Merge an array of size n into another array of size m+n

Asked by Binod
Question:
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[]).
MergemPlusNNA => Value is not filled/available in array mPlusN[]. There should be n such array blocks.

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

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

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[].

Implementation:

C/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

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.

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