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LeetCode 2414: Length of the Longest Alphabetical Continuous Substring

String

LeetCode 2414 Solution Explanation

Explanation:

To solve this problem, we can iterate through the input string s and keep track of the current alphabetical continuous substring's length. We reset the length whenever we encounter a character that breaks the alphabetical order. We then update the maximum length as we iterate through the string.

Algorithm:

  1. Initialize maxLen and currLen to 1.
  2. Iterate through the input string s starting from the second character.
  3. If the current character is in alphabetical order with the previous character, increment currLen.
  4. If not, update maxLen if currLen is greater and reset currLen to 1.
  5. Return the maximum length found.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the length of the input string s.

Space Complexity:

The space complexity is O(1) since we are using only a constant amount of extra space.

:

LeetCode 2414 Solutions in Java, C++, Python

class Solution {
    public int longestAlphabeticalContinuousSubstring(String s) {
        if (s.length() == 0) return 0;
        
        int maxLen = 1;
        int currLen = 1;
        
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) - s.charAt(i - 1) == 1) {
                currLen++;
            } else {
                maxLen = Math.max(maxLen, currLen);
                currLen = 1;
            }
        }
        
        return Math.max(maxLen, currLen);
    }
}

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