Given a positive integer . Find the number of steps required to minimize it to 1. In a single step N either got reduced to half if it is power of 2 else N is reduced to difference of N and its nearest power of 2 which is smaller than N.
Input : N = 2 Output : 1 Input : N = 20 Output : 3
Simple Approach: As per question a very simple and brute force approach is to iterate over N until it got reduced to 1, where reduction involve two cases:
- N is power of 2 : reduce n to n/2
- N is not power of 2: reduce n to n – (2^log2(n))
Efficient approach: Before proceeding to actual result lets have a look over bit representation of an integer n as per problem statement.
- When an integer is power of 2: In this case bit -representation includes only one set bit and that too is left most. Hence log2(n) i.e. bit-position minus One is the number of step required to reduce it to n. Which is also equal to number of set bit in n-1.
- When an integer is not power of 2:The remainder of n – 2^(log2(n)) is equal to integer which can be obtained by un-setting the left most set bit. Hence, one set bit removal count as one step in this case.
Hence the actual answer for steps required to reduce n is equal to number of set bits in n-1. Which can be easily calculated either by using the loop or any of method described in the post: Count Set bits in an Integer.
Below is the implementation of the above approach:
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