Merge 3 Sorted Arrays - GeeksforGeeks

26 Apr, 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.
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)
Below is the implementation of the above approach:
C++
Java
Python
Javascript
// 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;} // Driver Codeint 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;}
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 */
# 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))
<script>// Javascript program to merge three sorted arrays// by merging two at a time.    function mergeTwo(A, B){    // Get sizes of vectors    let m = A.length;    let n = B.length;     // Vector for storing Result    let D = [];     let i = 0, j = 0;    while (i < m && j < n) {         if (A[i] <= B[j])            D.push(A[i++]);        else            D.push(B[j++]);    }     // B has exhausted    while (i < m)        D.push(A[i++]);     // A has exhausted    while (j < n)        D.push(B[j++]);     return D;} // Driver Code     let A = [ 1, 2, 3, 5 ];    let B = [ 6, 7, 8, 9 ];    let C = [ 10, 11, 12 ];     // First Merge A and B    let T = mergeTwo(A, B);     // Print Result after merging T with C    document.write(mergeTwo(T, C));</script>
[1 2 3 5 6 7 8 9 10 11 12 ]

Time Complexity: O(m+n+o) where m, n, o are the lengths of the 1st, 2nd, 3rd Array.Auxiliary Space: O(m + n + o).
Method 2 (Three arrays at a time) The Space complexity of method 1 can be improved if 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
Time Complexity: O(m+n+o) where m, n, o are the lengths of the 1st, 2nd, and 3rd array.Auxiliary Space: O(1).
Below is the implementation of the above approach:
C++
Java
Python
Javascript
// 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 < o) {        if (A[i] <= C[k]) {            D.push_back(A[i]);            i++;        }        else {            D.push_back(C[k]);            k++;        }    }     // A has exhausted    while (j < n && k < o) {        if (B[j] <= C[k]) {            D.push_back(B[j]);            j++;        }        else {            D.push_back(C[k]);            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;} // Driver Codeint 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;}
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}
# 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))
<script>// Javascript program to merger three sorted arrays// by merging three simultaneously.function printVector(a) {    document.write("[");    for (let e of a) {        document.write(e + " ");    }    document.write("]" + "<br>");} function mergeThree(A, B, C){    let m, n, o, i, j, k;         // Get Sizes of three vectors    m = A.length;    n = B.length;    o = C.length;     // Vector for storing output    let D = [];    i = j = k = 0;    while (i < m && j < n && k < o)    {         // Get minimum of a, b, c        let m = Math.min(Math.min(A[i], B[j]), C[k]);         // Put m in D        D.push(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(A[i]);            i++;        }        else {            D.push(B[j]);            j++;        }    }     // B has exhausted    while (i < m && k < o) {        if (A[i] <= C[k]) {            D.push(A[i]);            i++;        }        else {            D.push(C[k]);            k++;        }    }     // A has exhausted    while (j < n && k < o) {        if (B[j] <= C[k]) {            D.push(B[j]);            j++;        }        else {            D.push(C[k]);            k++;        }    }     // A and B have exhausted    while (k < o)        D.push(C[k++]);     // B and C have exhausted    while (i < m)        D.push(A[i++]);     // A and C have exhausted    while (j < n)        D.push(B[j++]);     return D;} // Driver Code let A = [1, 2, 41, 52, 84];let B = [1, 2, 41, 52, 67];let C = [1, 2, 41, 52, 67, 85]; // Print ResultprintVector(mergeThree(A, B, C)); // This code is contributed by gfgking.</script>
[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 .
Another Approach (Without caring about the exhausting array):The code written above can be shortened by the below code. Here we do not need to write code if any array gets exhausted.
Below is the implementation of the above approach:
Java
/*package whatever //do not write package name here */ import java.io.*;import java.lang.*;import java.util.*; // Java program to merge three sorted arrays// by merging two at a time. // This code is contributed by Animesh Nag class Solution {    // A[], B[], C[]: input arrays    // Function to merge three sorted lists into a single    // list.    static ArrayList<Integer> merge3sorted(int A[], int B[],                                           int C[])    {        ArrayList<Integer> ans = new ArrayList<Integer>();        int l1 = A.length;        int l2 = B.length;        int l3 = C.length;        int i = 0, j = 0, k = 0;        while (i < l1 || j < l2 || k < l3) {            // Assigning a, b, c with max values so that if            // any value is not present then also we can sort            // the array.            int a = Integer.MAX_VALUE,                b = Integer.MAX_VALUE,                c = Integer.MAX_VALUE;             // a, b, c variables are assigned only if the            // value exist in the array.            if (i < l1)                a = A[i];            if (j < l2)                b = B[j];            if (k < l3)                c = C[k];             // Cheking if 'a' is the minimum            if (a <= b && a <= c) {                ans.add(a);                i++;            }            // Cheking if 'b' is the minimum            else if (b <= a && b <= c) {                ans.add(b);                j++;            }            // Cheking if 'c' is the minimum            else {                if (c <= a && c <= b) {                    ans.add(c);                    k++;                }            }        }        return ans;    }} class GFG {    // Driver program to test above functions    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 };         Solution sol = new Solution();        ArrayList<Integer> final_ans            = sol.merge3sorted(A, B, C);        printeSorted(final_ans);    }    // A utility function to print array list    static void printeSorted(ArrayList<Integer> list)    {        for (Integer x : list)            System.out.print(x + " ");    }}
1 1 1 2 2 2 41 41 41 52 52 52 67 67 84 85 
Time Complexity: O(m+n+o) where m, n, o are the lengths of the 1st, 2nd, and 3rd array.Space Complexity: O(1)
princiraj1992
Devarshi_Singh
suprinilomous
sharmaadityansh
_saurabh_jaiswal
gfgking
AnimeshNag
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