Minimum Cost of ropes | Queue | Heap | GFG POTD | Dry Run & Explanation

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CODE-
class Solution {
public:
long long minCost(vector& arr) {
priority_queue pq;
for(int i=0;i 1){
long long first = pq.top();
pq.pop();
long long second = pq.top();
pq.pop();
sum += first+second;
pq.push(first+second);
}
return sum;
}
};