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Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal of the code is to determine the total number of laser beams that can be formed
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of the code is to determine the total number of laser beams that can be formed between security devices in a bank. Each security device can potentially emit a laser beam, and the conditions for laser beam connection involve the presence of devices in different rows with no devices in the rows between them.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Iteration through Rows**:The code begins by iterating through each row of the bank (represented by the 2D matrix).\n2. **Counting Security Devices**:For each row, it counts the number of security devices (\'1\'s) present. This is done using the inner loop that iterates through the columns of the matrix.\n3. **Accumulating Laser Beams**:If the count of security devices in the current row is not zero, it accumulates the product of the current count and the count of devices in the previous row. This implies that each security device in the current row can potentially form a laser beam with every security device in the previous row.\n4. **Updating Previous Count**:After processing a row, the count of security devices in that row becomes the "previous" count for the next iteration.\n5. **Calculation of Total Beams**:The cumulative sum of these products represents the total number of laser beams that can be formed in the bank.\n6. **Returning the Result**:Finally, the function returns the total count of laser beams.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n1. The code uses nested loops to iterate through the rows and columns of the input matrix.\n2. Let n be the number of rows and m be the number of columns.\n3. The time complexity is **O(n*m)**, where \'n\' is the number of rows and \'m\' is the number of columns in the bank matrix.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n1. The code uses a constant amount of space for integer variables (curr, prev, n, m, ans1) regardless of the size of the input matrix.\n1. Therefore, the space complexity is **O(1)**, constant space.\n\n![download.jpeg](https://assets.leetcode.com/users/images/67562be9-a62d-40a5-9cc4-7b813e4b5f6e_1704256007.0353346.jpeg)\n\n# Code\nC++\n```\nclass Solution {\npublic:\n    int numberOfBeams(vector<string>& bank) {\n        int curr, prev = 0;\n        int n = bank.size();    // Number of rows in the bank\n        int m = bank[0].size(); // Number of columns in the bank\n        int ans1 = 0;           // Variable to store the total number of laser beams\n\n        // Iterate through each row of the bank\n        for (int i = 0; i < n; i++) {\n            curr = 0; // Reset the current count of security devices for each row\n\n            // Iterate through each column of the current row\n            for (int j = 0; j < m; j++) {\n                if (bank[i][j] == \'1\') {\n                    curr++; // Count the number of security devices in the current row\n                }\n            }\n\n            // If there are security devices in the current row\n            if (curr != 0) {\n                // Accumulate the product of current count and previous count\n                ans1 += curr * prev;\n                prev = curr; // Update the previous count for the next iteration\n            }\n        }\n\n        // Return the total number of laser beams\n        return ans1;\n    }\n};\n\n```\nPython3\n```\nclass Solution:\n    def numberOfBeams(self, bank: List[str]) -> int:\n        curr, prev = 0, 0\n        n, m = len(bank), len(bank[0])\n        ans1 = 0\n\n        # Iterate through each row of the bank\n        for i in range(n):\n            curr = 0  # Reset the current count of security devices for each row\n\n            # Iterate through each column of the current row\n            for j in range(m):\n                if bank[i][j] == \'1\':\n                    curr += 1  # Count the number of security devices in the current row\n\n            # If there are security devices in the current row\n            if curr != 0:\n                # Accumulate the product of current count and previous count\n                ans1 += curr * prev\n                prev = curr  # Update the previous count for the next iteration\n\n        # Return the total number of laser beams\n        return ans1\n\n```\nJava\n```\npublic class Solution {\n    public int numberOfBeams(String[] bank) {\n        int curr, prev = 0;\n        int n = bank.length;    // Number of rows in the bank\n        int m = bank[0].length(); // Number of columns in the bank\n        int ans1 = 0;           // Variable to store the total number of laser beams\n\n        // Iterate through each row of the bank\n        for (int i = 0; i < n; i++) {\n            curr = 0; // Reset the current count of security devices for each row\n\n            // Iterate through each column of the current row\n            for (int j = 0; j < m; j++) {\n                if (bank[i].charAt(j) == \'1\') {\n                    curr++; // Count the number of security devices in the current row\n                }\n            }\n\n            // If there are security devices in the current row\n            if (curr != 0) {\n                // Accumulate the product of current count and previous count\n                ans1 += curr * prev;\n                prev = curr; // Update the previous count for the next iteration\n            }\n        }\n\n        // Return the total number of laser beams\n        return ans1;\n    }\n}\n\n\n```\nJavascript\n```\n/**\n * @param {string[]} bank\n * @return {number}\n */\nvar numberOfBeams = function(bank) {\n    let curr = 0, prev = 0;\n    const n = bank.length;    // Number of rows in the bank\n    const m = bank[0].length; // Number of columns in the bank\n    let ans1 = 0;             // Variable to store the total number of laser beams\n\n    // Iterate through each row of the bank\n    for (let i = 0; i < n; i++) {\n        curr = 0; // Reset the current count of security devices for each row\n\n        // Iterate through each column of the current row\n        for (let j = 0; j < m; j++) {\n            if (bank[i][j] === \'1\') {\n                curr++; // Count the number of security devices in the current row\n            }\n        }\n\n        // If there are security devices in the current row\n        if (curr !== 0) {\n            // Accumulate the product of current count and previous count\n            ans1 += curr * prev;\n            prev = curr; // Update the previous count for the next iteration\n        }\n    }\n\n    // Return the total number of laser beams\n    return ans1;\n};\n```\n\n\n
['Array', 'Math', 'String', 'Matrix', 'Python', 'C++', 'Java', 'Python3', 'JavaScript']