# Longest Substring Without Repeating Characters Leetcode Solution

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Longest Substring Without Repeating Characters | Given a string `s`, find the length of the longest substring without repeating characters.

Example 1:

```Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
```

Example 2:

```Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
```

Example 3:

```Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
```

Constraints:

• `0 <= s.length <= 5 * 104`
• `s` consists of English letters, digits, symbols and spaces.

### Solutions

Time: O(n)
Space: O(128)=O(1)

### C++

``````class Solution {
public:
int lengthOfLongestSubstring(string s) {
int ans = 0;
vector<int> count(128);

for (int l = 0, r = 0; r < s.length(); ++r) {
++count[s[r]];
while (count[s[r]] > 1)
--count[s[l++]];
ans = max(ans, r - l + 1);
}

return ans;
}
};
``````

### Java

``````class Solution {
public int lengthOfLongestSubstring(String s) {
int ans = 0;
int[] count = new int[128];

for (int l = 0, r = 0; r < s.length(); ++r) {
++count[s.charAt(r)];
while (count[s.charAt(r)] > 1)
--count[s.charAt(l++)];
ans = Math.max(ans, r - l + 1);
}

return ans;
}
}
``````

### Python

``````class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
ans = 0
count = Counter()

l = 0
for r, c in enumerate(s):
count[c] += 1
while count[c] > 1:
count[s[l]] -= 1
l += 1
ans = max(ans, r - l + 1)

return ans

``````

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