Minimum Operations to Halve Array Sum LeetCode solution in C++

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class Solution {
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 public:
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  int halveArray(vector<int> &nums) {
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    priority_queue<double> pq;
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    double totalSum = 0;
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    double requiredSum = 0;
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    for (auto x : nums) {
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      totalSum += x;
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      pq.push(x);
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    }
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    requiredSum = totalSum / 2;
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    int minOps = 0;
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    while (totalSum > requiredSum) {
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      double currtop = pq.top();
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      pq.pop();
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      currtop = currtop / 2;
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      totalSum -= currtop;
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      pq.push(currtop);
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      minOps++;
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    }
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    return minOps;
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  }
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}