Sort Integers by The Number of 1 Bits

class Solution {
public:
    static int calculateWeight(int number) {
        int sum = 0;
    
        while (number > 0) {
            sum++;
            number &= (number - 1);
        }
    
        return sum;
    }
    
    static bool customSort(int x, int y) {
        int wx = calculateWeight(x);
        int wy = calculateWeight(y);
    
        if (wx == wy) return x < y;
    
        return wx < wy;
    }
    
    vector<int> sortByBits(vector<int>& values) {
        sort(values.begin(), values.end(), customSort);
        return values;
    }
};

This article summarizes the solution to the problem of sorting integers by the number of 1 bits found in their binary representation, using C++ as the programming language. The implementation is within a class named Solution.

• Implement a helper function calculateWeight which counts the number of 1 bits in the binary form of its input, using number &= (number - 1) to clear the least significant 1 bit on each iteration.

• Define another function customSort, which compares two integers based on their 1-bit count (calculated by calling calculateWeight). If two integers have the same count, they are sorted by their value.

• Define the sortByBits function inside the Solution class. This function uses the sort function from the C++ Standard Library. It employs customSort as its comparison function.

• The sortByBits function receives a vector of integers, sorts them using the defined criteria, and returns the sorted vector.

Ensure the given C++ code is properly implemented with thorough tests to handle varied input cases, and validate results to ensure accuracy in sorting based on the count of 1-bits, with secondary sorting by integer values in case of ties.

class Solution {
    public int[] sortOnBitCount(int[] array) {
        Integer[] integers = Arrays.stream(array).boxed().toArray(Integer[]::new);
        Comparator<Integer> bitComparator = new BitBasedComparator();
        Arrays.sort(integers, bitComparator);
        return Arrays.stream(integers).mapToInt(Integer::intValue).toArray();
    }
}
    
class BitBasedComparator implements Comparator<Integer> {
    private int calculateBitWeight(int number) {
        int count = 0;
            
        while (number > 0) {
            count++;
            number &= (number - 1);
        }
            
        return count;
    }
        
    @Override
    public int compare(Integer x, Integer y) {
        if (calculateBitWeight(x) == calculateBitWeight(y)) {
            return x - y;
        }
            
        return calculateBitWeight(x) - calculateBitWeight(y);
    }
}

The provided Java solution addresses the problem of sorting an array of integers based on the number of 1 bits they contain, with ties broken by the values of the integers themselves. Here's a breakdown of how the solution works:

• Conversion of Primitive Array: The solution starts by converting the primitive int array to an Integer array using Java Streams. This conversion is essential because the Arrays.sort() method in Java uses a Comparator, which can only be applied to objects, not primitives.

• Custom Comparator: A custom comparator, BitBasedComparator, is created for sorting. This comparator defines the sorting behavior based on the bit count of the integers. The compare method of this comparator calculates the number of 1s in the binary representation of each integer using the calculateBitWeight method, which employs Brian Kernighan's algorithm to efficiently count the number of 1 bits.

• Sorting Process: The Arrays.sort() method is employed, using the custom comparator to rank the integers. After sorting, the Integer array is converted back to a primitive int array to maintain the original data type consistency.

• Counting 1s in Binary Representation: The calculateBitWeight function iteratively counts the bits by continuously stripping the lowest set bit using the expression number &= (number - 1) until all bits are zeroed out. Each operation effectively removes the lowest set bit, and the count of these operations is the bit weight.

This solution efficiently sorts the integers using bit counts, providing a direct way to organize data based on binary characteristics which can be advantageous in various applications such as optimization algorithms, cryptographic functions, or memory management systems, where understanding bit-level data can be crucial.

class Solution:
    def bitSort(self, arr: List[int]) -> List[int]:
        def bit_weight(x):
            count = 0
                
            while x:
                count += 1
                x &= x - 1
                
            return count
            
        arr.sort(key=lambda x: (bit_weight(x), x))
        return arr

Sort integers by the number of 1 bits in their binary representation using Python. This example employs a custom sorting strategy.

First, define a function bit_weight that calculates the number of 1's in the binary representation of an integer. This function utilizes a bitwise operation where the integer is continually bitwise ANDed with its immediate predecessor, incrementing a counter until the integer becomes zero.

Next, sort the array with a key that uses a lambda function. The lambda function sorts primarily by the count of 1 bits (computed via bit_weight), and secondarily by the integer's value itself in cases where multiple integers have the same number of 1 bits. This dual sorting ensures a stable and accurate sort order.

The steps involved in sorting the integers in the list arr by the number of 1 bits are as follows:

1. Define the bit_weight helper function to count the number of 1's for any given integer.
2. Use the sort method on the array, specifying a lambda function as the key that applies bit_weight to each element for primary sorting and defaults to the integer's value for tie-breaking.

This method provides a straightforward approach to sorting where binary properties of numbers are considered, beneficial in various applications such as coding challenges or data processing tasks where binary representations are crucial.