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The Coding Puzzle 🧩
Last week's puzzle:
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 != i and nums[j] < nums[i].
Return the answer in an array.
Example 1:
Input: nums = [6,5,4,8]Output: [2,1,0,3]Example 2:
Input: nums = [7,7,7,7]Output: [0,0,0,0]Solution for last week's problem:
from typing import List
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
result = {}
for i, n in enumerate(sorted(nums)):
if n not in result:
result[n] = i
return [result[n] for n in nums]This week's puzzle:
You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
Example 1:
Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3Output: [1,2,2,3,5,6]Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
Example 2:
Input: nums1 = [1], m = 1, nums2 = [], n = 0Output: [1]Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].
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