Merge Sorted Array LeetCode Solution | Easy Approach

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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?

 Merge Sorted Array Solutions

Time: OO(m+n))
Space: O(1)

C++

 Will be updated Soonclass 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--];
  }
};
 Will be updated Soon

Java

 
 Will be updated Soonclass 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--];
  }
}
 
 Will be updated Soon

Python

  Will beclass 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

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