LeetCode 88. Merge Sorted Array

Description

https://leetcode.com/problems/merge-sorted-array/

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?

Explanation

The size of nums1 is m + n. nums1 is enough to store all numbers in nums1 and nums2. We only need to first shift all numbers in nums1 by n positions. Then we compare and put numbers in nums1 and nums2 into nums1 in ascending order.

Python Solution

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        
        for i in range(m - 1, -1, -1):
            nums1[i + n] = nums1[i] 

        i = n
        j = 0
        k = 0
        while i < m + n and j < n:
            if nums1[i] < nums2[j]:
                nums1[k] = nums1[i]
                i += 1
            else:
                nums1[k] = nums2[j]
                j += 1
            k += 1

        while j < n:
            nums1[k] = nums2[j]
            j += 1
            k += 1
  • Time complexity: O(m + n).
  • Space complexity: O(m + n).

3 Thoughts to “LeetCode 88. Merge Sorted Array”

  1. class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
    for (int i = m; i < (m+n); i++) {
    nums1[i] = nums2[i – m];
    }

    Arrays.sort(nums1);
    }
    }

  2. /** Java solution */

    public void merge(int[] nums1, int m, int[] nums2, int n) {
    int insertP = m + n – 1;
    int nums1P = m – 1;
    int nums2P = n – 1;
    while (nums1P >= 0 && nums2P >= 0){
    if (nums1[nums1P] > nums2[nums2P]) {
    nums1[insertP–] = nums1[nums1P–];
    }
    else {
    nums1[insertP–] = nums2[nums2P–];
    }
    }
    while (nums2P >= 0) {
    nums1[insertP–] = nums2[nums2P–];
    }
    }

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