en/lcci/1.1/

[giscus[bot]]

en/lcci/1.1/

LeetCode solutions in any programming language

https://leetcode.doocs.org/en/lcci/1.1/

[jumael99]

Hi! Without bit manipulation can you give some other easy solution?

[devmustafa03]

const isCheckingSimilarCharacter = (word) => {
const currentWord = new Set();

for (let char of word) {
    if (currentWord.has(char)) {
        return false;
    }
    currentWord.add(char);
}

return true;

};

This is for javascript

[rajputanisha]

#include <bits/stdc++.h>
using namespace std;
bool fun(string s,int n ){
unordered_map<char,int>mp;
for(auto it :s){
mp[it]++;
}
for(auto it :mp){
if(it.second>1){
return false;
}

}
return true;
}
int main()
{
string s = "lswtcde";
int n = s.size();
bool ans = fun(s, n);
cout << ans << endl;

return 0;

}

[rajputanisha]

In this we use map to store the key value pair. If count greater than one means there are duplicate present . we simply return false. otherwise we will simply return true if count of all elements are equal to one. i e all elements are unique
map store element like this for example string example = "Leetcode"
L ->1
e ->2
t ->1
c ->1
o ->1
d ->1
e ->1
this is simple representation of unordered_map

[ADITYAPIMPALE11127]

import java.util.HashSet;
import java.util.Scanner;
public class UniqueString {
public static boolean UniqueChars(String str){
HashSet seen = new HashSet<>();
for(int i = 0; i < str.length();i++){
char ch = str.charAt(i);
if(seen.contains(ch)){
return false;
} seen.add(ch);
}
return true;
}
public static void main(String[] args){
Scanner input = new Scanner(System.in);
String str1 = input.nextLine();

    System.out.print(str1+" and the string is "+UniqueChars(str1));
}

}