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 + nnums2.length == n0 <= m, n <= 2001 <= 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?
The most efficient way to merge two sorted arrays without using extra space is by using a two-pointer approach from the end. You start comparing elements from the end of both arrays and fill `nums1` from the back. This way you avoid needing additional space for a temporary array and can overwrite existing data in `nums1`.
This C implementation uses a two-pointer technique starting from the back of the arrays `nums1` and `nums2`. It compares elements of both arrays from the end and places the larger element at the current end position in `nums1`.
C++
Java
Python
C#
JavaScript
Time Complexity: O(m + n); Space Complexity: O(1) since we do not use extra space.
This is a less efficient approach that involves first combining the contents of `nums1` and `nums2` into a new array. Then, the new array is sorted, and its contents are copied back into `nums1`. Although straightforward, this approach involves extra memory allocation.
In C, we first append the elements of `nums2` onto `nums1`, and then employ the Quick Sort function `qsort` to sort the entire array.
C++
Java
Python
C#
JavaScript
Time Complexity: O((m + n) log (m + n)); Space Complexity: O(1) additional space aside from function call stack during sorting.
| Approach | Complexity |
|---|---|
| Start from the End with Two Pointers | Time Complexity: O(m + n); Space Complexity: O(1) since we do not use extra space. |
| Use Extra Space to Combine and Sort | Time Complexity: O((m + n) log (m + n)); Space Complexity: O(1) additional space aside from function call stack during sorting. |
Merge Two Sorted Lists - Leetcode 21 - Python • NeetCode • 438,617 views views
Watch 9 more video solutions →Practice Merge Sorted Array with our built-in code editor and test cases.
Practice on FleetCodePractice this problem
Open in Editor