Given a positive integer n, print the next smallest and the previous largest number that have the same number of 1 bits in their binary representation.

Examples :

Input : n = 5 Output : Closest Greater = 6 Closest Smaller = 3 Note that 5, 6 and 3 have same number of set bits. Input : n = 11 Output : Closest Greater = 13 Closest Smaller = 7

**The Brute Force Approach**

An easy approach is simply brute force: count the number of 1s in n, and then increment (or decrement) until we find a number with the same number of 1s.

**Optimal Approaches**

Let’s start with the code for getNext, and then move on to getPrev.

**Bit Manipulation Approach for Get Next Number**

If we think about what the next number should be, we can observe the following. Given the number 13948, the binary representation looks like:

1 1 0 1 1 001 1 1 1 1 0 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

We want to make this number bigger (but not too big). We also need to keep the same number of ones.

Observation: Given a number n and two bit locations i and j, suppose we flip bit i from a 1 to a 0, and bit j from a 0 to a 1. If i > j, then n will have decreased. If i < j, then n will have increased.

**We know the following:**

- If we flip a zero to a one, we must flip a one to a zero.
- The number (After two flips) will be bigger if and only if the zero-to-one bit was to the left of the one to zero bit.
- We want to make the number bigger, but not unnecessarily bigger. Therefore, we need to flip the rightmost zero which has ones on the right of it.

To put this in a different way, we are flipping the rightmost non-trailing zero. That is, using the above example, the trailing zeros are in the 0th and 1st spot. The rightmost non-trailing zero is at bit 7. Let's call this position p.

p==> Position of rightmost non-trailing 0.

**Step 1: Flip rightmost non-trailing zero
**

1 1 0 1 1 011 1 1 1 1 0 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

With this change, we have increased the number of 1s of n. We can shrink the number by rearranging all the bits to the right of bit p such that the 0s are on the left and the 1s are on the right. As we do this, we want to replace one of the 1s with a 0.

A relatively easy way of doing this is to count how many ones are to the right of p, clear all the bits from 0 until p, and then add back in c1-1 ones. Let c1 be the number of ones to the right of p and c0 be the number of zeros to the right of p.

Let’s walk through this with an example.

c1==> Number of ones to the right of pc0==> Number of zeros to the right of p. p = c0 + c1

**Step 2: Clear bits to the right of p. From before, c0 = 2. c1 = 5. p = 7.**

1 1 0 1 1 0 1 0 0 0 0 0 0 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

To clear these bits, we need to create a mask that is a sequence of ones, followed by p zeros. We can do this as follows:

// all zeros except for a 1 at position p. a = 1 << p; // all zeros, followed by p ones. b = a - 1; // all ones, followed by p zeros. mask = ~b; // clears rightmost p bits. n = n & mask; Or, more concisely, we do: n &= ~((1 << p) - 1).

**Step 3: Add one c1 – 1 ones.**

1 1 0 1 1 0 1 0 0 0 1 1 1 1 13 12 11 10 9 8 7 6 5 4 3 2 1 0

To insert c1 – 1 ones on the right, we do the following:

// 0s with a 1 at position c1– 1 a = 1 << (c1 - 1); // 0s with 1s at positions 0 through c1-1 b = a - 1; // inserts 1s at positions 0 through c1-1 n = n | b; Or, more concisely: n | = (1 << (c1 - 1)) - 1;

We have now arrived at the smallest number bigger than n with the same number of ones. The Implementation of code for getNext is below in C++

// C++ implementation of getNext with // same number of bits 1's is below #include <bits/stdc++.h> using namespace std; // Main Function to find next smallest // number bigger than n int getNext(int n) { /* Compute c0 and c1 */ int c = n; int c0 = 0; int c1 = 0; while (((c & 1) == 0) && (c != 0)) { c0 ++; c >>= 1; } while ((c & 1)==1) { c1++; c >>= 1; } // If there is no bigger number with the // same no. of 1's if (c0 +c1 == 31 || c0 +c1== 0) return -1; // position of rightmost non-trailing zero int p = c0 + c1; // Flip rightmost non-trailing zero n |= (1 << p); // Clear all bits to the right of p n &= ~((1 << p) - 1); // Insert (c1-1) ones on the right. n |= (1 << (c1 - 1)) - 1; return n; } // Driver Code int main() { int n = 5; // input 1 cout << getNext(n) << endl; n = 8; // input 2 cout << getNext(n); return 0; }

Output:

6 16

**Optimal Bit Manipulation Approach for Get Previous Number**

To implement getPrev, we follow a very similar approach.

- Compute c0 and c1. Note that
**c1**is the number of trailing ones, and**c0**is the size of the block of zeros immediately to the left of the trailing ones. - Flip the rightmost non-trailing one to a zero. This will be at position p = c1 + c0.
- Clear all bits to the right of bit p.
- Insert c1 + 1 ones immediately to the right of position p.

Note that Step 2 sets bit p to a zero and Step 3 sets bits 0 through p-1 to a zero. We can merge these steps.

Let’s walk through this with an example.

c1==> number of trailing onesc0==> size of the block of zeros immediately to the left of the trailing ones. p = c1 + c0

**Step 1: Initial Number: p = 7. c1 = 2. c0 = 5.**

1 0 0 1 1 1 1 0 0 0 0 0 1 1 13 12 11 10 9 8 7 6 5 4 3 2 1 0

**Steps 2 & 3: Clear bits 0 through p.**

1 0 0 1 1 1 0 0 0 0 0 0 0 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

**We can do this as follows:**

// Sequence of 1s int a = ~0; // Sequence of 1s followed by p + 1 zeros. int b = a << (p + 1); // Clears bits 0 through p. n & = b;

**Steps 4: Insert c1 + 1 ones immediately to the right of position p.
**

1 0 0 1 1 1 0 1 1 1 0 0 0 0 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Note that since p =c1 + c0, then (c1 + 1) ones will be followed by (c0 – 1)zeros.

**We can do this as follows:**

// 0s with 1 at position (c1 + 1) int a = 1 << (c1 + 1); // 0s followed by c1 + 1 ones int b = a - 1; // c1 + 1 ones followed by c0 - 1 zeros. int c = b << (c0 - 1); n |= c;

The code to implement this is below.

// C++ Implementation of getPrev in // Same number of bits 1's is below #include <bits/stdc++.h> using namespace std; // Main Function to find next Bigger number // Smaller than n int getPrev(int n) { /* Compute c0 and c1 and store N*/ int temp = n; int c0 = 0; int c1= 0; while ((temp & 1) == 1) { c1++; temp = temp >> 1; } if (temp == 0) return -1; while (((temp & 1) == 0) && (temp!= 0)) { c0++; temp = temp >> 1; } // position of rightmost non-trailing one. int p = c0 + c1; // clears from bit p onwards n = n & ((~0) << (p + 1)); // Sequence of (c1+1) ones int mask = (1 << (c1 + 1)) - 1; n = n | mask << (c0 - 1); return n; } // Driver Code int main() { int n = 6; // input 1 cout << getPrev(n); n = 16; // input 2 cout << endl; cout << getPrev(n); return 0; }

Output:

5 8

**Arithmetic Approach to Get Next Number**

If c0 is the number of trailing zeros, c1 is the size of the one block immediately following, and p = c0 + c1, we can form our solution from earlier as follows:

- Set the p-th bit to 1.
- Set all bits following p to 0.
- Set bits 0 through c1 – 2 to 1. This will be c1 – 1 total bits.

A quick way to perform steps 1 and 2 is to set the trailing zeros to 1 (giving us p trailing ones), and then add 1. Adding one will flip all trailing ones, so we wind up with a 1 at bit p followed by p zeros. We can perform this arithmetically.

// Sets trailing 0s to 1, giving us p trailing 1s.

n += 2^{c0} – 1 ;

// Flips first p ls to 0s, and puts a 1 at bit p.

n += 1;

Now, to perform Step 3 arithmetically, we just do:

// Sets trailing c1 – 1 zeros to ones.

n += 2^{c1 – 1} – 1;

This math reduces to:

next = n + (2^{c0} – 1) + 1 + (2^{c1 – 1} – 1)

= n + 2^{c0} + 2^{c1 – 1} – 1

The best part is that, using a little bit manipulation, it’s simple to code.

Below Implementation in C++

// C++ Implementation of getNext with // Same number of bits 1's is below #include <bits/stdc++.h> using namespace std; // Main Function to find next smallest number // bigger than n int getNext(int n) { /* Compute c0 and c1 */ int c = n; int c0 = 0; int c1 = 0; while (((c & 1) == 0) && (c != 0)) { c0 ++; c >>= 1; } while ((c & 1)==1) { c1++; c >>= 1; } // If there is no bigger number with the // same no. of 1's if (c0 +c1 == 31 || c0 +c1== 0) return -1; return n + (1 << c0) + (1 << (c1 - 1)) - 1; } // Driver Code int main() { int n = 5; // input 1 cout << getNext(n); n = 8; // input 2 cout << endl; cout << getNext(n); return 0; }

Output:

6 16

**Arithmetic Approach to Get Previous Number**

If c1 is the number of trailing ones, c0 is the size of the zero block immediately following, and p =c0 + c1, we can word the initial getPrev solution as follows:

- Set the pth bit to 0
- Set all bits following p to 1
- Set bits 0 through c0 – 1 to 0.

We can implement this arithmetically as follows. For clarity in the example, we assume n = 10000011. This makes c1 = 2 and c0 = 5.

// Removes trailing 1s. n is now 10000000. n -= 2^{c1}– 1; // Flips trailing 0s. n is now 01111111. n -= 1; // Flips last (c0-1) 0s. n is now 01110000. n -= 2^{c0 - 1}- 1; This reduces mathematically to: next = n - (2^{c1}- 1) - 1 - ( 2^{c0-1}- 1) . = n - 2^{c1}- 2^{c0-1}+ 1;

Again, this is very easy to implement.

// C++ Implementation of Arithmetic Approach to // getPrev with Smae number of bits 1's is below #include <bits/stdc++.h> using namespace std; // Main Function to find next Bigger number // Smaller than n int getPrev(int n) { /* Compute c0 and c1 and store N*/ int temp = n; int c0 = 0; int c1 = 0; while ((temp & 1) == 1) { c1++; temp = temp >> 1; } if (temp == 0) return -1; while (((temp & 1) == 0) && (temp!= 0)) { c0++; temp = temp >> 1; } return n - (1 << c1) - (1 << (c0 - 1)) + 1; } // Driver Code int main() { int n = 6; // input 1 cout << getPrev(n); n = 16; // input 2 cout << endl; cout << getPrev(n); return 0; }

Output:

5 8

This article is contributed by **Mr. Somesh Awasthi**. 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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