Valid Anagram

PHOTO

Tue Sep 21 2021 14:51:47 GMT+0000 (Coordinated Universal Time)

Saved by @sangelici94 #javascript #strings #regex #sort #anagram #loops #unsolved

// SOLUTION 1 - MY (DISCOVERED) SOLUTION (RUNTIME - 104MS, 44.9 MB)
var isAnagram = function(s, t) {
    
    // set function to split, sort, and rejoin characters in a string
    const sortString = (str) => {
        return str.split("").sort().join("");
    }
    
    // regex removes any non-alphabet characters in the string and makes it lowercase
    s = s.replace(/[^\w]/g, '').toLowerCase()
    t = t.replace(/[^\w]/g, '').toLowerCase()

    // final comparison
    return sortString(s) === sortString(t)
  
  // ATTEMPT 1
  //     let anagram = []
    
  // return false is lengths don't match
  //     if (s.length !== t.length) return false;
  //     else {
  //         for (let i = 0; i < s.length; i++) {
  //             let arrT = t.split("")
  //             let newArrT;
  //             if (arrT.includes(s.charAt(i))) {
  //                 let index = arrT.indexOf(s.charAt(i));
  //                 console.log("s char: ", s.charAt(i));
  //                 console.log("index: ", index);

  //                 anagram.push(s.charAt(i))

  //                 console.log("splice: ", arrT.splice(index, 1))
  //                 arrT.splice(index, 0)

  //                 newArrT = arrT
  //                 console.log("newArr: ", newArrT)
  //             };
  //         };
  //     }
  //     console.log(anagram)
  //     return anagram.join("") === s;
}

// SOLTUION 2 - (BEST RUNTIME - 60MS)
var isAnagram = function(s, t) {
    const key = w => Object.entries([...w].reduce((a, c) => {
        if (!(c in a)) a[c] = 0;
        a[c] += 1;
      
        return a;
    }, {})).sort(([c1], [c2]) => c1.localeCompare(c2)).flat().join('');
  
    return key(s) === key(t);
};

// SOLUTION 3 - (RUNTIME - 72MS)
var isAnagram = function(s, t) {
    if(s.length !== t.length) return false;
  
    let map = {};
  
    for(let item of s) {
        map[item] = map[item] + 1 || 1;
    }
    
    for(let item of t) {
        if(!map[item]) return false;
        else map[item]--;
    }
  
    return true;
};

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Solutions to check if a string is an anagram of the second string using loops, reduce(), and sorting methods

https://leetcode.com/submissions/detail/558649741/?from=explore&item_id=882

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Valid Anagram

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