Merge Sorted Array#

Problem statement#

[1]You’re given two integer arrays, nums1 and nums2, both sorted in non-decreasing order. Additionally, you have two integers, m and n, representing the number of elements in nums1 and nums2, respectively.

Your task is to merge the elements from nums2 into nums1 in a way that the resulting array is sorted in non-decreasing order.

However, the sorted array should not be returned as a separate result. Instead, the merged elements should be stored inside the nums1 array. Here’s the setup for that purpose:

  • nums1 has a total length of m + n, where the first m elements represent the elements that should be merged, and the last n elements are initialized to 0 and should be ignored.

  • The nums2 array has a length of n, representing the elements to be merged from nums2 into the final nums1 array.

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.

Follow up#

  • Can you come up with an algorithm that runs in O(m + n) time?

Solution 1: Store the result in a new container#

Code#

#include <iostream>
#include <vector>
using namespace std;
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
{
    vector<int> result;
    int i = 0;
    int j = 0;
    while (i < m || j < n) {
        if (j == n) { 
            // nums2 is done, only nums1 still runs
            result.push_back(nums1[i++]);
        } else if (i == m) {
            // nums1 is done, only nums2 still runs
            result.push_back(nums2[j++]);
        } else if (nums1[i] < nums2[j]) {
            result.push_back(nums1[i++]);
        } else {
            result.push_back(nums2[j++]);
        }
    }
    nums1.swap(result);
}
void printResult(const vector<int>& nums1) {
    cout << "[";
    for (auto& n : nums1) {
        cout << n << ",";
    }
    cout << "]\n";
}
int main() {
    vector<int> nums1 = {1,2,3,0,0,0};
    vector<int> nums2 = {2,5,6};
    merge(nums1, 3, nums2, 3);
    printResult(nums1);
    nums1 = {1};
    nums2 = {};
    merge(nums1, 1, nums2, 0);
    printResult(nums1);
    nums1 = {0};
    nums2 = {1};
    merge(nums1, 0, nums2, 1);
    printResult(nums1);
}

Output:
[1,2,2,3,5,6,]
[1,]
[1,]

This solution merges two sorted arrays nums1 and nums2 into nums1 while maintaining sorted order. It iterates through both arrays, comparing elements and adding them to a temporary result vector. After the merging is complete, it replaces the contents of nums1 with the merged result.

Complexity#

  • Runtime: O(m+n), where m = nums1.length and n = nums2.length.

  • Extra space: O(m+n).

Solution 2: Reassigning nums1 backward#

Code#

#include <iostream>
#include <vector>
using namespace std;
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) 
{    
    int k = m + n - 1;
    int i = m - 1;
    int j = n - 1;
    while (k >= 0) {
        if (j < 0) { 
            // nums2 is done
            nums1[k--] = nums1[i--];
        } else if (i < 0) { 
            // nums1 is done
            nums1[k--] = nums2[j--];
        } else if (nums1[i] > nums2[j]) {
            nums1[k--] = nums1[i--]; 
        } else {
            nums1[k--] = nums2[j--]; 
        }
    }
}
void printResult(const vector<int>& nums1) {
    cout << "[";
    for (auto& n : nums1) {
        cout << n << ",";
    }
    cout << "]\n";
}
int main() {
    vector<int> nums1 = {1,2,3,0,0,0};
    vector<int> nums2 = {2,5,6};
    merge(nums1, 3, nums2, 3);
    printResult(nums1);
    nums1 = {1};
    nums2 = {};
    merge(nums1, 1, nums2, 0);
    printResult(nums1);
    nums1 = {0};
    nums2 = {1};
    merge(nums1, 0, nums2, 1);
    printResult(nums1);
}

Output:
[1,2,2,3,5,6,]
[1,]
[1,]

Complexity#

  • Runtime: O(m+n), where m = nums1.length and n = nums2.length.

  • Extra space: O(1).

Conclusion#

Solution 2 efficiently merges two sorted arrays, nums1 and nums2, into nums1 while preserving the sorted order. It uses three pointers (k, i, and j) to perform the merge in reverse order, which helps avoid the need for additional space.

Exercise#

  • Squares of a Sorted Array[2].