Pop string from left to right if matches in JavaScript

user_16400843 0
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If you see two consecutive characters that are the same, you pop them from left to right, until you cannot pop any more characters. Return the resulting string. let str = "abba"

"abba" - pop the two b's -> "aa"

"aa" - pop the two a's -> ""

return ""

Here's what i have tried so far:

function match(str){
    
 for (let i = 0; i< str.length; i++){
   if (str[i] === str[i+1]){
     return str.replace(str[i], ""); 
   
    } 
  }
};
match('abba');

It removes one b only. On first loop, i want it to remove two b's and console the output. On second i want the remaining two a's to remove and console the output. Also It would be great if good time complexity is maintained.

javascript
1 Answers

Use recursion.

function match(str){
    if(str.length==1 || str=="") {
        return str;
    }
    for (let i = 0; i< str.length-1; i++) {
        if(str[i] === str[i+1] && i+2==str.length) {
            return str.substr(0,i);
        } else if(i+2==str.length) {
            return str;
        } else if (str[i] === str[i+1]) {
            return match(str.substr(0,i)+str.substr(i+2,(str.length)-(i+2))); 
        }
    }
};

console.log('aa = '+ match('aa'));
console.log('abba = '+ match('abba'));
console.log('abccba = '+ match('abccba'));
console.log('abdccba = '+ match('abdccba'));
console.log('aabdccba = '+ match('aabdccba'));
console.log('abdccbaa = '+ match('abdccbaa'));
console.log('abbac = '+ match('abbac'));
console.log('abbcc = '+ match('abbcc'));

user_44236950
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