0088. Merge Sorted Array

Easy | Array + Two Pointers | 16ms (98.22%), 13.3 MB (94.73%)

❓ Problem Statement

Source: LeetCode - Merge Sorted Array GitHub: Solution / Performance

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.

✍🏻 Constraints & Example

Constraints:

• nums1.length == m + n

• nums2.length == n

• 0 <= m, n <= 200

• 1 <= m + n <= 200

• -10^9 <= nums1[i], nums2[j] <= 10^9

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.

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].

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.

💡 Ideas

Three-pointer is required for placing numbers reversely from the end of array1:

1. place-pointer: starting from the end, which value is m + n - 1.

2. array1-pointer: starting from the last element in array 1.

3. array2-pointer: starting from the last element in array 2.

Note that since the above pointers move forward (index is decreasing in each iteration), if array2-pointer reaches the beginning p2 < 0, we could return directly because array1 is already sorted correctly. On the other hand, we also need to check whether array1-pointer reaches the beginning by p1 >= 0.

🤖 Python3

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        #  (base case)
        if n == 0: return
        if m == 0: 
            for i in range(n): nums1[i] = nums2[i]
        
        # ==================================================
        #  Array + Three Pointers        (Start from End)  =
        # ==================================================        
        # time  : O(m+n)
        # space : O(1)
        
        p1, p2, placeP = m - 1, n - 1, m + n - 1
        while placeP >= 0:
            if p2 < 0: return
            
            if p1 >= 0 and nums1[p1] >= nums2[p2]:
                nums1[placeP] = nums1[p1]
                p1 -= 1
            else:
                nums1[placeP] = nums2[p2]
                p2 -= 1
                
            placeP -= 1

🤖 Java

class Solution {
    /**  
     * @time  : O(m+n)
     * @space : O(1)
     */

    public void merge(int[] nums1, int m, int[] nums2, int n) {
        /* base case */
        if(n == 0) return;
        if(m == 0) {
            for(int i=0 ; i<n ; i++) nums1[i] = nums2[i];
            return;
        }
        
        int p1 = m - 1, p2 = n - 1, placeP = m + n - 1;
        while(placeP >= 0) {
            if(p2 < 0) return;
            
            if(p1 >= 0 && nums1[p1] >= nums2[p2]) {
                nums1[placeP] = nums1[p1];
                p1--;
            } else {
                nums1[placeP] = nums2[p2];
                p2--;
            }
            
            placeP--;
        }
    }
}