Merge Sorted Array Leetcode Solution

In this post, you will know how to solve the Merge Sorted Array Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Problem

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?

Merge Sorted Array Leetcode Solutions in Python

class Solution:
  def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
    i = m - 1      # nums1's index (actual nums)
    j = n - 1      # nums2's index
    k = m + n - 1  # nums1's index (next filled position)
    while j >= 0:
      if i >= 0 and nums1[i] > nums2[j]:
        nums1[k] = nums1[i]
        k -= 1
        i -= 1
      else:
        nums1[k] = nums2[j]
        k -= 1
        j -= 1

Merge Sorted Array Leetcode Solutions in CPP

class Solution {
 public:
  void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
    int i = m - 1;      // nums1's index (actual nums)
    int j = n - 1;      // nums2's index
    int k = m + n - 1;  // nums1's index (next filled position)
    while (j >= 0)
      if (i >= 0 && nums1[i] > nums2[j])
        nums1[k--] = nums1[i--];
      else
        nums1[k--] = nums2[j--];
  }
};

Merge Sorted Array Leetcode Solutions in Java

class Solution {
  public void merge(int[] nums1, int m, int[] nums2, int n) {
    int i = m - 1;     // nums1's index (actual nums)
    int j = n - 1;     // nums2's index
    int k = m + n - 1; // nums1's index (next filled position)
    while (j >= 0)
      if (i >= 0 && nums1[i] > nums2[j])
        nums1[k--] = nums1[i--];
      else
        nums1[k--] = nums2[j--];
  }
}

Note: This problem Merge Sorted Array is generated by Leetcode but the solution is provided by  BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

Next: Scramble String Leetcode Solution