => result is 0 (divisible by 25)
    // Cost = delete everything except that one '0' = n - 1 (if any '0' exists)
    let has_zero = b.iter().any(|&c| c == b'0');
    let mut ans = if has_zero { n as i32 - 1 } else { n as i32 }; // n => delete all => 0

    for &(d2, d1) in endings {
      // find rightmost d1, then rightmost d2 before it
      let mut j = n;
      loop {
        if j == 0 { break; }
        j -= 1;
        if b[j] == d1 {
          // deletions after j
          let del_right = (n - 1 - j) as i32;
          // find d2 to the left of j
          let mut i = j;
          loop {
            if i == 0 { break; }
            i -= 1;
            if b[i] == d2 {
              let del_mid = (j - 1 - i) as i32;
              ans = ans.min(del_right + del_mid);
              break;
            }
          }
          break;
        }
      }
    }
    ans
  }
}