Merge Sorted Array

Problem Statement

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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 = 3 Output: [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 = 0 Output: [1] Explanation: The arrays we are merging are [1] and []. The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1 Output: [1] Explanation: The arrays we are merging are [] and [1]. The result of the merge is [1]. Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

Constraints:

• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -109 <= nums1[i], nums2[j] <= 109

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

Solution

Time : O(n.log(n)), Space: 1

• Merge nums2 at the back of nums1
• Sort nums1 after merging

Time: O(m+n), Space 1

• Loop from len(nums1) to 0 on nums1
• use 2 pointers to Loop nums1 from m-1 to 0 and nums2 from n-1 to 0 look for maximum number
• Add maximum number at the back of nums1 and reduce m or n (whichever has maximum number at that index)
• At last copy numbers from 0 to n to nums1

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        if n == 0:
            return
       

        len1 = len(nums1)
        for i in range(1, len1+1):
            if (m-1 >=0 and n-1 >= 0):
                if (nums1[m-1] > nums2[n-1]):
                    nums1[-i] = nums1[m-1]
                    m -= 1
                else:
                    nums1[-i] = nums2[n-1]
                    n -= 1
            else:
                break

        if m==0:
            for i in range(0, n):
                nums1[i] = nums2[i]

        # nums1[0] = min(nums1[0], nums2[0])