Using the same structure as a bubblesort algorithm
Solution to this problem resembles the same structure as a bubble-sort algorithm, which would be:
done = false
while(not done)
done = true
try to swap at least one pair. if success, done = false
It is an N^2 solution, for the problem below with N=1000, it works very quickly. Problem and solution down below, cheers, ACC.
2027. Minimum Moves to Convert String
Easy
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'sinsto convert.
Constraints:
3 <= s.length <= 1000s[i]is either'X'or'O'.
Accepted
7,165
Submissions
14,440
public int MinimumMoves(string s)
{
StringBuilder sb = new StringBuilder(s);
int count = 0;
bool done = false;
while (!done)
{
done = true;
for (int i = 0; i < sb.Length; i++)
{
if (sb[i] == 'X')
{
for (int j = 0; j < 3; j++)
{
if (i + j < sb.Length)
{
sb[i + j] = 'O';
}
}
count++;
done = false;
}
}
}
return count;
}
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