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Time Complexity of a Loop when Loop variable “Expands or Shrinks” exponentially

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For such cases, time complexity of the loop is O(log(log(n))).The following cases analyse different aspects of the problem. Case 1 :
for (int i = 2; i <=n; i = pow(i, k)) 
    // some O(1) expressions or statements
}

                    
In this case, i takes values 2, 2k, (2k)k = 2k2, (2k2)k = 2k3, …, 2klogk(log(n)). The last term must be less than or equal to n, and we have 2klogk(log(n)) = 2log(n) = n, which completely agrees with the value of our last term. So there are in total logk(log(n)) many iterations, and each iteration takes a constant amount of time to run, therefore the total time complexity is O(log(log(n))). Case 2 :
// func() is any constant root function
for (int i = n; i > 1; i = func(i)) 
   // some O(1) expressions or statements
}

                    
In this case, i takes values n, n1/k, (n1/k)1/k = n1/k2, n1/k3, …, n1/klogk(log(n)), so there are in total logk(log(n)) iterations and each iteration takes time O(1), so the total time complexity is O(log(log(n))). Refer below article for analysis of different types of loops. https://www.geeksforgeeks.org/analysis-of-algorithms-set-4-analysis-of-loops/

Last Updated : 08 May, 2019
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