There are two sorted arrays. First one is of size m+n containing only m elements. Another one is of size n and contains n elements. Merge these two arrays into the first array of size m+n such that the output is sorted.

Input: array with m+n elements (mPlusN[]).

NA => Value is not filled/available in array mPlusN[]. There should be n such array blocks.

Input: array with n elements (N[]).

Output: N[] merged into mPlusN[] (Modified mPlusN[])

**Algorithm:**

Let first array be mPlusN[] and other array be N[] 1) Move m elements of mPlusN[] to end. 2) Start from nth element of mPlusN[] and 0th element of N[] and merge them into mPlusN[].

Below is the implementation of the above algorithm :

## C++

// C++ program to Merge an array of // size n into another array of size m + n #include <bits/stdc++.h> using namespace std; /* Assuming -1 is filled for the places where element is not available */ #define NA -1 /* Function to move m elements at the end of array mPlusN[] */ void moveToEnd(int mPlusN[], int size) { int j = size - 1; for (int i = size - 1; i >= 0; i--) if (mPlusN[i] != NA) { mPlusN[j] = mPlusN[i]; j--; } } /* Merges array N[] of size n into array mPlusN[] of size m+n*/ int merge(int mPlusN[], int N[], int m, int n) { int i = n; /* Current index of i/p part of mPlusN[]*/ int j = 0; /* Current index of N[]*/ int k = 0; /* Current index of of output mPlusN[]*/ while (k < (m + n)) { /* Take an element from mPlusN[] if a) value of the picked element is smaller and we have not reached end of it b) We have reached end of N[] */ if ((i < (m + n) && mPlusN[i] <= N[j]) || (j == n)) { mPlusN[k] = mPlusN[i]; k++; i++; } else // Otherwise take element from N[] { mPlusN[k] = N[j]; k++; j++; } } } /* Utility that prints out an array on a line */ void printArray(int arr[], int size) { for (int i = 0; i < size; i++) cout << arr[i] << " "; cout << endl; } /* Driver function to test above functions */ int main() { /* Initialize arrays */ int mPlusN[] = {2, 8, NA, NA, NA, 13, NA, 15, 20}; int N[] = {5, 7, 9, 25}; int n = sizeof(N) / sizeof(N[0]); int m = sizeof(mPlusN) / sizeof(mPlusN[0]) - n; /*Move the m elements at the end of mPlusN*/ moveToEnd(mPlusN, m + n); /*Merge N[] into mPlusN[] */ merge(mPlusN, N, m, n); /* Print the resultant mPlusN */ printArray(mPlusN, m+n); return 0; }

## C

#include <stdio.h> /* Assuming -1 is filled for the places where element is not available */ #define NA -1 /* Function to move m elements at the end of array mPlusN[] */ void moveToEnd(int mPlusN[], int size) { int i = 0, j = size - 1; for (i = size-1; i >= 0; i--) if (mPlusN[i] != NA) { mPlusN[j] = mPlusN[i]; j--; } } /* Merges array N[] of size n into array mPlusN[] of size m+n*/ int merge(int mPlusN[], int N[], int m, int n) { int i = n; /* Current index of i/p part of mPlusN[]*/ int j = 0; /* Current index of N[]*/ int k = 0; /* Current index of of output mPlusN[]*/ while (k < (m+n)) { /* Take an element from mPlusN[] if a) value of the picked element is smaller and we have not reached end of it b) We have reached end of N[] */ if ((i < (m+n) && mPlusN[i] <= N[j]) || (j == n)) { mPlusN[k] = mPlusN[i]; k++; i++; } else // Otherwise take element from N[] { mPlusN[k] = N[j]; k++; j++; } } } /* Utility that prints out an array on a line */ void printArray(int arr[], int size) { int i; for (i=0; i < size; i++) printf("%d ", arr[i]); printf("\n"); } /* Driver function to test above functions */ int main() { /* Initialize arrays */ int mPlusN[] = {2, 8, NA, NA, NA, 13, NA, 15, 20}; int N[] = {5, 7, 9, 25}; int n = sizeof(N)/sizeof(N[0]); int m = sizeof(mPlusN)/sizeof(mPlusN[0]) - n; /*Move the m elements at the end of mPlusN*/ moveToEnd(mPlusN, m+n); /*Merge N[] into mPlusN[] */ merge(mPlusN, N, m, n); /* Print the resultant mPlusN */ printArray(mPlusN, m+n); return 0; }

## Java

class MergeArrays { /* Function to move m elements at the end of array mPlusN[] */ void moveToEnd(int mPlusN[], int size) { int i, j = size - 1; for (i = size - 1; i >= 0; i--) { if (mPlusN[i] != -1) { mPlusN[j] = mPlusN[i]; j--; } } } /* Merges array N[] of size n into array mPlusN[] of size m+n*/ void merge(int mPlusN[], int N[], int m, int n) { int i = n; /* Current index of i/p part of mPlusN[]*/ int j = 0; /* Current index of N[]*/ int k = 0; /* Current index of of output mPlusN[]*/ while (k < (m + n)) { /* Take an element from mPlusN[] if a) value of the picked element is smaller and we have not reached end of it b) We have reached end of N[] */ if ((i < (m + n) && mPlusN[i] <= N[j]) || (j == n)) { mPlusN[k] = mPlusN[i]; k++; i++; } else // Otherwise take element from N[] { mPlusN[k] = N[j]; k++; j++; } } } /* Utility that prints out an array on a line */ void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) System.out.print(arr[i] + " "); System.out.println(""); } public static void main(String[] args) { MergeArrays mergearray = new MergeArrays(); /* Initialize arrays */ int mPlusN[] = {2, 8, -1, -1, -1, 13, -1, 15, 20}; int N[] = {5, 7, 9, 25}; int n = N.length; int m = mPlusN.length - n; /*Move the m elements at the end of mPlusN*/ mergearray.moveToEnd(mPlusN, m + n); /*Merge N[] into mPlusN[] */ mergearray.merge(mPlusN, N, m, n); /* Print the resultant mPlusN */ mergearray.printArray(mPlusN, m + n); } } // This code has been contributed by Mayank Jaiswal

## Python3

# Assuming -1 is filled # for the places where element # is not available NA = -1 # Function to move m elements # at the end of array mPlusN[] def moveToEnd(mPlusN, size): i = 0 j = size - 1 for i in range(size-1, -1, -1): if (mPlusN[i] != NA): mPlusN[j] = mPlusN[i] j-=1 # Merges array N[] # of size n into array mPlusN[] # of size m+n def merge(mPlusN, N, m, n): i = n # Current index of i/p part of mPlusN[] j = 0 # Current index of N[] k = 0 # Current index of of output mPlusN[] while (k < (m+n)): # Take an element from mPlusN[] if # a) value of the picked # element is smaller and we have # not reached end of it # b) We have reached end of N[] */ if ((i < (m+n) and mPlusN[i] <= N[j]) or (j == n)): mPlusN[k] = mPlusN[i] k+=1 i+=1 else: # Otherwise take element from N[] mPlusN[k] = N[j] k+=1 j+=1 # Utility that prints # out an array on a line def printArray(arr,size): for i in range(size): print(arr[i], " ", end="") print() # Driver function to # test above functions # Initialize arrays mPlusN = [2, 8, NA, NA, NA, 13, NA, 15, 20] N = [5, 7, 9, 25] n = len(N) m = len(mPlusN) - n # Move the m elements # at the end of mPlusN moveToEnd(mPlusN, m+n) # Merge N[] into mPlusN[] merge(mPlusN, N, m, n) # Print the resultant mPlusN printArray(mPlusN, m+n) # This code is contributed # by Anant Agarwal.

Output:

2 5 7 8 9 13 15 20 25

**Time Complexity:** O(m+n)

Please write comment if you find any bug in the above program or a better way to solve the same problem.

**Practice Tags :**