So with the sort function you have O(nlog(n)). Then how about this solution for O(n). You have a dictionary with the char as key, and counter (how many times char appeared) as value. You iterate with a for over both arrays at the same time (O(n)) and each char in the first string adds to that key, while the same char in the second string subtracts. In the end you want to have 0 for all keys:
function anagram(str1, str2){
//Prepare a dictionary to count appearances
const dict = {}
const arr1 = str1.split('');
const arr2 = str2.split('');
// if lenght is different return false, think it's O(1)
if(arr1.length != arr2.length) return false
for(let i = 0; i < str1.length; ++i){
// increase number for keys in arr1 and decrease for keys in arr 2
// if the same, result should be 0 in the end
if(dict[arr1[i]] != undefined){
dict[arr1[i]] = 1
}else{
dict[arr1[i]] += 1
}
if(dict[arr2[i]] != undefined){
dict[arr2[i]] = -1
}else{
dict[arr2[i]] -= 1
}
}
// check the dict
for(let key in dict){
if(dict[key] != 0){
return false;
}
}
// if reatched, it means all keys are 0
return true
}
For further actions, you may consider blocking this person and/or reporting abuse
We're a place where coders share, stay up-to-date and grow their careers.
So with the sort function you have
O(nlog(n)). Then how about this solution forO(n). You have a dictionary with the char as key, and counter (how many times char appeared) as value. You iterate with a for over both arrays at the same time(O(n))and each char in the first string adds to that key, while the same char in the second string subtracts. In the end you want to have 0 for all keys: