What is the time complexity of below code?
void fun(int n) { int j = 1, i = 0; while (i < n) { // Some O(1) task i = i + j; j++; } } |
The loop variable ‘i’ is incremented by 1, 2, 3, 4, … until i becomes greater than or equal to n.
The value of i is x(x+1)/2 after x iterations. So if loop runs x times, then x(x+1)/2 < n. Therefore time complexity can be written as Θ(√n).
This article is contributed by Piyush Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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