leetcode-477-Total Hamming Distance

最新推荐文章于 2024-11-30 07:15:00 发布

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本文探讨了如何通过计算每对数组中不同位的数量来优化算法复杂度，避免使用O(n^2)的时间复杂度。具体方法是积累每比特位的不同数量，并计算总数组合，从而等效于计算每对数的不同位。更新内容包括使用基数排序时的注意事项。

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Instead of counting different of the pair of array which has O(n^2) time complexity, we decide to count different of each bit, i.e. accumulate different of each bit. More specify, res += (n - count) * count.

Why it works?

for each bit, we are counting the total combination, e.g. in first bit, we have first and second number has 0 bit and third number has 1 bit, so the total combination is 2 * 1(first&third, second&third). It is equal to count the pair of each number.

update:
radix sort, remember, DO NOT judge bit & to 1 explicitly, i.e., do write ((x & 1) == 1)