Question

2027. Minimum Moves to Convert String

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Description

You are given a string s consisting of n characters which are either 'X' or 'O'.

A move is defined as selecting three consecutive characters of s and converting them to 'O'. Note that if a move is applied to the character 'O', it will stay the same.

Return _the minimum number of moves required so that all the characters of _s_ are converted to _'O'.

 

Example 1:

Input: s = "XXX"
Output: 1
Explanation: XXX -> OOO
We select all the 3 characters and convert them in one move.

Example 2:

Input: s = "XXOX"
Output: 2
Explanation: XXOX -> OOOX -> OOOO
We select the first 3 characters in the first move, and convert them to 'O'.
Then we select the last 3 characters and convert them so that the final string contains all 'O's.

Example 3:

Input: s = "OOOO"
Output: 0
Explanation: There are no 'X's in s to convert.

 

Constraints:

• 3 <= s.length <= 1000
• s[i] is either 'X' or 'O'.

Solutions

Solution 1: Greedy Algorithm

Traverse the string $s$. Whenever you encounter 'X', move the pointer $i$ three steps forward and add $1$ to the answer; otherwise, move the pointer $i$ one step forward.

The time complexity is $O(n)$, where $n$ represents the length of the string $s$.

class Solution:
  def minimumMoves(self, s: str) -> int:
    ans = i = 0
    while i < len(s):
      if s[i] == "X":
        ans += 1
        i += 3
      else:
        i += 1
    return ans

class Solution {
  public int minimumMoves(String s) {
    int ans = 0;
    for (int i = 0; i < s.length(); ++i) {
      if (s.charAt(i) == 'X') {
        ++ans;
        i += 2;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int minimumMoves(string s) {
    int ans = 0;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == 'X') {
        ++ans;
        i += 2;
      }
    }
    return ans;
  }
};

func minimumMoves(s string) (ans int) {
  for i := 0; i < len(s); i++ {
    if s[i] == 'X' {
      ans++
      i += 2
    }
  }
  return
}

function minimumMoves(s: string): number {
  const n = s.length;
  let ans = 0;
  let i = 0;
  while (i < n) {
    if (s[i] === 'X') {
      ans++;
      i += 3;
    } else {
      i++;
    }
  }
  return ans;
}

impl Solution {
  pub fn minimum_moves(s: String) -> i32 {
    let s = s.as_bytes();
    let n = s.len();
    let mut ans = 0;
    let mut i = 0;
    while i < n {
      if s[i] == b'X' {
        ans += 1;
        i += 3;
      } else {
        i += 1;
      }
    }
    ans
  }
}

int minimumMoves(char* s) {
  int n = strlen(s);
  int ans = 0;
  int i = 0;
  while (i < n) {
    if (s[i] == 'X') {
      ans++;
      i += 3;
    } else {
      i++;
    }
  }
  return ans;
}