why Binary Operator gives TLE while compound assignment operator is faster?

Question

I am trying to solve Sort Characters By Frequency on LeetCode. The first used Binary Operator gives me a TLE, but when I use a compound assignment operator then it works fine. But I don't understand why.

Is there any logic behind it? I'm attaching both codes below so you can try it for yourself.

This gives TLE

class Solution {
public:
    string frequencySort(string s) {
        
        unordered_map<char, int> m;
        for(int i = 0; i < s.length(); i++)
            m[s[i]]++;
        
        priority_queue<pair<int, char>> pq; // maxheap based on frequency
        for(auto x = m.begin(); x != m.end(); x++)
            pq.push(make_pair(x->second, x->first));
        
        string ans = "";
        while (!pq.empty())
        {
            for(int i = 0; i < pq.top().first; i++)
                ans = ans + pq.top().second;
            pq.pop();
        }
        
        return ans;
    }
};

This works fine

class Solution {
public:
    string frequencySort(string s) {
        
        unordered_map<char, int> m;
        for(int i = 0; i < s.length(); i++)
            m[s[i]]++;
        
        priority_queue<pair<int, char>> pq; // maxheap based on frequency
        for(auto x = m.begin(); x != m.end(); x++)
            pq.push(make_pair(x->second, x->first));
        
        string ans = "";
        while (!pq.empty())
        {
            for(int i = 0; i < pq.top().first; i++)
                ans += pq.top().second;
            pq.pop();
        }
        
        return ans;
    }
};

Answer 1

ans = ans + pq.top().second; creates a temporary string which is then moved to ans. Unless small string optimization applies, each operation involves memory allocation and de-allocation.

ans += pq.top().second; does not create a temporary, and memory is allocated only if the memory previously reserved by ans is exhausted - which happens far less often.