Previous smaller integer having one less number of set bits

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)10 = (1000)2
is having 1 set bit.

(0)10 = (0)2
is having 0 set bit and is the previous smaller.

Input : 25
Output : 24
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Following are the steps:

1. 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.
2. 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
```

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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