2.4 KiB
Leetcode Longest-Substring-Without-Repeating-Characters
2022-07-14 09:33
Algorithms:
#algorithm #Kadane_s_algorithm #sliding_window
Data structures:
#DS #string
Difficulty:
#coding_problem #difficulty-medium
Additional tags:
#leetcode
Revisions:
N/A
Related topics:
tag:#Kadane_s_algorithm OR tag:#sliding_window
Links:
Problem
Given a string s, find the length of the longest substring without repeating characters.
Examples
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 * 104sconsists of English letters, digits, symbols and spaces.
Thoughts
[!summary] This is a #Kadane_s_algorithm
Initially, I thought of a kadane's algo, but implemented wrong.
Then I figured out kadane's algorithm.
similar to this one. Leetcode Best-Time-To-Buy-And-Sell-Stock
The goal is making cur as small as possible, without duplicating
And the localMax is max(localMax, i - cur)
Solution
Kadane's algorithm
class Solution {
public:
int lengthOfLongestSubstring(string s) {
// Kadane's algorithm
vector<int> hMap(255, -1);
int cur = -1;
int localMax = 0;
for (int i = 0; i < s.length(); i++) {
if (hMap[s[i]] > cur) {
// If the element occurs again, reset
cur = hMap[s[i]];
}
hMap[s[i]] = i;
// The char at cur is ignored
localMax = max(localMax, i - cur);
}
return localMax;
}
};
Initial solution
class Solution {
public:
int lengthOfLongestSubstring(string s) {
unordered_map<char, int> hMap;
int cur = 0;
int localMax = 0;
for (int i = 0; i < s.size(); i++) {
if (hMap.find(s[i]) == hMap.end()) {
cur++;
localMax = max(cur, localMax);
hMap[s[i]] = i;
} else {
i = hMap[s[i]];
hMap.clear();
cur = 0;
}
}
return localMax;
}
};