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Java Program for Merge 3 Sorted Arrays

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  • Last Updated : 17 Jan, 2022

Given 3 arrays (A, B, C) which are sorted in ascending order, we are required to merge them together in ascending order and output the array D. 

Examples: 

Input : A = [1, 2, 3, 4, 5] 
        B = [2, 3, 4]
        C = [4, 5, 6, 7]
Output : D = [1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 7]

Input : A = [1, 2, 3, 5]
        B = [6, 7, 8, 9 ]
        C = [10, 11, 12]
Output: D = [1, 2, 3, 5, 6, 7, 8, 9. 10, 11, 12]

Method 1 (Two Arrays at a time) 
We have discussed at Merging 2 Sorted arrays . So we can first merge two arrays and then merge the resultant with the third array. Time Complexity for merging two arrays O(m+n). So for merging the third array, the time complexity will become O(m+n+o). Note that this is indeed the best time complexity that can be achieved for this problem. 
Space Complexity: Since we merge two arrays at a time, we need another array to store the result of the first merge. This raises the space complexity to O(m+n). Note that space required to hold the result of 3 arrays is ignored while calculating complexity. 

Algorithm 

function merge(A, B)
    Let m and n be the sizes of A and B
    Let D be the array to store result
   
    // Merge by taking smaller element from A and B
    while i < m and j < n
        if A[i] <= B[j]
            Add A[i] to D and increment i by 1
        else Add B[j] to D and increment j by 1

    // If array A has exhausted, put elements from B
    while j < n
        Add B[j] to D and increment j by 1
   
    // If array B has exhausted, put elements from A
    while i < n
        Add A[j] to D and increment i by 1
   
    Return D

function merge_three(A, B, C)
    T = merge(A, B)
    return merge(T, C)

The Implementations are given below

Java




import java.util.*;
// Java program to merge three sorted arrays
// by merging two at a time.
class GFG {
  
    static ArrayList<Integer> mergeTwo(List<Integer> A,
                                       List<Integer> B)
    {
        // Get sizes of vectors
        int m = A.size();
        int n = B.size();
  
        // ArrayList for storing Result
        ArrayList<Integer> D = new ArrayList<Integer>(m + n);
  
        int i = 0, j = 0;
        while (i < m && j < n) {
  
            if (A.get(i) <= B.get(j))
                D.add(A.get(i++));
            else
                D.add(B.get(j++));
        }
  
        // B has exhausted
        while (i < m)
            D.add(A.get(i++));
  
        // A has exhausted
        while (j < n)
            D.add(B.get(j++));
  
        return D;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        Integer[] a = { 1, 2, 3, 5 };
        Integer[] b = { 6, 7, 8, 9 };
        Integer[] c = { 10, 11, 12 };
        List<Integer> A = Arrays.asList(a);
        List<Integer> B = Arrays.asList(b);
        List<Integer> C = Arrays.asList(c);
  
        // First Merge A and B
        ArrayList<Integer> T = mergeTwo(A, B);
  
        // Print Result after merging T with C
        System.out.println(mergeTwo(T, C));
    }
}
  
/* This code contributed by PrinciRaj1992 */


Output: 
 

[1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12]

Method 2 (Three arrays at a time) 
The Space complexity of method 1 can be improved we merge the three arrays together. 
 

function merge-three(A, B, C)
    Let m, n, o be size of A, B, and C
    Let D be the array to store the result
    
    // Merge three arrays at the same time
    while i < m and j < n and k < o
        Get minimum of A[i], B[j], C[i]
        if the minimum is from A, add it to 
           D and advance i
        else if the minimum is from B add it 
                to D and advance j
        else if the minimum is from C add it 
                to D and advance k
    
   // After above step at least 1 array has 
   // exhausted. Only C has exhausted
   while i < m and j < n
       put minimum of A[i] and B[j] into D
       Advance i if minimum is from A else advance j 
   
   // Only B has exhausted
   while i < m and k < o
       Put minimum of A[i] and C[k] into D
       Advance i if minimum is from A else advance k
 
   // Only A has exhausted
   while j < n and k < o
       Put minimum of B[j] and C[k] into D
       Advance j if minimum is from B else advance k

   // After above steps at least 2 arrays have 
   // exhausted
   if A and B have exhausted take elements from C
   if B and C have exhausted take elements from A
   if A and C have exhausted take elements from B
   
   return D

Complexity: The Time Complexity is O(m+n+o) since we process each element from the three arrays once. We only need one array to store the result of merging and so ignoring this array, the space complexity is O(1).

 Implementation of the algorithm is given below:

Java




import java.util.*;
import java.io.*;
import java.lang.*;
class Sorting {
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 41, 52, 84 };
        int B[] = { 1, 2, 41, 52, 67 };
        int C[] = { 1, 2, 41, 52, 67, 85 };
  
        // call the function to sort and print the sorted numbers
        merge3sorted(A, B, C);
    }
  
    // Function to merge three sorted arrays
    // A[], B[], C[]: input arrays
    static void merge3sorted(int A[], int B[], int C[])
    {
        // creating an empty list to store sorted numbers
        ArrayList<Integer> list = new ArrayList<Integer>();
        int i = 0, j = 0, k = 0;
  
        // using merge concept and trying to find
        // smallest of three while all three arrays
        // contains at least one element
        while (i < A.length && j < B.length && k < C.length) {
            int a = A[i];
            int b = B[j];
            int c = C[k];
            if (a <= b && a <= c) {
                list.add(a);
                i++;
            }
            else if (b <= a && b <= c) {
                list.add(b);
                j++;
            }
            else {
                list.add(c);
                k++;
            }
        }
        // next three while loop is to sort two
        // of arrays if one of the three gets exhausted
        while (i < A.length && j < B.length) {
            if (A[i] < B[j]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(B[j]);
                j++;
            }
        }
        while (j < B.length && k < C.length) {
            if (B[j] < C[k]) {
                list.add(B[j]);
                j++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }
        while (i < A.length && k < C.length) {
            if (A[i] < C[k]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }
  
        // if one of the array are left then
        // simply appending them as there will
        // be only largest element left
        while (i < A.length) {
            list.add(A[i]);
            i++;
        }
        while (j < B.length) {
            list.add(B[j]);
            j++;
        }
        while (k < C.length) {
            list.add(C[k]);
            k++;
        }
        
        // finally print the list
        for (Integer x : list)
            System.out.print(x + " ");
    } // merge3sorted closing braces
}


Output

[1 1 1 2 2 2 41 41 41 52 52 52 67 67 84 85 ]

Note: While it is relatively easy to implement direct procedures to merge two or three arrays, the process becomes cumbersome if we want to merge 4 or more arrays. In such cases, we should follow the procedure shown in Merge K Sorted Arrays .
 

Please refer complete article on Merge 3 Sorted Arrays for more details!


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