Explained || reverse iteration - O(N)
EXPLANATION\nWe start at the bottom-right corner of the grid.\n\nFor each cell, we check if it is a row operation or a column operation.\n\nIf it\'s a row opera
# EXPLANATION\nWe start at the bottom-right corner of the grid.\n\nFor each cell, we check if it is a row operation or a column operation.\n\nIf it\'s a row operation, we multiply the cell value by the number of remaining rows. This value represents how many times this cell\'s value will contribute to the final answer.\n\nIf it\'s a column operation, we multiply the cell value by the number of remaining columns. This value represents how many times this cell\'s value will contribute to the final answer.\n\nAfter each row operation, we decrement the count of remaining rows by 1 because the next column operation will have one less row to work with.\n\nAfter each column operation, we decrement the count of remaining columns by 1 because the next row operation will have one less column to work with.\n\nWe continue this process until we reach the top-left corner of the grid.\n\nFinally, we return the sum of all the values obtained from the above steps as the answer.\n# CODE\n\n    long long matrixSumQueries(int n, vector<vector<int>>& queries) {\n    vector<bool> rowFlag(n, true), colFlag(n, true);\n    long long totalSum = 0, remainingRows = n, remainingCols = n;\n\n    for (int i = queries.size() - 1; i >= 0; --i) {\n        int queryType = queries[i][0];\n        int index = queries[i][1];\n        int value = queries[i][2];\n\n        if (queryType == 0 && rowFlag[index]) {\n            totalSum += remainingCols * value;\n            rowFlag[index] = false;\n            remainingRows--;\n        }\n        if (queryType == 1 && colFlag[index]) {\n            totalSum += remainingRows * value;\n            colFlag[index] = false;\n            remainingCols--;\n        }\n    }\n\n    return totalSum;\n    }\n
['C++']