Go solution | O(mn) | With Explaination
Step 1: Initialize Count Variables\nStart by initializing two integers, prev and ans. prev will hold the count of \'1\'s from the previous row in the bank, and 
**Step 1: Initialize Count Variables**\nStart by initializing two integers, prev and ans. prev will hold the count of \'1\'s from the previous row in the bank, and ans will accumulate the total number of beams formed.\n\n**Step 2: Iterate Through Bank**\nLoop through each string (str) in the bank array. For each string, convert it into a byte slice to optimize the iteration process.\nInside this loop, initialize a counter count to track the number of \'1\'s in the current string.\n\n**Step 3: Count \'1\'s and Calculate Beams**\nIterate over the byte slice (byteStr). Increment count for each occurrence of \'1\' (b == \'1\').\nIf count is greater than 0 (indicating that the current string has at least one \'1\'), calculate the number of beams formed with the previous row by multiplying prev with count and add it to ans. Update prev to be the current count.\n\n**Step 4: Return the Total Number of Beams**\nAfter processing all strings in the bank, ans contains the total number of beams formed between security devices in different rows. \n\n```\nfunc numberOfBeams(bank []string) int {\n\tvar prev, ans int\n\n\tfor _, str := range bank {\n\t\tcount := 0\n\n\t\tbyteStr := []byte(str)\n\t\tfor _, b := range byteStr {\n\t\t\tif b == \'1\' {\n\t\t\t\tcount++\n\t\t\t}\n\t\t}\n\n\t\tif count > 0 {\n\t\t\tans += (prev * count)\n\t\t\tprev = count\n\t\t}\n\t}\n\n\treturn ans\n}\n```\n\nIf you found it helpful, please consider giving it an upvote! It\'ll motivate me to share more solutions in the future. If you have any questions or suggestions, please let me know. Happy coding!
['Go']