This question is not too difficult.
Thought process:
- First, we can find all occurrences of s based on order and replace with a space since we've used it
- Replace the spaces and append the remaining characters to the result set.
Time complexity: O(n^2)
Link:Custom Sort String
/**
* @param {string} order
* @param {string} s
* @return {string}
*/
var customSortString = function(order, s) {
let res = "";
for(let i=0;i<order.length;i++){
for(let j=0; j<s.length;j++){
if(s[j] == order[i]){
res += s[j];
s = s.replace(s[j], ' ');
}
}
}
s = s.replace(/\s/g, '');
res += s.trim();
return res;
};