88. Merge Sorted Array

Hint - Sort

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
    {
        // Loop through each element in nums2 and copy it to the end of nums1
        for ()
        {
            // Place the element from nums2 into the correct position in nums1
        }
        // Sort nums1 to merge the two arrays
    }
};

• Sort

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
    {
        for (int i = 0; i < n; ++i)
        {
            nums1[m + i] = nums2[i];
        }
        sort(nums1.begin(), nums1.end());
    }
};

• T: $O(N \cdot \log N)$
• S: $O(N)$

Hint - Two Pointers

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
    {
        // Initialize three pointers:
        // i - points to the last element of the initialized part of nums1
        // j - points to the last element of nums2
        // k - points to the last element of the merged array in nums1

        // Merge nums1 and nums2 from the end to the beginning
        while ()
        {
            // Compare the elements pointed by i and j
            // Place the larger element at position k in nums1
                // Copy element from nums1 and move pointers i and k
                // Copy element from nums2 and move pointers j and k
        }

        // If there are remaining elements in nums2, copy them
        // This is needed because the remaining elements of nums1 are already in place
    }
};

• Two Pointers

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
    {
        int i = m - 1, j = n - 1, k = m + n - 1;
        while (i >= 0 && j >= 0)
        {
            if (nums1[i] > nums2[j]) nums1[k--] = nums1[i--];
            else nums1[k--] = nums2[j--];
        }
        while (j >= 0) nums1[k--] = nums2[j--];
    }
};

• T: $O(N + M)$
• S: $O(M)$