Run time analysis of inner loop

Asked Jan 13 '19 at 11:08

Active Jan 13 '19 at 11:57

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What is the run time of the following piece of code in Big-Oh notation? The first loop runs n times in the worst case. But I am having difficulty in finding run time of nested loop which runs V / deno[i] times to be exact. So I think the total runtime would be O(n* V/deno[i])? Please correct me if I am wrong.

int deno[] = {1, 2, 5, 10, 20, 50, 100, 500, 1000};

int n = sizeof(deno)/sizeof(deno[0]);

// Driver program void findMin(int V) { // Initialize result vector ans;

// Traverse through all denomination 
for (int i=n-1; i>=0; i--) 
{ 
    // Find denominations 
    while (V >= deno[i]) 
    { 
       V -= deno[i]; 
       ans.push_back(deno[i]); 
    } 
}

edited Jan 13 '19 at 11:57

asked Jan 13 '19 at 11:08

Xashi

Without knowing what values deno[i] might take and what ans.push_back() does, this is pretty much unanswerable. – David Richerby Jan 13 '19 at 11:20
deno[i] can take value from 1000, 500, 100, 50, 20 10, 5, 2 , 1 – Xashi Jan 13 '19 at 11:24
The array deno is given as input. push_back() appends the the argument given to it at the end of the vector 'ans' – Xashi Jan 13 '19 at 11:26
Your code runs in $O(1)$. This is always the case when there is no input. – Yuval Filmus Jan 16 '19 at 09:23

Run time analysis of inner loop

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