Calculate the run-time efficiency of the following program segment:for (i = 1; i <= n; i++)printf("%d ", i);
Question
Calculate the run-time efficiency of the following program segment:
for (i = 1; i <= n; i++)
printf("%d ", i);
Solution
To calculate the run-time efficiency of the given program segment, we need to analyze the time complexity of the code.
The program segment consists of a for loop that iterates from 1 to n, where n is the input size. Inside the loop, the printf function is called to print the value of i.
Let's break down the steps and analyze their time complexity:
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Initialization: The loop starts with initializing the variable i to 1. This step takes constant time, denoted as O(1).
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Condition check: The loop checks the condition i <= n before each iteration. This condition check also takes constant time, O(1).
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Body execution: Inside the loop, the printf function is called to print the value of i. The time complexity of the printf function depends on the implementation, but for simplicity, let's assume it takes constant time, O(1).
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Increment: After each iteration, the loop increments the value of i by 1. This step also takes constant time, O(1).
Now, let's consider the number of iterations the loop will perform. Since the loop iterates from 1 to n, it will execute n times.
Therefore, the overall time complexity of the program segment can be calculated as follows:
Time complexity = (Initialization + Condition check + Body execution + Increment) * Number of iterations = (O(1) + O(1) + O(1) + O(1)) * n = O(1) * n = O(n)
Hence, the run-time efficiency of the given program segment is O(n), where n is the input size.
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