Simplest & Efficient C Code Time[O(n^2)], Space[O(1)]🔥
Intuition\nThe code aims to compare the count of each digit in the string with the actual digit itself.\n\n# Approach\nThe code iterates through each digit in t
# Intuition\nThe code aims to compare the count of each digit in the string with the actual digit itself.\n\n# Approach\nThe code iterates through each digit in the given string and compares its count with the actual digit itself. If any count doesn\'t match, it returns false; otherwise, it returns true. \n\n# Complexity\n- Time complexity: Let\'s consider the length of the string num as n. The code uses two nested loops, with each loop iterating over the entire string. Therefore, the time complexity of the code is $O(n^2)$.\n\n- Space complexity: The code uses a constant amount of additional space to store variables l, i, j, and count. Hence, the space complexity is $O(1)$.\n\n# Code\n```\nbool digitCount(char * num)\n{\n    int l = strlen(num);\n    for(int i=0;i<l;i++)\n    {\n        int count = 0;\n        for(int j = 0;j<l;j++)\n        {\n            if(i==num[j]-\'0\')\n            count++;\n        }\n        if(count != num[i]-\'0\')\n        return false;\n    }\n    return true;\n}\n```
['C']