88. Merge Sorted Array(Solution || Leetcode easy || Java)

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?

SOLUTION:

class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
//making a variable for counting index
int j = 0;

//adding elements of nums2 in nums1
for(int i = m; i < nums1.length; i++){
nums1[i] = nums2[j++];
}

//sorting them using insertion sort
for(int i = 0; i < nums1.length — 1; i++){
for(int k = i+1; k > 0; k — ){

//swapping the element if it is not sorted
if(nums1[k] < nums1[k-1]){
int temp = nums1[k];
nums1[k] = nums1[k-1];
nums1[k-1] = temp;
}else{
break;
}
}
}
}
}

Thank you for reading. If you have any queries then, please let me know in the comment section, I will surely be responsive toward it.