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Given an integer array arr, you need to sort the array in a specific order:

  1. First, sort the integers by the number of 1 bits in their binary representation (in ascending order).
  2. If two integers have the same number of 1 bits, sort them by their decimal value (in ascending order).

Return the sorted array.

Clarifying Questions

  1. Input Origin: Should we assume that the input array is taken directly from the user, or will it always be passed as a function argument?
    • Assumption: The input is passed as a function argument.
  2. Range of Values: Are there any constraints on the size of the array or the range of integer values?
    • Assumption: Follow general leetcode constraints which usually involve reasonable limits for programming contests (e.g., 1 <= arr.length <= 500 and 0 <= arr[i] <= 10^4).
  3. Output: Is the output supposed to be returned or printed?
    • Assumption: The output should be returned.

Strategy

  1. Counting 1 Bits: Use Python’s in-built functions to convert an integer to its binary representation and count the number of 1 bits. For example, bin(x).count('1').
  2. Sorting: Use Python’s sorted() function with a custom key that first sorts by the number of 1 bits and then by the integer value itself.

Time Complexity

  1. Counting 1 Bits: The binary representation of an integer up to 10^4 can have at most 14 bits. Counting the bits is O(1) for each number.
  2. Sorting: Sorting an array of length n is O(n log n).

The overall time complexity will be dominated by the sorting step: O(n log n).

Code

def sortByBits(arr):
    # Define the custom sorting key
    def sort_key(x):
        return (bin(x).count('1'), x)
    
    # Return the sorted array
    return sorted(arr, key=sort_key)

# Example usage
arr = [0,1,2,3,4,5,6,7,8]
sorted_array = sortByBits(arr)
print(sorted_array)  # Output: [0, 1, 2, 4, 8, 3, 5, 6, 7]

This code correctly implements the strategy and efficiently sorts the array based on the provided rules.

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