Optimize for Isogram - string

Question

Code wars Question:(Isogram)
An Isogram is a word that has no repeating letters,consecutive or non-consecutive.Implement a function that determines whether a string that contains only letters is an Isogram. Assume the empty string is an Isogram.Ignore letter case.
Test Cases:
"Dermatoglyphics"-->true
"aba" --> false
"balLoon"--> false
My Code:
#include <bits/stdc++.h>
using namespace std;
bool isIsogram(string s)
{
for (int i = 0; i < s.length(); i++)
{
for (int j = i + 1; j < s.length(); j++)
{
if (s[i] == s[j] || s[i] == s[j] + 32 || s[i] == s[j] - 32)
{
return false;
}
return true;
}
}
}
int main()
{
cout << isIsogram("abd") << endl;
return 0;
}
If there is an optimized solution for this problem?
Can we reduce time complexity of this problem?

Answer 1

Your code is of O(N^2) Time Complexity.
Yes, there is an Optimized Way of doing it.
Here's the optimized solution: O(N)
#include <bits/stdc++.h>
using namespace std;
bool Isogram(string s)
{
int N = 125;
bool check[N];
for (int i = 0; i < 125; i++)
{
check[i] = false;
}
for (int i = 0; i < s.size(); i++)
{
if (check[s[i]] != false)
{
return false;
}
check[s[i]] = true;
if (s[i] + 32 < N)
check[s[i] + 32] = true;
if (s[i] - 32 >= 0)
check[s[i] - 32] = true;
}
return true;
}
int main()
{
string s;
cin >> s;
if (Isogram(s) == true)
cout << "The word " << s << " is an Isogram" << endl;
else
cout << "The word " << s << " is not an Isogram" << endl;
return 0;
}
We made an array of 125 elements because the ASCII value of lowercase particles lies in the range from 65 to 90, and for uppercase, it is 97 to 122.
And we are adding 32 and subtracting 32 to cover the uppercase alphabet (if the element is in lowercase) and vice versa.
Hope you get it...