C++ Program to Merge Two Sorted Arrays
Given two sorted arrays, the task is to merge them in a sorted manner.
Examples:
Input: arr1[] = { 1, 3, 4, 5}, arr2[] = {2, 4, 6, 8}
Output: arr3[] = {1, 2, 3, 4, 4, 5, 6, 8}Input: arr1[] = { 5, 8, 9}, arr2[] = {4, 7, 8}
Output: arr3[] = {4, 5, 7, 8, 8, 9}
Naive Approach:
It is the brute force method to do the same. Take all the elements of arr1 and arr2 in arr3. Then simply sort the arr3.
The implementation of above approach is:
// C++ program to merge two sorted arrays/
#include<bits/stdc++.h>
using namespace std;
void mergeArrays(int arr1[], int arr2[], int n1,
int n2, int arr3[])
{
int i = 0, j = 0, k = 0;
// traverse the arr1 and insert its element in arr3
while(i < n1){
arr3[k++] = arr1[i++];
}
// now traverse arr2 and insert in arr3
while(j < n2){
arr3[k++] = arr2[j++];
}
// sort the whole array arr3
sort(arr3, arr3+n1+n2);
}
// Driver code
int main()
{
int arr1[] = {1, 3, 5, 7};
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int arr2[] = {2, 4, 6, 8};
int n2 = sizeof(arr2) / sizeof(arr2[0]);
int arr3[n1+n2];
mergeArrays(arr1, arr2, n1, n2, arr3);
cout << "Array after merging" <<endl;
for (int i=0; i < n1+n2; i++)
cout << arr3[i] << " ";
return 0;
}
Output
Array after merging 1 2 3 4 5 6 7 8
Time Complexity : O((m+n) log(m+n)) , the whole size of arr3 is m+n
Auxiliary Space: O(1), No extra space is used
Method 2 (O(n1 * n2) Time and O(n1+n2) Extra Space)
- Create an array arr3[] of size n1 + n2.
- Copy all n1 elements of arr1[] to arr3[]
- Traverse arr2[] and one by one insert elements (like insertion sort) of arr3[] to arr1[]. This step take O(n1 * n2) time.
We have discussed implementation of above method in Merge two sorted arrays with O(1) extra space
Method 3 (O(n1 + n2) Time and O(n1 + n2) Extra Space)
The idea is to use Merge function of Merge sort.
- Create an array arr3[] of size n1 + n2.
- Simultaneously traverse arr1[] and arr2[].
- Pick smaller of current elements in arr1[] and arr2[], copy this smaller element to next position in arr3[] and move ahead in arr3[] and the array whose element is picked.
- If there are remaining elements in arr1[] or arr2[], copy them also in arr3[].
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
// C++ program to merge two sorted arrays/
#include<iostream>
using namespace std;
// Merge arr1[0..n1-1] and arr2[0..n2-1] into
// arr3[0..n1+n2-1]
void mergeArrays(int arr1[], int arr2[], int n1,
int n2, int arr3[])
{
int i = 0, j = 0, k = 0;
// Traverse both array
while (i<n1 && j <n2)
{
// Check if current element of first
// array is smaller than current element
// of second array. If yes, store first
// array element and increment first array
// index. Otherwise do same with second array
if (arr1[i] < arr2[j])
arr3[k++] = arr1[i++];
else
arr3[k++] = arr2[j++];
}
// Store remaining elements of first array
while (i < n1)
arr3[k++] = arr1[i++];
// Store remaining elements of second array
while (j < n2)
arr3[k++] = arr2[j++];
}
// Driver code
int main()
{
int arr1[] = {1, 3, 5, 7};
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int arr2[] = {2, 4, 6, 8};
int n2 = sizeof(arr2) / sizeof(arr2[0]);
int arr3[n1+n2];
mergeArrays(arr1, arr2, n1, n2, arr3);
cout << "Array after merging" <<endl;
for (int i=0; i < n1+n2; i++)
cout << arr3[i] << " ";
return 0;
}
Output
Array after merging 1 2 3 4 5 6 7 8
Output:
Array after merging 1 2 3 4 5 6 7 8
Time Complexity : O(n1 + n2)
Auxiliary Space : O(n1 + n2)
Method 4: Using Maps (O(nlog(n) + mlog(m)) Time and O(N) Extra Space)
- Insert elements of both arrays in a map as keys.
- Print the keys of the map.
Below is the implementation of above approach.
// C++ program to merge two sorted arrays
//using maps
#include<bits/stdc++.h>
using namespace std;
// Function to merge arrays
void mergeArrays(int a[], int b[], int n, int m)
{
// Declaring a map.
// using map as a inbuilt tool
// to store elements in sorted order.
map<int, int> mp;
// Inserting values to a map.
for(int i = 0; i < n; i++)mp[a[i]]++;
for(int i = 0;i < m;i++)mp[b[i]]++;
// Printing keys of the map.
for(auto j: mp)
{
for(int i=0; i<j.second;i++)cout<<j.first<<" ";
}
}
// Driver Code
int main()
{
int a[] = {1, 3, 5, 7}, b[] = {2, 4, 6, 8};
int size = sizeof(a)/sizeof(int);
int size1 = sizeof(b)/sizeof(int);
// Function call
mergeArrays(a, b, size, size1);
return 0;
}
//This code is contributed by yashbeersingh42
Output
1 2 3 4 5 6 7 8
Time Complexity: O( nlog(n) + mlog(m) )
Auxiliary Space: O(N)
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