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++
// 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; } |
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++
// 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; } |
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.
CPP
// 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 |
1 2 3 4 5 6 7 8
Time Complexity: O( nlog(n) + mlog(m) )
Auxiliary Space: O(N)
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