LeetCode 1750: Minimum Length of String After Deleting Similar Ends
LeetCode 1750 Solution Explanation
Explanation:
To solve this problem, we can use a two-pointer approach. We start with two pointers, one at the beginning and one at the end of the string. We keep moving these pointers towards each other as long as the characters at these positions are the same. Whenever we find different characters at these positions, we delete the common prefix and suffix and continue the process.
The algorithm can be summarized as follows:
- Initialize two pointers,
startat index 0 andendat indexs.length()-1. - While
start < end, check ifs[start] == s[end], if so, incrementstartand decrementend. - If
s[start] != s[end], delete the common prefix and suffix by incrementingstartand decrementingend. - Repeat steps 2 and 3 until
start >= end. - Return the remaining length of
s.
Time Complexity: O(n) where n is the length of the input string s. Space Complexity: O(1) as we are using constant extra space.
:
LeetCode 1750 Solutions in Java, C++, Python
class Solution {
public int minimumLength(String s) {
int start = 0, end = s.length() - 1;
while (start < end && s.charAt(start) == s.charAt(end)) {
char common = s.charAt(start);
while (start <= end && s.charAt(start) == common) start++;
while (end >= start && s.charAt(end) == common) end--;
}
return end - start + 1;
}
}Interactive Code Editor for LeetCode 1750
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