2027. Minimum number of operations to convert strings

Address ：

Power button https://leetcode-cn.com/problems/minimum-moves-to-convert-string/

subject ：

Give you a string s , from n Characters make up , Each character is not 'X' Namely 'O' .

once operation Defined as from s Selected from Three consecutive characters And convert each selected character to 'O' . Be careful , If the character is already 'O' , Just keep unchanged .

Return to s All characters in are converted to 'O' executable   least   Operating frequency .

Example 1：

Input ：s = "XXX" Output ：1 explain ：XXX -> OOO One operation , Select All 3 Characters , And turn them into 'O' .

Example 2：

Input ：s = "XXOX" Output ：2 explain ：XXOX -> OOOX -> OOOO The first operation , Choose the former 3 Characters , And convert these characters to 'O' . then , After selection 3 Characters , And perform the conversion . The resulting string consists entirely of characters 'O' form .

Example 3：

Input ：s = "OOOO" Output ：0 explain ：s There is no need to convert 'X' .

Tips ：

3 <= s.length <= 1000 s[i] by 'X' or 'O'

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/minimum-moves-to-convert-string
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas ：

Flop once yes 3 Elements , If the first element is 'O', It's a waste of opportunity , So looking for elements directly is 'X'  The beginning of

Every time 3 One for a group , Till the end , Pay attention to the lower boundary

Method 1 、 find 'X' At the beginning 3 Tuples

int minimumMoves(char * s){
	int slen = strlen(s);
	int moves = 0;
	
	int i=0;
	while(s[i])
	{
		if(s[i] == 'O')
			i++;
		else
		{
			if( (i+3) <= slen )
			{
				moves++;
				i += 3;
			}
			else
			{
				moves++;
				break;
			}
		}
	}
	
	return moves;
}

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