LeetCode 1750: Minimum Length of String After Deleting Similar Ends Solution

Master LeetCode problem 1750 (Minimum Length of String After Deleting Similar Ends), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1750. Minimum Length of String After Deleting Similar Ends

Medium

Two Pointers String

Problem 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, start at index 0 and end at index s.length()-1.
While start < end, check if s[start] == s[end], if so, increment start and decrement end.
If s[start] != s[end], delete the common prefix and suffix by incrementing start and decrementing end.
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.

Solution Code

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;
    }
}

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