LeetCode 2414: Length of the Longest Alphabetical Continuous Substring
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:
- Initialize
maxLen
andcurrLen
to 1. - Iterate through the input string
s
starting from the second character. - If the current character is in alphabetical order with the previous character, increment
currLen
. - If not, update
maxLen
ifcurrLen
is greater and resetcurrLen
to 1. - 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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