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Leetcode 1365 solution

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	`1`	`+# Link to question: https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/`
	`2`	`+def smallerNumbersThanCurrent(nums):`
	`3`	`+ count = [0]*101`
	`4`	`+`
	`5`	`+ for num in nums:`
	`6`	`+ count[num] += 1`
	`7`	`+`
	`8`	`+ # Maintain a running sum in the count array`
	`9`	`+ for i in range(1, 101):`
	`10`	`+ count[i] += count[i-1]`
	`11`	`+`
	`12`	`+ final_ans = [0] * len(nums)`
	`13`	`+`
	`14`	`+ # Format to fit output answer`
	`15`	`+ for i in range(len(nums)):`
	`16`	`+ curr_num = nums[i]`
	`17`	`+ # If the number is 0, there will never be elements less than it`
	`18`	`+ if curr_num == 0:`
	`19`	`+ final_ans[i] = 0`
	`20`	`+ else:`
	`21`	`+ # Set it to the running sum up to the previous element in nums`
	`22`	`+ final_ans[i] = count[curr_num-1]`
	`23`	`+`
	`24`	`+ return final_ans`

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