Brick Wall LeetCode solution in C++

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class Solution {
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 public:
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  int leastBricks(vector<vector<int>> &wall) {
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    unordered_map<int, int> edge_frequency; // HashMap to store the number of
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                                            // common edges among the rows
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    int max_frequency = 0;                  // Variable to store the frequency of most occuring edge
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    for (int row = 0; row < wall.size(); row++) // Iterating through each row
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    {
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      int edge_postion = 0; // Variable to store different edge postion
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      for (int brick_no = 0; brick_no < wall[row].size() - 1;
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           brick_no++) // Iterating through each brick inside a row
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      {
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        int current_brick_length = wall[row][brick_no]; // Length of the current brick
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        edge_postion =
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            edge_postion + current_brick_length; // Next Edge Position = Previous Edge Position
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                                                 // + Current Brick's Length
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        edge_frequency[edge_postion]++;          // Incrementing the Frequency of just
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                                                 // calculated Edge Postion
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        max_frequency = max(edge_frequency[edge_postion],
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                            max_frequency); // Comparing the "Frequency of just calculated
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                                            // Edge Postion" with "Max Frequency seen till
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                                            // now" & storing whichever is greater.
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      }
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    }
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    return wall.size() - max_frequency; // returning (Number of Bricks Crossed
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                                        // by Line) i.e. (Number of Rows in Wall
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                                        // - Frequency of Most Occuring Edge)
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  }
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};