Consider the C code Below.void function(int n){ if (n == 1) return; for (int i = 0; i

Question

Consider the C code Below.void function(int n){  if (n == 1)   return;  for (int i = 0; i<n; i++)  {    for (int j = 1; j< = n; j++)    {      printf("*");            break;    }  }}Which of the following is the tightest upper bound on time complexity of above function.a)O(n2)b)O(n)c)O(n log n)d)O(1)Correct answer is option 'B'. Can you explain this answer?

Answer

The code seems to be incomplete as it ends abruptly with a for-loop without any closing braces. However, based on the available code, it appears to be a recursive function that takes an integer parameter 'n' and performs some operation on it.

The if-statement checks if the value of 'n' is equal to 1, and if so, it immediately returns from the function. This is the base case for the recursion.

If the value of 'n' is not 1, then the function enters a for-loop that initializes an integer variable 'i' to 0 and continues the loop as long as 'i' is less than 'n'. The loop performs some operation on 'i', but since the code is incomplete, we cannot determine what that operation is.

After the for-loop, the function calls itself with the parameter value of 'n-1'. This is where the recursion takes place, as the function keeps calling itself with decreasing values of 'n' until it reaches the base case where 'n' is equal to 1.

Overall, the purpose of the function and the operation it performs are unclear without more context or complete code.

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