Longest Substring Without Repeating Characters

From the software interviewing series, we bring the following String manipulation excersice from leetcode

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"
Input: s = "pwwkabc"
j -> 2
i -> 0
a -> 2
S -> a,b
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.

Example 4:

Input: s = ""
Output: 0

Constraints:

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

Solution

The above solution requires at most 2n steps. In fact, it could be optimized to require only n steps. Instead of using a set to tell if a character exists or not, we could define a mapping of the characters to its index. Then we can skip the characters immediately when we found a repeated character.

The reason is that if s[j]s[j] have a duplicate in the range [i, j)[i,j) with index j’j′, we don’t need to increase ii little by little. We can skip all the elements in the range [i, j’][i,j′] and let ii to be j’ + 1j′+1 directly.

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        Map<Character, Integer> map = new HashMap<>(); // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            if (map.containsKey(s.charAt(j))) {
                i = Math.max(map.get(s.charAt(j)), i);
            }
            ans = Math.max(ans, j - i + 1);
            map.put(s.charAt(j), j + 1);
        }
        return ans;
    }
}