LeetCode 2027. Minimum Moves to Convert String Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScriptRustC

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(code-box)

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; } }(code-box)

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; } };(code-box)

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

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; }(code-box)

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 } }(code-box)

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; }(code-box)