Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != iandnums[j] < nums[i].
Return the answer in an array.
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Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums=1 does not exist any smaller number than it. For nums=2 there exist one smaller number than it (1). For nums=2 there exist one smaller number than it (1). For nums=3 there exist three smaller numbers than it (1, 2 and 2).
Input: nums = [6,5,4,8] Output: [2,1,0,3]
It’s a sorting question. We have to know the position of each element in array. So, first method I used would be sort it and used HashMap to track the relationship. In addition, we can use bucket sort to reduce the time complexity.