C Solution with 0ms runtime
Intuition\nThe problem requires counting the number of seniors (aged 60 or older) in a given list of details. Each detail is a string, and the age is encoded in
# Intuition\nThe problem requires counting the number of seniors (aged 60 or older) in a given list of details. Each detail is a string, and the age is encoded in a specific position within these strings.\n\n# Approach\n1.Iterate through each detail: We loop through the list of details.\n2.Extract the age information: From each detail string, extract the characters that represent the tens and units place of the age.\n3.Check if the person is a senior: If the tens place character is \'6\' and the units place character is \'0\' or higher, or if the tens place character is greater than \'6\', increment the senior count.\n4.Return the count: After the loop, return the count of seniors.\n\n# Complexity\n- Time complexity:\nO(n) where n is the number of details. We are iterating through the list once.\n\n- Space complexity:\nO(1) as we are using a constant amount of extra space for counting.\n\n# Code\n```\nint countSeniors(char ** details, int detailsSize){\n    int count=0;\n    for(int i=0;i<detailsSize;i++){\n        if(details[i][11]==\'6\' && details[i][12]>\'0\' || (details[i][11]>\'6\')){\n            count++;\n        }\n    }\n    return count;\n}\n```
['Array', 'String', 'C', 'C++']