Find relative complement of two sorted arrays
Last Updated :
02 Dec, 2023
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 15
The elements 6, 12 and 15 are present
in arr[], but not present in arr2[]
Input : arr1[] = {10, 20, 36, 59}
arr2[] = {5, 10, 15, 59}
Output : 20 36
- Take two pointers i and j which traverse through arr1 and arr2 respectively.
- If arr1[i] element is smaller than arr2[j] element print this element and increment i.
- If arr1 element is greater than arr2[j] element then increment j.
- otherwise increment i and j.
Implementation:
C++
// CPP program to find all those
// elements of arr1[] that are not
// present in arr2[]
#include <iostream>
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[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
relativeComplement(arr1, arr2, n, m);
return 0;
}
// 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.
# 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# 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
<script>
// JavaScript program to find all those
// elements of arr1[] that are not
// present in arr2[]
function relativeComplement(arr1, arr2,
n, m)
{
let 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])
{
document.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)
document.write(arr1[i] + " ");
}
// Driver Code
let arr1 = [3, 6, 10, 12, 15];
let arr2 = [1, 3, 5, 10, 16];
let n = arr1.length;
let m = arr2.length;
relativeComplement(arr1, arr2, n, m);
// This code is contributed by splevel62.
</script>
<?php
// PHP program to find all those
// elements of arr1[] that are not
// present in arr2[]
function relativeComplement($arr1, $arr2,
$n, $m)
{
$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])
{
echo $arr1[$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)
echo $arr1[$i] , " ";
}
// Driver code
{
$arr1 = array(3, 6, 10, 12, 15);
$arr2 = array(1, 3, 5, 10, 16);
$n = sizeof($arr1) / sizeof($arr1[0]);
$m = sizeof($arr2) / sizeof($arr2[0]);
relativeComplement($arr1, $arr2, $n, $m);
return 0;
}
// This code is contributed by nitin mittal
?>
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 <iostream>
#include <unordered_set>
#include <vector>
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[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
relativeComplement(arr1, arr2, n, m);
return 0;
}
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<Integer> set = new HashSet<>();
// Initializing our answer ArrayList
ArrayList<Integer> 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);
}
}
def relative_complement(arr1, arr2):
# initializing our set
s = set()
# initializing our ans list
ans = []
# storing elements of the second array in the set
for num in arr2:
s.add(num)
# traversing the first array
for num in arr1:
# if the element is not found in the set, add it to the ans list
if num not in s:
ans.append(num)
# printing the answer list
for x in ans:
print(x, end=" ")
if __name__ == "__main__":
arr1 = [3, 6, 10, 12, 15]
arr2 = [1, 3, 5, 10, 16]
relative_complement(arr1, arr2)
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);
}
}
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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