What is the time complexity of this function / algorithm?void f(unsigned int n){ int i; for (i = 1; i < n; i = i * 2) { printf("[%d]\n", i); }}O(n!)O(2^n)O(1)O(n)O(nlog(n))O(n^2)O(log(n))
Question
What is the time complexity of this function / algorithm?void f(unsigned int n){ int i; for (i = 1; i < n; i = i * 2) { printf("[%d]\n", i); }}O(n!)O(2^n)O(1)O(n)O(nlog(n))O(n^2)O(log(n))
Solution
The time complexity of the given function is O(log(n)). This is because the loop counter 'i' is doubled in each iteration, which means the number of iterations will be proportional to the logarithm (base 2) of 'n'.
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What is the time complexity of this function / algorithm?void f(int n){ int i; for (i = 0; i < n; i += 98) { printf("[%d]\n", i); }}O(n!)O(2^n)O(1)O(n)O(nlog(n))O(n^2)O(log(n))
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What is the time complexity of the following code :int i, j, k = 0;for (i = n/2; i <= n; i++) { for (j = 2; j <= n; j = j * 2) { k = k + n/2; }}
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