Merge_Sorted_Array.cpp

PROBLEM STATEMENT :- 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.
  
  
  SOLUTION 1:- (8 ms )
    class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
    {
        priority_queue<int> pq;
        for(int i=0;i<m;i++)
            pq.push(nums1[i]);
        for(int i=0;i<n;i++)
            pq.push(nums2[i]);
        for(int i=m+n-1;i>=0;i--)
        {
            nums1[i]=pq.top();
            pq.pop();
        }
    }
};

SOLUTION 2 :- (0 ms)
 //More efficient in terms of time and space
 class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
    {
        int tail = nums1.size() - 1;
        --m; --n;
        while(tail >= 0)
        {
        while(m >= 0 && (n < 0 || nums1[m] >= nums2[n]))
        {
            nums1[tail--] = nums1[m--];
        }
        while(n >= 0 && (m < 0 || nums2[n] > nums1[m]))
        {
            nums1[tail--] = nums2[n--];
        }
        }
    }
};