# 5 Best Ways to Find the Next Higher Number with the Same Number of Set Bits as n in Python

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π‘ Problem Formulation: The challenge is to write a Python program that finds the smallest number greater than `n` which has the exact same number of `1` bits in its binary representation. For example, if `n` is `5` (binary `101`), the next higher number with two set bits is `6` (binary `110`).

## Method 1: Bit Manipulation Method

This technique executes specific bit manipulation operations that follow the property of set bits. In this method, we manipulate the position of the ‘1’ bits to find the consecutive higher number with the same amount of ‘1’ bits. This is a direct mathematical way which efficiently achieves the target.

Here’s an example:

```def next_higher_with_same_bits(n):
c = n
c0 = c1 = 0
while (c & 1) == 0 and (c != 0):
c0 += 1
c >>= 1
while (c & 1) == 1:
c1 += 1
c >>= 1
if c0 + c1 == 31 or c0 + c1 == 0:
return -1
p = c0 + c1
n |= (1 << p)
n &= ~((1 << p) - 1)
n |= (1 << (c1 - 1)) - 1
return n

print(next_higher_with_same_bits(5))```

Output:

`6`

This function `next_higher_with_same_bits` uses bitwise operations to find the next higher integer with the same number of set bits. It counts the trailing zeros, the number of ones following them, and then reconstructs the binary number so that it forms the next higher number with the same number of ‘1’s. The function returns -1 if no such number exists.

## Method 2: Use the Python ‘bin’ Function

This approach converts the number `n` to its binary string representation using Python’s `bin` function. It then repeatedly increments the number by 1 until a number with the same number of set bits is found. This method is straightforward but less efficient for larger numbers.

Here’s an example:

```def next_higher_num_set_bits(n):
count_set_bits = bin(n).count('1')
next_num = n + 1
while bin(next_num).count('1') != count_set_bits:
next_num += 1
return next_num

print(next_higher_num_set_bits(5))```

Output:

`6`

The function `next_higher_num_set_bits` counts the number of set bits in the binary representation of the given number and iterates through the successive integers until an integer with the same count of set bits is found.

## Method 3: Brian Kernighan’s Algorithm

Brian Kernighan’s algorithm is used for counting the number of set bits in a given integer. We can utilize this algorithm by incrementing our number and using the algorithm to check if the number of set bits matches the original one. This method is a bit more efficient than method 2 but may also face performance issues with large numbers.

Here’s an example:

```def count_set_bits(n):
count = 0
while n:
n &= (n - 1)
count += 1
return count

def next_higher_same_set_bits(n):
num_of_set_bits = count_set_bits(n)
next_num = n + 1
while count_set_bits(next_num) != num_of_set_bits:
next_num += 1
return next_num

print(next_higher_same_set_bits(5))```

Output:

`6`

The `count_set_bits` function uses Brian Kernighan’s algorithm to count set bits of an integer. The `next_higher_same_set_bits` function finds the next higher number with the same number of set bits as `n` by iterating through consecutive numbers and applying the `count_set_bits` function each time.

## Method 4: Precompute Set Bits

For performance reasons, especially when handling multiple queries for different numbers, precomputation comes in handy. This method involves precalculating the number of set bits for a given range of numbers and then querying this for faster comparisons.

Here’s an example:

```set_bits_count = {i: bin(i).count('1') for i in range(256)}

def quick_next_higher_num_set_bits(n):
target_set_bits = set_bits_count[n & 255]
next_num = n + 1
while set_bits_count[next_num & 255] != target_set_bits:
next_num += 1
return next_num

print(quick_next_higher_num_set_bits(5))```

Output:

`6`

In the code snippet, a dictionary `set_bits_count` is created to hold the number of set bits for all 8-bit numbers. This enables a quick lookup for the number of set bits in the least significant byte of the number in question. The `quick_next_higher_num_set_bits` function then only needs to check the last 8 bits to match the target set bits count.

## Bonus One-Liner Method 5: Using itertools

`itertools` is a powerful Python module providing various functions creating iterators for efficient looping. We can leverage `itertools.count` to generate numbers and combine it with list comprehension for a compact solution. However, understand that the readability can decrease as code complexity increases.

Here’s an example:

```from itertools import count

def next_higher_with_set_bits_itertools(n):
return next(x for x in count(n+1) if bin(x).count('1') == bin(n).count('1'))

print(next_higher_with_set_bits_itertools(5))```

Output:

`6`

The function `next_higher_with_set_bits_itertools` uses generator expression inside `next()` to find the immediate next number with the same number of set bits. It generates numbers starting from `n+1` and checks if the count of set bits matches that of `n`.

## Summary/Discussion

• Method 1: Bit Manipulation Method. Most efficient for any number. Complexity arises from understanding bitwise operations.
• Method 2: Python ‘bin’ Function. Easiest to understand and implement. Inefficient for large numbers due to linear search.
• Method 3: Brian Kernighan’s Algorithm. Offers a balance between simplicity and efficiency. Iterative approach can be slow for very large numbers.
• Method 4: Precompute Set Bits. Very efficient for repeated queries within a range. Requires additional setup and memory for storing precomputed results.
• Method 5: Using itertools. Elegant one-liner, but is a syntactic sugar masking a similar approach to Method 2, thus shares its weaknesses.