# 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

 `// C++ program to merge three sorted arrays ` `// by merging two at a time. ` `#include ` `#include ` `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; ` `} `

## 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(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` `}; ` `        ``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)) `

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).

The C++ and Python 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<``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; ` `} `

## 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)) `

## 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 list = ``new` `ArrayList(); ` `        ``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 .

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