Given a positive integer ‘n’ having ‘x’ number of set bits in its binary representation. The problem is to find the previous smaller integer(greatest integer smaller than n), having (x-1) number of set bits in its binary representation.

**Note:** 1 <= **n**

Examples:

Input : 8 Output : 0(8)= (1000)_{10}_{2}is having 1 set bit.(0)= (0)_{10}_{2}is having 0 set bit and is the previous smaller. Input : 25 Output : 24

Following are the steps:

- Find the position of the rightmost set bit(considering last bit at position 1, second last bit at position 2 and so on) in the binary representation of
**n**. Let the position be represented by**pos**. Refer this post. - Turn off or unset the bit at position
**pos**. Refer this post.

// C++ implementation to find the previous // smaller integer with one less number of // set bits #include<bits/stdc++.h> using namespace std; // function to find the position of // rightmost set bit. int getFirstSetBitPos(int n) { return log2(n & -n) + 1; } // function to find the previous smaller // integer int previousSmallerInteger(int n) { // position of rightmost set bit of n int pos = getFirstSetBitPos(n); // turn off or unset the bit at // position 'pos' return (n & ~(1 << (pos - 1))); } // Driver program int main() { int n = 25; cout << previousSmallerInteger(n); return 0; }

Output:

Previous smaller integer = 24

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