Max Length Substring without any repeating character - c++

Question

Given a String, find the length of longest substring without any repeating character.
Example 1:
Input: s = ”abcabcbb”
Output: 3
Explanation: The answer is abc with length of 3.
Example 2:
Input: s = ”bbbbb”
Output: 1
Explanation: The answer is b with length of 1 units.
My solution works, but it isn't optimised. How can this be done in O(n) time?
#include<bits/stdc++.h>
using namespace std;
int solve(string str) {
if(str.size()==0)
return 0;
int maxans = INT_MIN;
for (int i = 0; i < str.length(); i++) // outer loop for traversing the string
{
unordered_set < int > set;
for (int j = i; j < str.length(); j++) // nested loop for getting different string starting with str[i]
{
if (set.find(str[j]) != set.end()) // if element if found so mark it as ans and break from the loop
{
maxans = max(maxans, j - i);
break;
}
set.insert(str[j]);
}
}
return maxans;
}
int main() {
string str = "abcsabcds";
cout << "The length of the longest substring without repeating characters is " <<
solve(str);
return 0;
}

Answer 1

Use a two pointer approach along with a hashmap here.
Initialise two pointers i = 0, j = 0 (i and j denote the left and right boundary of the current substring)
If the j-th character is not in the map, we can extend the substring. Add the j-th char to the map and increment j.
If the j-th character is in the map, we can not extend the substring without removing the earlier occurrence of the character. Remove the i-th char from the map and increment i.
Repeat this while j < length of string
This will have a time and space complexity of O(n).

Answer 2

#include <string>
#include <iostream>
#include <vector>
int main() {
// 1
std::string s;
std::cin >> s;
// 2
std::vector<int> lut(127, -1);
int i, beg{ 0 }, len_curr{ 0 }, len_ans{ 0 };
for (i = 0; i != s.size(); ++i) {
if (lut[s[i]] == -1 || lut[s[i]] < beg) {
++len_curr;
}
else {
if (len_curr > len_ans) {
len_ans = len_curr;
}
beg = lut[s[i]] + 1;
len_curr = i - lut[s[i]];
}
lut[s[i]] = i;
}
if (len_curr > len_ans) {
len_ans = len_curr;
}
// 3
std::cout << len_ans << '\n';
return 0;
}
In // 1 you:
Define and read your string s.
In // 2 you:
Define your look up table lut, which is a vector of int and consists of 127 buckets each initialized with -1. As per this article there are 95 printable ASCII characters numbered 32 to 126 hence we allocated 127 buckets. lut[ch] is the position in s where you found the character ch for the last time.
Define i (index variable for s), beg (the position in s where your current substring begin at), len_curr (the length of your current substring), len_ans (the length you are looking for).
Loop through s. If you have never found the character s[i] before OR you have found it but at a position BEFORE beg (It belonged to some previous substring in s) you increment len_curr. Otherwise you have a repeating character ! You compare len_curr against len_ans and If needed you assign. Your new beg will be the position AFTER the one you found your repeating character for the last time at. Your new len_curr will be the difference between your current position in s and the position that you found your repeating character for the last time at.
You assign i to lut[s[i]] which means that you found the character s[i] for the last time at position i.
You repeat the If clause when you fall through the loop because your longest substring can be IN the end of s.
In // 3 you:
Print len_ans.