Problem
Given two strings s and t, return true if t is an anagram of s, and false otherwise.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
- Example
- Input: s = "anagram", t = "nagaram"
- Output: true
- Input: s = "rat", t = "car"
- Output: false
- Constraints
- 1 <= s.length, t.length <= 5 * 104
- s and t consist of lowercase English letters.
Solution
There are two ways of solving this problem.
Method One
/** * Since t is compared to s, * convert all to lowercase, * sort t and s, * if t is exactly equal to s, then it is an anagram, else it isn't * Runtime O(n) */ // const s = "anagram"; // const t = "nagaram" const s = "rat", t = "car" function anagram(s, t){ const sortedS = s.split('').sort().join(''); const sortedT = t.split('').sort().join(''); if (sortedS === sortedT) return true; return false; } Method Two
Using Map
/** * Create a map object * For each char in S, add the first occurrence with a value of 1. For subsequent occurrence, increase the value by +1. * For each char in T, if it doesn't exist in map i.e undefined or value is <= 0, return false. If it does exist, do value -1, return true outside loop. * @param {*} s * @param {*} t */ function anagramTwo(s,t){ // for our answer to be an anagram, the word must be rearranged, hence, if the length of the two words are not same, it is not an anagram. if(s.length !== t.length) return false; const map = {}; for (let i of s) { if(map[i] !== undefined) { map[i]++ } else map[i] = 1 } for (let i of t){ if (map[i] === undefined || map[i] <= 0) return false; map[i]--; } return true } console.log(anagramTwo("ab", "a")) If you find this helpful, like, share and bookmark. :)
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