# Find relative complement of two sorted arrays

Given two sorted arrays arr1 and arr2 of size m and n respectively. We need to find relative complement of two array i.e, arr1 – arr2 which means that we need to find all those elements which are present in arr1 but not in arr2.

Examples:

`Input : arr1[] = {3, 6, 10, 12, 15}        arr2[] = {1, 3, 5, 10, 16}Output : 6 12 15The elements 6, 12 and 15 are presentin arr[], but not present in arr2[]         Input : arr1[] = {10, 20, 36, 59}        arr2[] = {5, 10, 15, 59}Output : 20 36`
1. Take two pointers i and j which traverse through arr1 and arr2 respectively.
2. If arr1[i] element is smaller than arr2[j] element print this element and increment i.
3. If arr1 element is greater than arr2[j] element then increment j.
4. otherwise increment i and j.

Implementation:

## C++

 `// CPP program to find all those ` `// elements of arr1[] that are not` `// present in arr2[]` `#include ` `using` `namespace` `std;`   `void` `relativeComplement(``int` `arr1[], ``int` `arr2[],` `                               ``int` `n, ``int` `m) {`   `  ``int` `i = 0, j = 0;` `  ``while` `(i < n && j < m) {`   `    ``// If current element in arr2[] is` `    ``// greater, then arr1[i] can't be ` `    ``// present in arr2[j..m-1]` `    ``if` `(arr1[i] < arr2[j]) {` `      ``cout << arr1[i] << ``" "``;` `      ``i++;`   `    ``// Skipping smaller elements of` `    ``// arr2[]` `    ``} ``else` `if` `(arr1[i] > arr2[j]) {` `      ``j++;`   `    ``// Equal elements found (skipping` `    ``// in both arrays)` `    ``} ``else` `if` `(arr1[i] == arr2[j]) {` `      ``i++;` `      ``j++;` `    ``}` `  ``}`   `  ``// Printing remaining elements of` `  ``// arr1[]` `  ``while` `(i < n) ` `    ``cout << arr1[i] << ``" "``;  ` `}`   `// Driver code` `int` `main() {` `  ``int` `arr1[] = {3, 6, 10, 12, 15};` `  ``int` `arr2[] = {1, 3, 5, 10, 16};` `  ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1);` `  ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2);` `  ``relativeComplement(arr1, arr2, n, m);` `  ``return` `0;` `}`

## Java

 `// Java program to find all those ` `// elements of arr1[] that are not` `// present in arr2[]`   `class` `GFG` `{` `    ``static` `void` `relativeComplement(``int` `arr1[], ``int` `arr2[],` `                                             ``int` `n, ``int` `m) ` `    ``{` `    `  `        ``int` `i = ``0``, j = ``0``;` `        ``while` `(i < n && j < m) ` `        ``{` `        `  `            ``// If current element in arr2[] is` `            ``// greater, then arr1[i] can't be ` `            ``// present in arr2[j..m-1]` `            ``if` `(arr1[i] < arr2[j]) ` `            ``{` `                ``System.out.print(arr1[i] + ``" "``);` `                ``i++;` `        `  `            ``// Skipping smaller elements of` `            ``// arr2[]` `            ``} ``else` `if` `(arr1[i] > arr2[j]) ` `            ``{` `                ``j++;` `        `  `            ``// Equal elements found (skipping` `            ``// in both arrays)` `            ``} ` `            ``else` `if` `(arr1[i] == arr2[j]) ` `            ``{` `                ``i++;` `                ``j++;` `            ``}` `        ``}` `        `  `        ``// Printing remaining elements of` `        ``// arr1[]` `        ``while` `(i < n){ ` `            ``System.out.print(arr1[i] + ``" "``);` `            ``i++;` `        ``}    ` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `arr1[] = {``3``, ``6``, ``10``, ``12``, ``15``};` `        ``int` `arr2[] = {``1``, ``3``, ``5``, ``10``, ``16``};` `        ``int` `n = arr1.length;` `        ``int` `m = arr2.length;` `        ``relativeComplement(arr1, arr2, n, m);` `     ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to find all those ` `# elements of arr1[] that are not` `# present in arr2[]`   `def` `relativeComplement(arr1, arr2, n, m):` ` `  `    ``i ``=` `0` `    ``j ``=` `0` `    ``while` `(i < n ``and` `j < m):` ` `  `        ``# If current element in arr2[] is` `        ``# greater, then arr1[i] can't be ` `        ``# present in arr2[j..m-1]` `        ``if` `(arr1[i] < arr2[j]):` `            ``print``(arr1[i] , ``" "``, end``=``"")` `            ``i ``+``=` `1` ` `  `            ``# Skipping smaller elements of` `            ``# arr2[]` `        ``elif` `(arr1[i] > arr2[j]):` `            ``j ``+``=` `1` ` `  `            ``# Equal elements found (skipping` `            ``# in both arrays)` `        ``elif` `(arr1[i] ``=``=` `arr2[j]):` `            ``i ``+``=` `1` `            ``j ``+``=` `1` `    `  `    ``# Printing remaining elements of` `    ``# arr1[]` `    ``while` `(i < n): ` `        ``print``(arr1[i] , ``" "``, end``=``"")` ` `  `# Driver code` `arr1``=` `[``3``, ``6``, ``10``, ``12``, ``15``]` `arr2 ``=` `[``1``, ``3``, ``5``, ``10``, ``16``]` `n ``=` `len``(arr1)` `m ``=` `len``(arr2)` `relativeComplement(arr1, arr2, n, m)`   `# This code is contributed` `# by Anant Agarwal.`

## C#

 `// C# program to find all those ` `// elements of arr1[] that are not` `// present in arr2[]` `using` `System;`   `namespace` `Complement` `{` `    ``public` `class` `GFG` `    ``{     ` `                `  `        ``static` `void` `relativeComplement(``int` `[]arr1, ``int` `[]arr2,` `                                                   ``int` `n, ``int` `m)` `        ``{` `    `  `        ``int` `i = 0, j = 0;` `        ``while` `(i < n && j < m) ` `        ``{` `        `  `            ``// If current element in arr2[] is` `            ``// greater, then arr1[i] can't be ` `            ``// present in arr2[j..m-1]` `            ``if` `(arr1[i] < arr2[j]) ` `            ``{` `                ``Console.Write(arr1[i] + ``" "``);` `                ``i++;` `        `  `            ``// Skipping smaller elements of` `            ``// arr2[]` `            ``} ``else` `if` `(arr1[i] > arr2[j]) ` `            ``{` `                ``j++;` `        `  `            ``// Equal elements found (skipping` `            ``// in both arrays)` `            ``} ` `            ``else` `if` `(arr1[i] == arr2[j]) ` `            ``{` `                ``i++;` `                ``j++;` `            ``}` `        ``}` `        `  `        ``// Printing remaining elements of` `        ``// arr1[]` `        ``while` `(i < n) ` `            ``Console.Write(arr1[i] + ``" "``); ` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]arr1 = {3, 6, 10, 12, 15};` `        ``int` `[]arr2 = {1, 3, 5, 10, 16};` `        ``int` `n = arr1.Length;` `        ``int` `m = arr2.Length;` `        ``relativeComplement(arr1,arr2, n, m);` `    ``}` `    ``}` `}`   `// This code is contributed by Sam007`

## Javascript

 ``

## PHP

 ` ``\$arr2``[``\$j``]) ` `        ``{` `            ``\$j``++;` `        `  `            ``// Equal elements found (skipping` `            ``// in both arrays)` `        ``} ` `        ``else` `if` `(``\$arr1``[``\$i``] == ``\$arr2``[``\$j``])` `        ``{` `            ``\$i``++;` `            ``\$j``++;` `        ``}` `    ``}`   `    ``// Printing remaining elements of` `    ``// arr1[]` `    ``while` `(``\$i` `< ``\$n``) ` `        ``echo` `\$arr1``[``\$i``] , ``" "``; ` `}`   `// Driver code` `{` `    ``\$arr1` `= ``array``(3, 6, 10, 12, 15);` `    ``\$arr2` `= ``array``(1, 3, 5, 10, 16);` `    ``\$n` `= sizeof(``\$arr1``) / sizeof(``\$arr1``);` `    ``\$m` `= sizeof(``\$arr2``) / sizeof(``\$arr2``);` `    ``relativeComplement(``\$arr1``, ``\$arr2``, ``\$n``, ``\$m``);` `    ``return` `0;` `}`   `// This code is contributed by nitin mittal` `?>`

Output

```6 12 15

```

Time Complexity : O(m + n)
Auxiliary Space: O(1)

Another Approach:

Using an unordered_set we can do the same by following these steps.

• store all the elements of the second array in the set.
• Now traverse the second array and for each element check whether it is present in the set or not
• If the element is not present in the map we add it to our answer array.

Below is the implementation for the same

## C++

 `#include ` `#include ` `#include ` `using` `namespace` `std;`   `void` `relativeComplement(``int` `arr1[], ``int` `arr2[], ``int` `n,` `                        ``int` `m)` `{` `    ``// initializing our set` `    ``unordered_set<``int``> s;` `    ``// initialixing our ans vector` `    ``vector<``int``> ans;` `    ``// storing elements of the second array in the set` `    ``for` `(``int` `i = 0; i < m; i++)` `        ``s.insert(arr2[i]);` `    ``// traversing the second array` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// if the element is not found in the set add it to` `        ``// the ans vector` `        ``if` `(s.find(arr1[i]) == s.end())` `            ``ans.push_back(arr1[i]);` `    ``}` `    ``// printing the answer vector.` `    ``for` `(``auto` `x : ans)` `        ``cout << x << ``" "``;` `}`   `int` `main()` `{` `    ``int` `arr1[] = { 3, 6, 10, 12, 15 };` `    ``int` `arr2[] = { 1, 3, 5, 10, 16 };` `    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1);` `    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``relativeComplement(arr1, arr2, n, m);` `    ``return` `0;` `}`

## Java

 `import` `java.io.*;` `import` `java.util.*;`   `public` `class` `GFG {` `    ``public` `static` `void` `relativeComplement(``int``[] arr1, ``int``[] arr2, ``int` `n, ``int` `m) {` `        ``// Initializing our set` `        ``HashSet set = ``new` `HashSet<>();` `        ``// Initializing our answer ArrayList` `        ``ArrayList ans = ``new` `ArrayList<>();` `        ``// Storing elements of the second array in the set` `        ``for` `(``int` `i = ``0``; i < m; i++) {` `            ``set.add(arr2[i]);` `        ``}` `        ``// Traversing the first array` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// If the element is not found in the set, add it to the answer ArrayList` `            ``if` `(!set.contains(arr1[i])) {` `                ``ans.add(arr1[i]);` `            ``}` `        ``}` `        ``// Printing the answer ArrayList.` `        ``for` `(``int` `x : ans) {` `            ``System.out.print(x + ``" "``);` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr1 = { ``3``, ``6``, ``10``, ``12``, ``15` `};` `        ``int``[] arr2 = { ``1``, ``3``, ``5``, ``10``, ``16` `};` `        ``int` `n = arr1.length;` `        ``int` `m = arr2.length;` `        ``relativeComplement(arr1, arr2, n, m);` `    ``}` `}`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `Program` `{` `    ``// Function to find the relative complement of two integer arrays` `    ``static` `void` `RelativeComplement(``int``[] arr1, ``int``[] arr2)` `    ``{` `        ``// Initializing a HashSet to store elements of the second array` `        ``HashSet<``int``> ``set` `= ``new` `HashSet<``int``>();` `        `  `        ``// Initializing a List to store the result` `        ``List<``int``> result = ``new` `List<``int``>();` `        `  `        ``// Storing elements of the second array in the HashSet` `        ``foreach` `(``int` `num ``in` `arr2)` `        ``{` `            ``set``.Add(num);` `        ``}` `        `  `        ``// Traversing the first array` `        ``foreach` `(``int` `num ``in` `arr1)` `        ``{` `            ``// If the element is not found in the HashSet, add it to the result list` `            ``if` `(!``set``.Contains(num))` `            ``{` `                ``result.Add(num);` `            ``}` `        ``}` `        `  `        ``// Printing the result` `        ``foreach` `(``int` `num ``in` `result)` `        ``{` `            ``Console.Write(num + ``" "``);` `        ``}` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``int``[] arr1 = { 3, 6, 10, 12, 15 };` `        ``int``[] arr2 = { 1, 3, 5, 10, 16 };` `        `  `        ``// Call the function to find the relative complement` `        ``RelativeComplement(arr1, arr2);` `    ``}` `}`

## Javascript

 `function` `relativeComplement(arr1, arr2) {` `    ``// initializing our set` `    ``let s = ``new` `Set();` `    ``// initializing our ans array` `    ``let ans = [];`   `    ``// storing elements of the second array in the set` `    ``for` `(let i = 0; i < arr2.length; i++) {` `        ``s.add(arr2[i]);` `    ``}`   `    ``// traversing the first array` `    ``for` `(let i = 0; i < arr1.length; i++) {` `        ``// if the element is not found in the set, add it to the ans array` `        ``if` `(!s.has(arr1[i])) {` `            ``ans.push(arr1[i]);` `        ``}` `    ``}`   `    ``// printing the answer array.` `    ``console.log(ans.join(``' '``));` `}`   `// Driver Code` `let arr1 = [3, 6, 10, 12, 15];` `let arr2 = [1, 3, 5, 10, 16];`   `relativeComplement(arr1, arr2);`

Output:

`6 12 15 `

Time Complexity: O(G) where G is the size of the bigger array.
Auxiliary Space: O(m), we are storing elements of the second array in the set.

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