Given a string, find the length of the longest substring without repeating characters.
Input: A single string s
.
Output: An integer representing the length of the longest substring without repeating characters.
Constraints:
0 ≤ s.length ≤ 5 * 104
s
consists of English letters, digits, symbols, and spaces.Example:
Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3.
The core challenge of this problem is to identify the longest substring within a given string that does not contain any repeating characters. This problem is significant in various applications such as data validation, parsing, and text processing where unique sequences are required.
Potential pitfalls include misunderstanding the definition of a substring (which must be contiguous) and not efficiently handling the search for repeating characters, leading to suboptimal solutions.
To solve this problem, we can use a sliding window technique. The idea is to use two pointers to represent the current window of characters being considered and a set to track the characters within this window.
Here is a step-by-step approach:
left
and right
, both set to the start of the string.right
pointer and adding characters to the set until a duplicate character is found.left
pointer to the right until the duplicate is removed from the set.This approach ensures that each character is processed at most twice, resulting in a linear time complexity of O(n)
.
Here is a detailed breakdown of the algorithm:
left
and right
pointers to 0, and an empty set charSet
.right
pointer over the string.right
is not in charSet
, add it to the set and update the maximum length.left
from the set and move left
to the right.right
pointer reaches the end of the string.public class Solution {
public int lengthOfLongestSubstring(String s) {
// Initialize the set to store unique characters
Set<Character> charSet = new HashSet<>();
int left = 0, right = 0, maxLength = 0;
// Iterate over the string with the right pointer
while (right < s.length()) {
// If the character is not in the set, add it and update maxLength
if (!charSet.contains(s.charAt(right))) {
charSet.add(s.charAt(right));
maxLength = Math.max(maxLength, right - left + 1);
right++;
} else {
// If the character is in the set, remove the leftmost character and move left pointer
charSet.remove(s.charAt(left));
left++;
}
}
return maxLength;
}
}
The time complexity of this approach is O(n)
because each character is processed at most twice (once by the right
pointer and once by the left
pointer). The space complexity is O(min(n, m))
, where n
is the length of the string and m
is the size of the character set (which is constant for English letters, digits, symbols, and spaces).
Consider the following edge cases:
These cases can be tested to ensure the robustness of the solution.
To test the solution comprehensively, consider the following test cases:
""
(Empty string)"bbbbbb"
(All identical characters)"abcdef"
(All unique characters)"pwwkew"
(Mixed characters)JUnit or any other testing framework can be used to automate these tests.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to solve the problem of finding the longest substring without repeating characters using a sliding window technique. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills.
For further reading and practice, consider the following resources: