Problem Description

The problem is related to the concept of Hamming distance, which is a measure of difference between two integers represented in binary. More specifically, it is the number of bit positions in which the two bits are different. The problem asks us to calculate the sum of Hamming distances between all pairs of integers in an array of integers, nums.

To tackle this problem, we are given an array of integers and we want to find the pair-wise Hamming distances and then sum all those distances together. If we have N integers in our array, there are (N*(N-1))/2 pairs, where each pair's Hamming distance contributes to the sum.

Consider an example with smaller numbers and do this task manually. For [4, 14, 2], represented in binary as [0100, 1110, 0010], the Hamming distances between each pair of numbers would be calculated and summed up like this:

• The Hamming distance between 4 (0100) and 14 (1110) is 3.
• The Hamming distance between 4 (0100) and 2 (0010) is 1.
• The Hamming distance between 14 (1110) and 2 (0010) is 2.

The sum of all the Hamming distances is 3 + 1 + 2 = 6.

Intuition

The brute-force approach to calculating the sum of all Hamming distances between pairs in the array nums would be to compare every pair and count the bits that are different. However, calculating pairwise Hamming distances for large arrays would be an inefficient solution with a time complexity of O(n^2 * k), where n is the number of integers in nums and k is the number of bits.

A clever way to tackle this problem is to count, for each bit position, the number of integers in the array that have a '1' in that position and the number that have a '0'. This is based on the concept that the Hamming distance for a particular bit position across all pairs is simply the number of times '1' occurs at that position multiplied by the number of times '0' occurs at the same position, because a difference occurs only if one is 1 and the other is 0.

For instance, if at the ith bit position, we have a numbers with '1' and b numbers with '0', the total Hamming distance contributed from the ith bit position would be a * b. This is true because each of the a numbers with '1' will have a Hamming distance of 1 with each of the b numbers with '0'. Therefore, by doing this for all bit positions (0 to 30 for 31-bit integers, since the problem mention doesn't include negative numbers that would need the 32nd bit for the sign) and summing up the results, we get the total Hamming distance for all pairs.

This method is more efficient because it only requires us to pass through all the numbers for each bit position, making it O(n*k), which is significantly faster than the brute-force method, especially for a large number of integers in nums.

The given solution uses this efficient method to calculate the sum of all Hamming distances in the array nums using a loop that iterates over each bit position, followed by another loop to count occurrences of '1's and '0's, and then combining those counts to calculate the contribution of each bit position to the total Hamming distance.

Learn more about Math patterns.

Solution Approach

The implementation of the solution involves a bit manipulation technique and an understanding of combinatorics.

Here are the steps of the algorithm used in the provided Python code:

1. Initialize ans to 0. This variable will hold the sum of Hamming distances.
2. Use a loop to iterate over each bit position. In this case, we loop from 0 to 30, inclusive, because an integer is 32 bits in size and we assume we're dealing with positive numbers only (no need for the sign bit).
3. Inside this loop, initialize two counters a and b to 0. Counter a will keep track of the number of 1's and b of the number of 0's for the current bit position across all numbers in nums.
4. Inner loop through each number v in nums:
   • Shift v right by i bits and perform a bitwise AND with 1 ((v >> i) & 1) to isolate the bit at the current bit position.
   • If the result is 1, increment counter a (since this number contributes a 1 to the current bit position).
   • If the result is 0, increment counter b (since this number contributes a 0 to the current bit position).
5. Outside the inner loop but still inside the first loop, multiply a and b and add the result to ans. This is based on the observation that each pair of numbers contributes 1 to the Hamming distance sum for this bit position if one of them has a 1 and the other has a 0. The total contribution from this bit position is therefore the product of the numbers of 1's and 0's.
6. After completing the loops, ans contains the sum of the Hamming distances, and we return this value.

In terms of data structures, the algorithm uses a list to store the input numbers and two integers as counters. There are no complex data structures required. The primary pattern used is bit manipulation, specifically shifting and masking to access individual bits of integers. The algorithm's time complexity is O(n*k), where n is the number of integers in the input array and k is the number of bits we are considering (31 in this case).

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using the array [4, 14, 2]. Follow along with the steps of the provided algorithm.

Step 1: Initialize ans to 0.

Step 2: We will iterate over each bit position from 0 to 30. For simplicity, let's assume we only deal with 4-bit binary representations since the highest number here is 14. So, we loop from 0 to 3.

Step 3 & 4: Let's start with the least significant bit (LSB), i.e., bit position 0.

• Initialize counters a and b to 0 for this bit position.
• Inner loop over the numbers in nums:
  • For 4 (0100): (4 >> 0) & 1 equals 0, so we increment b.
  • For 14 (1110): (14 >> 0) & 1 equals 0, so we increment b.
  • For 2 (0010): (2 >> 0) & 1 equals 0, so we increment b.
• After loop, a is 0 and b is 3 for this bit position.

Step 5: Multiply a and b and add the result to ans.

• a * b is 0 because there are no 1s in this bit position across all numbers. So, ans remains 0.

Step 6: Continue this process for the next bit positions (1, 2, 3).

• Bit position 1:

  • a would be 2 (4 and 2 have 1 in this bit).
  • b would be 1 (14 has 0 in this bit).
  • a * b gives us 2 * 1 = 2, so ans updates to 2.

• Bit position 2:

  • a would be 1 (14 has 1 in this bit).
  • b would be 2 (4 and 2 have 0 in this bit).
  • a * b gives us 1 * 2 = 2, so ans updates to 4.

• Bit position 3:

  • a would be 1 (14 has 1 in this bit).
  • b would be 2 (4 and 2 have 0 in this bit).
  • a * b gives us 1 * 2 = 2, so ans updates to 6.

Step 7: After completing the loops for all bit positions, ans is 6, which is the sum of the Hamming distances for all pairs in the array [4, 14, 2].

Result: The final sum of Hamming distances computed is 6. This result corresponds to adding the pairwise Hamming distances: 3 (from 4 and 14), 1 (from 4 and 2), and 2 (from 14 and 2), as was calculated manually at the beginning of the problem description.

Solution Implementation

Time and Space Complexity

The provided Python code calculates the total Hamming distance between all pairs of integers in an input array. Each integer is represented in 32-bit binary form (assuming a maximum of 31 usable bits due to the problem constraints). Here's the analysis of the complexities:

Time Complexity:

To analyze the time complexity, we observe that the code consists of two nested loops:

• The outer loop runs a fixed 31 times, corresponding to the number of bits in a 32-bit integer (minus the sign bit).
• The inner loop iterates through every element in the input list nums.

Since the outer loop is constant, we are mostly concerned with the inner loop, which has a runtime proportional to n, where n is the length of nums. Iterating through nums happens once for each bit, so the overall time complexity is O(n * 31), which simplifies to O(n).

Thus, the time complexity of the code is O(n).

Space Complexity:

The space complexity is determined by the amount of extra space used by the program, which does not grow with the input. The variables ans, a, b, and t use a fixed amount of space regardless of the input size. No additional data structures that scale with the input size are being used.

Hence, the space complexity of the code is O(1).

Learn more about how to find time and space complexity quickly using problem constraints.