Merge 3 Sorted Arrays

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

CPP

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// C++ program to merge three sorted arrays
// by merging two at a time.
#include <iostream>
#include <vector>
using namespace std;
  
using Vector = vector<int>;
  
void printVector(const Vector& a)
{
    cout << "[";
    for (auto e : a)
        cout << e << " ";
    cout << "]" << endl;
}
  
Vector mergeTwo(Vector& A, Vector& B)
{
    // Get sizes of vectors
    int m = A.size();
    int n = B.size();
  
    // Vector for storing Result
    Vector D;
    D.reserve(m + n);
  
    int i = 0, j = 0;
    while (i < m && j < n) {
  
        if (A[i] <= B[j]) 
            D.push_back(A[i++]);
        else
            D.push_back(B[j++]);
    }
  
    // B has exhausted
    while (i < m) 
        D.push_back(A[i++]);
      
    // A has exhausted
    while (j < n) 
        D.push_back(B[j++]);    
  
    return D;
}
  
int main()
{
    Vector A = { 1, 2, 3, 5 };
    Vector B = { 6, 7, 8, 9 };
    Vector C = { 10, 11, 12 };
  
    // First Merge A and B
    Vector T = mergeTwo(A, B);
  
    // Print Result after merging T with C
    printVector(mergeTwo(T, C));
    return 0;
}

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Java

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import java.util.*;
  
// Java program to merge three sorted arrays
// by merging two at a time.
class GFG 
{
  
static void printVector(Vector<Integer> a)
{
    System.out.print("[");
    for (Integer e : a)
        System.out.print(e + " ");
    System.out.println("]");
}
  
static Vector<Integer> mergeTwo(Vector<Integer> A, 
                                Vector<Integer> B)
{
    // Get sizes of vectors
    int m = A.size();
    int n = B.size();
  
    // Vector for storing Result
    Vector<Integer> D = new Vector<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(i++));
    }
  
    // 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 };
    Vector<Integer> A = new Vector<Integer>(Arrays.asList(a));
    Vector<Integer> B = new Vector<Integer>(Arrays.asList(b));
    Vector<Integer> C = new Vector<Integer>(Arrays.asList(c));
    // First Merge A and B
    Vector T = mergeTwo(A, B);
  
    // Print Result after merging T with C
    printVector(mergeTwo(T, C));
}
}
  
/* This code contributed by PrinciRaj1992 */

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Python

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# Python program to merge three sorted arrays
# by merging two at a time.
  
def merge_two(a, b):
    (m, n) = (len(a), len(b))
    i = j = 0
  
    # Destination Array
    d = []
  
    # Merge from a and b together
    while i < m and j < n:
        if a[i] <= b[j]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1
  
    # Merge from a if b has run out
    while i < m:
        d.append(a[i])
        i += 1
  
    # Merge from b if a has run out
    while j < n:
        d.append(b[j])
        j += 1
  
    return d
  
def merge(a, b, c):
    t = merge_two(a, b)
    return merge_two(t, c)
  
if __name__ == "__main__":
    A = [1, 2, 3, 5]
    B = [6, 7, 8, 9]
    C = [10, 11, 12]
    print(merge(A, B, C))

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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).

The C++ and Python Implementation of the algorithm is given below:

C++

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// C++ program to merger three sorted arrays
// by merging three simultaneously.
#include <iostream>
#include <vector>
using namespace std;
  
using Vector = vector<int>;
  
void printVector(const Vector& a)
{
    cout << "[";
    for (auto e : a) {
        cout << e << " ";
    }
    cout << "]" << endl;
}
  
Vector mergeThree(Vector& A, Vector& B,
                  Vector& C)
{
    int m, n, o, i, j, k;
    // Get Sizes of three vectors
    m = A.size();
    n = B.size();
    o = C.size();
  
    // Vector for storing output
    Vector D;
    D.reserve(m + n + o);
  
    i = j = k = 0;
  
    while (i < m && j < n && k < o) {
  
        // Get minimum of a, b, c
        int m = min(min(A[i], B[j]), C[k]);
  
        // Put m in D
        D.push_back(m);
  
        // Increment i, j, k
        if (m == A[i])
            i++;
        else if (m == B[j])
            j++;
        else
            k++;
    }
  
    // C has exhausted
    while (i < m && j < n) {
        if (A[i] <= B[j]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(B[j]);
            j++;
        }
    }
  
    // B has exhausted
    while (i < m && k < 0) {
        if (A[i] <= C[j]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(C[j]);
            k++;
        }
    }
  
    // A has exhausted
    while (j < n && k < 0) {
        if (B[j] <= C[k]) {
            D.push_back(B[j]);
            j++;
        }
        else {
            D.push_back(C[j]);
            k++;
        }
    }
  
    // A and B have exhausted
    while (k < o)
        D.push_back(C[k++]);
  
    // B and C have exhausted
    while (i < m)
        D.push_back(A[i++]);
  
    // A and C have exhausted
    while (j < n)
        D.push_back(B[j++]);
  
    return D;
}
  
int main()
{
    Vector A = { 1, 2, 41, 52, 84 };
    Vector B = { 1, 2, 41, 52, 67 };
    Vector C = { 1, 2, 41, 52, 67, 85 };
  
    // Print Result
    printVector(mergeThree(A, B, C));
    return 0;
}

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Python

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# Python program to merge three sorted arrays
# simultaneously. 
  
def merge_three(a, b, c):
    (m, n, o) = (len(a), len(b), len(c))
    i = j = k = 0
  
    # Destination array
    d = []
  
    while i < m and j < n and k < o:
  
        # Get Minimum element
        m = min(a[i], b[j], c[k])
  
        # Add m to D
        d.append(m)
  
        # Increment the source pointer which
        # gives m
        if a[i] == m:
            i += 1
        elif b[j] == m:
            j += 1
        elif c[k] == m:
            k += 1
  
    # Merge a and b in c has exhausted
    while i < m and j < n:
        if a[i] <= b[k]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1
  
    # Merge b and c if a has exhausted
    while j < n and k < o:
        if b[j] <= c[k]:
            d.append(b[j])
            j += 1
        else:
            d.append(c[k])
            k += 1
  
    # Merge a and c if b has exhausted
    while i < m and k < o:
        if a[i] <= c[k]:
            d.append(a[i])
            i += 1
        else:
            d.append(c[k])
            k += 1
  
    # Take elements from a if b and c
    # have exhausted
    while i < m:
        d.append(a[i])
        i += 1
  
    # Take elements from b if a and c 
    # have exhausted
    while j < n:
        d.append(b[j])
        j += 1
  
    # Take elements from c if a and 
    # b have exhausted
    while k < o:
        d.append(c[k])
        k += 1
  
    return d
  
if __name__ == "__main__":
    a = [1, 2, 41, 52, 84]
    b = [1, 2, 41, 52, 67]
    c = [1, 2, 41, 52, 67, 85]
  
    print(merge_three(a, b, c))

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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 .



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