Champagne Tower LeetCode solution in C++

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class Solution {
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 public:
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  double champagneTower(int poured, int query_row, int query_glass) {
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    vector<double> currRow(1, poured);
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    for (int i = 0; i <= query_row; i++) { // we need to make the dp matrix only till query row. No
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                                           // need to do after that
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      vector<double> nextRow(i + 2, 0); // If we are at row 0, row 1 will have 2 glasses. So next
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                                        // row will have currRow number + 2 number of glasses.
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      for (int j = 0; j <= i; j++) {    // each row will have currRow number + 1 number of glasses.
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        if (currRow[j] >= 1) { // if the champagne from the current glass is being overflowed.
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          nextRow[j] +=
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              (currRow[j] - 1) / 2.0; // fill the left glass with the overflowing champagne
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          nextRow[j + 1] +=
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              (currRow[j] - 1) / 2.0; // fill the right glass with the overflowing champagne
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          currRow[j] = 1;             // current glass will store only 1 cup of champagne
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        }
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      }
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      if (i != query_row)
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        currRow = nextRow; // change the currRow for the next iteration. But if
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                           // we have already reached the query_row, then the
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                           // next iteration will not even take place, so the
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                           // currRow is the query_row itself. So don't change
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                           // as we need the currRow only.
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    }
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    return currRow[query_glass];
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  }
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};