Checking if the characters in a string are all unique

Question

I am trying to solve this problem using JS by just using an array.

var str = 'abcdefgh';

for (i = 0; i < 255; i++) {
  arr[i] = false;
}

function check() {
  for (i = 0; i < str.length; i++) {
    if (arr[str.charCodeAt(i)] == true) {
      return false;
    }
    arr[str.charCodeAt(i)] = true;
  }
  return true;
}

I am initializing an array of fixed size 256 to have the boolean value false . Then i am setting the value for the corresponding ASCII index to true for characters in the string. And if i find the same character again, i am returning false .

While running the program, i am getting false returned even if the string doesn't have any duplicate characters.

Answer 1

Fill a Set with all characters and compare its size to the string's length:

 function isUnique(str) { return new Set(str).size == str.length; } console.log(isUnique('abc')); // true console.log(isUnique('abcabc')); // false

Answer 2

Use object for faster result

 function is_unique(str) { var obj = {}; for (var z = 0; z < str.length; ++z) { var ch = str[z]; if (obj[ch]) return false; obj[ch] = true; } return true; } console.log(is_unique("abcdefgh")); // true console.log(is_unique("aa")); // false

Answer 3

use .match() function for each of the character. calculate occurrences using length. Guess thats it.

(str.match(/yourChar/g) || []).length

Answer 4

You are using arr[str.charCodeAt(i)] which is wrong. It should be arr[str[i].charCodeAt(0)]

var arr = [];
var str="abcdefgh";
for (i=0;i<255;i++){
    arr[i]=false;
}
function check(){
    for (i=0;i<str.length;i++){
        if (arr[str[i].charCodeAt(0)]==true){
            return false;
        }
        arr[str[i].charCodeAt(0)]=true;
    }
    console.log(arr);
    return true;
}
check();

Answer 5

Time complexity = O(n) Space complexity = O(n)

const isUnique = (str) => {
      let charCount = {};
      for(let i = 0; i < str.length; i++) {
        if(charCount[str[i]]){
          return false;
        }

        charCount[str[i]] = true;
      }

      return true;
    }


    const isUniqueCheekyVersion = (str) => {
      return new Set(str).size === str.length;
    }

Solution 3: Transform string to chars array, sort them and then loop through them to check the adjacent elements, if there is a match return false else true

Solution 4: It's similar to Solution 1 except that we use a Set data structure which is introduced in recent versions of javascript

Answer 6

 // no additional Data structure is required. we can use naive solution // Time Complexity:O(n^2) function isUnique(str) { for (let i = 0; i < str.length; i++) { for (let j = 1 + i; j < str.length; j++) { if (str[i] === str[j]) { return false; } } } return true; } // if you can use additional Data structure // Time Complexity:O(n) function isUniqueSecondMethos(str) { let dup_str = new Set(); for (let i = 0; i < str.length; i++) { if (dup_str.has(str[i])) { return false; } dup_str.add(str[i]); } return true; } console.log(isUniqueSecondMethos('hello'));

Answer 7

Use an object as a mapper

function uniqueCharacterString(inputString) {
  const characterMap = {};

  let areCharactersUnique = true;

  inputString.trim().split("").map((ch)=>{
    if(characterMap[ch]===undefined) {
      characterMap[ch] = 1;
    } else {
      areCharactersUnique = false;
    }
  })
  return areCharactersUnique;
}

Answer 8

Algo

*1. step -first string is ->stack *

*2.step-string covert to CharArray *

3. step - use iteration in array ['s','t','a','c','k']

4. step - if(beginElement !== nextElement){return true}else{return false}

Implement code

function uniqueChars(string){
var charArray = Array.from(string) //convert charArray
  for(var i=0;i<charArray.length;i++){
    if(charArray[i] !== charArray[i+1]){ 
      return true
    }
    else{
      return false
    }
  }

}
var string ="stack"
console.log(uniqueChars(string))

Time complexity O(nlogn)

Answer 9

We can also try using indexOf and lastIndexOf method:

function stringIsUnique(input) {
  for (i = 0; i < input.length; i++) {
    if (input.indexOf(input[i]) !== input.lastIndexOf(input[i])) {
      return false;
    }
  }
  return true;
}

Answer 10

Algo

1. Counting frequency of alphabets. eg 'Mozilla' will returns Object{ M: 1, o: 1, z: 1, i: 1, l: 2, a: 1 } . Note that, the bitwise NOT operator (~) on -~undefined is 1 , -~1 is 2 , -~2 is 3 etc.
2. Return true when all occurrences appear only once .

Implement code

 var isUnique = (str) => { const hash = {}; for (const key of str) { hash[key] = -~hash[key]; } return Object.values(hash).every((t) => t === 1); }; console.log(isUnique('Mozilla')); console.log(isUnique('Firefox'));

---

Another alternative could be:

 var isUnique = (str) => { const hash = {}; for (const i in str) { if (hash[str[i]]) return false; hash[str[i]] = true; } return true; }; console.log(isUnique('Mozilla')); console.log(isUnique('Firefox'));

Answer 11

To make efficient one, you can use simple hash map

 let isUnique = (s) => { let ar = [...s]; let check = {}; for (let a of ar) { if (!check[a]) { check[a] = 1; } else { return false } } return true; } alert("isUnique : "+isUnique("kailu"));

Time complexity & Space complexity

using ES6

 let isUnique = (s)=>{ return new Set([...s]).size == s.length; } console.log("using ES6 : ",isUnique("kailu"));

Answer 12

We can use split method of string:

const checkString = (str) => {
let isUniq = true;

for (let i = 0; i < str.length; i++) {
  if (str.split(str[i]).length > 2) {
    isUniq = false;
    break;
  }
}
return isUniq;
};
console.log(checkString("abcdefgh")); //true
console.log(checkString("aa")); //false