88. Merge Sorted Arrays 🧩

Difficulty: Easy - Tags: Array, Two Pointers, Sorting

---

Description 📋

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.

Examples 🌟

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

• -10^9 <= nums1[i], nums2[j] <= 10^9

Solution 💡

Java

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n){
        int i = m - 1; // index of the last element in the sorted of nums1
        int j = n - 1; // index of the last element in nums2
        int k = m + n -1; // index of the last position in nums1 affter merging

        // loop until all elements from nums2 are merged into nums1
        while (j >= 0){
            // if nums1 has element left and the current element in nums1 is larger
            if (i >= 0 && nums1[i] > nums2[j]){
                nums1[k--] = nums1[i--]; // place nums1's element at position k
            } else {
                nums1[k--] = nums2[j--] // Otherwise, place nums2's element at position k
            }
        }
    }
}

You can find the full Solution.java file here.