What will be the time complexity of the following code?int fun(int n){ int count = 0; for (int i = n; i > 0; i /= 2){ for (int j = 0; j < i; j++){ count += 1;| } }return count;
Question
What will be the time complexity of the following code?int fun(int n){ int count = 0; for (int i = n; i > 0; i /= 2){ for (int j = 0; j < i; j++){ count += 1;| } }return count;
Solution
The time complexity of the given code is O(n log n).
Here's why:
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The outer loop runs log(n) times because i is halved in each iteration. This is a property of logarithmic functions.
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The inner loop runs i times, where i starts from n and is halved in each iteration of the outer loop. So, in the first iteration, it runs n times, in the second iteration it runs n/2 times, in the third iteration it runs n/4 times, and so on.
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The total number of times the inner loop runs is therefore n + n/2 + n/4 + ... + 1. This is a geometric series which sums up to 2n.
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Therefore, the total time complexity is the product of the number of times the outer loop runs and the average number of times the inner loop runs, which is O(n log n).
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What is the time complexity of the following code : int a = 0, i = N; while (i > 0) { a += i; i /= 2; }
The time complexity of the following code snippet isc=0; while (n>=1) {for(i=1;i>=n;i++){ c++;}n=n/2;}
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