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Last Updated: February 25, 2016
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· lukasz-madon
B2bc42f69510d6a1d2c9cde1c2214d24

Sum bitdiff between all pairs of numbers in a collection

There are many ways to compare two integers. One of them is to compare each bit of numbers e.g. comparing 2 and 1:

2 -> ...0010
1 -> ...0001

Two last bits are different, so the result is 2. Here is one of many possible implementations:

public static int bitDiff(int a, int b)
{
    int count = 0;
    for (int i = 0; i < 32; i++)
    {
        int aBit = (1 << i) & a;
        int bBit = (1 << i) & b;
        if (aBit != bBit)
        {
            count++;
        }
    }
    return count;
}

Now let's have a look at more complex problem. From a collection of integers calculate bitDiff of all pairs and then sum the results.

āˆ‘ bitDiff(array[i], array[j])

input: {1,2,3} -> pairs {(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)} -> output: 8

This can be optimize by generating just half of the pairs (bitDiff operation is transitive), but it's still O(n^2). Can we do O(n)? Scroll for a hint!

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What if we just have 0 and 1 in the array?

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Let's consider simple case

[0,1] ->  1
[1,1] ->  0
[0,0] ->  0
[1,0] ->  1

Clearly, the result is 1 for pairs that are different. We can count 1's and 0's and multiply each other to get the number of pairs that result is 1 and then multiply by 2 (permutation of a pair)

[0,1] = 1 * 1 * 2 = 2
[0,1,1] = 1 * 2 * 2 = 4
[0,1,0,1] = 2 * 2 * 2 = 8

How to get the generalization for ints?

We can threat numbers as 32 arrays of 1's and 0's

public static bool isBitSet(int number, int index)
{
    int bit = (1 << index) & number;
    return bit > 0;
}

public static int sumPairBitDiff(int[] numbers)
{
    int sum = 0;
    for (int i = 0; i < 32; i++)
    {
        int oneCount = 0;
        int zeroCount = 0;
        foreach (var num in numbers)
        {
            if (isBitSet(num, i)) oneCount++;
        }
        zeroCount = numbers.Length - oneCount;
        sum += oneCount * zeroCount * 2;
    }
    return sum;
}

Awesome!

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