Merge 3 Sorted Arrays

• Difficulty Level : Easy
• Last Updated : 02 Jul, 2021

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

C++

 // C++ program to merge three sorted arrays// by merging two at a time.#include #include using namespace std; using Vector = vector; 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;}

Java

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

Python

 # 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))

Javascript



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
else if the minimum is from B add it
else if the minimum is from C add it

// 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

// Only B has exhausted
while i < m and k < o
Put minimum of A[i] and C[k] into D

// Only A has exhausted
while j < n and k < o
Put minimum of B[j] and C[k] into D

// 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:

C++

 // C++ program to merger three sorted arrays// by merging three simultaneously.#include #include using namespace std; using Vector = vector; 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;}

Python

 # 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))

Javascript


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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