Common Elements in Two Arrays
Last Updated :
17 Oct, 2024
Given two arrays, the task is find common elements of the two arrays. The input arrays may contain duplicates and our job is to find all common elements We are allowed to return elements in any order.
Input : a[] = {1, 2, 1, 3, 1}, b[] = {3, 1, 3, 4, 1}
Output : {1, 3, 1}
1 appears two times in both the input arrays.
Input : a[] = {1, 1, 1}, b[] = {1, 1, 1, 1, 1}
Output : {1, 1, 1}
1 appears three times in both the input arrays.
Input : a[] = {1, 2, 3}, b[] = {4, 5, 6}
Output : {}
Naive Approach - O(n*m) Time and O(1) Space
- Initialize an empty array res[] to store result
- Traverse through a[] and for every element check if it is in b[]. If we find in b[] also, then add it to the result. Also, mark the current element of b[] as visited using a separate boolean array. We need visited array because we search the whole b[] for every a[i].
- Return res[]
C++
#include <iostream>
#include <vector>
using namespace std;
vector<int> findCommon(vector<int>& a, vector<int>& b) {
vector<int> res;
vector<bool> visited(b.size(), false);
// For every element in a[] check if we have
// matching not yet visited element in b[].
// If yes, add the item to result.
for (int i = 0; i < a.size(); i++) {
for (int j = 0; j < b.size(); j++) {
if (a[i] == b[j] && !visited[j]) {
res.push_back(a[i]);
visited[j] = true;
break;
}
}
}
return res;
}
int main() {
vector<int> a = {1, 2, 1, 3, 1};
vector<int> b = {3, 1, 3, 4, 1};
vector<int> result = findCommon(a, b);
for (int x : result) {
cout << x << " ";
}
return 0;
}
#include <stdio.h>
#include <stdbool.h>
void findCommon(int a[], int sizeA, int b[], int sizeB, int res[], int *res_size) {
bool visited[sizeB];
// Initialize visited array with false
for (int i = 0; i < sizeB; i++) {
visited[i] = false;
}
// For every element in a[] check if we have
// matching not yet visited element in b[].
// If yes, add the item to result.
for (int i = 0; i < sizeA; i++) {
for (int j = 0; j < sizeB; j++) {
if (a[i] == b[j] && !visited[j]) {
res[(*res_size)++] = a[i];
visited[j] = true;
break;
}
}
}
}
int main() {
int a[] = {1, 2, 1, 3, 1};
int b[] = {3, 1, 3, 4, 1};
int res[5];
int res_size = 0;
findCommon(a, 5, b, 5, res, &res_size);
for (int i = 0; i < res_size; i++) {
printf("%d ", res[i]);
}
return 0;
}
import java.util.ArrayList;
import java.util.List;
class GfG {
static List<Integer> findCommon(int[] a, int[] b) {
List<Integer> res = new ArrayList<>();
boolean[] visited = new boolean[b.length];
// For every element in a[] check if we have
// matching not yet visited element in b[].
// If yes, add the item to result.
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < b.length; j++) {
if (a[i] == b[j] && !visited[j]) {
res.add(a[i]);
visited[j] = true;
break;
}
}
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 2, 1, 3, 1};
int[] b = {3, 1, 3, 4, 1};
List<Integer> result = findCommon(a, b);
for (int x : result) {
System.out.print(x + " ");
}
}
}
def findCommon(a, b):
res = []
visited = [False] * len(b)
# For every element in a[] check if we have
# matching not yet visited element in b[].
# If yes, add the item to result.
for i in range(len(a)):
for j in range(len(b)):
if a[i] == b[j] and not visited[j]:
res.append(a[i])
visited[j] = True
break
return res
# Driver code
a = [1, 2, 1, 3, 1]
b = [3, 1, 3, 4, 1]
result = findCommon(a, b)
print(" ".join(map(str, result)))
using System;
using System.Collections.Generic;
class GfG
{
static List<int> findCommon(int[] a, int[] b) {
List<int> res = new List<int>();
bool[] visited = new bool[b.Length];
// For every element in a[] check if we have
// matching not yet visited element in b[].
// If yes, add the item to result.
for (int i = 0; i < a.Length; i++) {
for (int j = 0; j < b.Length; j++) {
if (a[i] == b[j] && !visited[j]) {
res.Add(a[i]);
visited[j] = true;
break;
}
}
}
return res;
}
static void Main() {
int[] a = {1, 2, 1, 3, 1};
int[] b = {3, 1, 3, 4, 1};
List<int> result = findCommon(a, b);
foreach (int x in result) {
Console.Write(x + " ");
}
}
}
function findCommon(a, b) {
const res = [];
const visited = new Array(b.length).fill(false);
// For every element in a[] check if we have
// matching not yet visited element in b[].
// If yes, add the item to result.
for (let i = 0; i < a.length; i++) {
for (let j = 0; j < b.length; j++) {
if (a[i] === b[j] && !visited[j]) {
res.push(a[i]);
visited[j] = true;
break;
}
}
}
return res;
}
// Driver code
const a = [1, 2, 1, 3, 1];
const b = [3, 1, 3, 4, 1];
const result = findCommon(a, b);
console.log(result.join(" "));
Expected Approach - O(n + m) Time and O(n) Space
- Create a hash map bm and store every item of b[] as key and its frequency as value.
- Create an empty array res[] to store result
- Traverse through all items of a[]. If an item is in the hash map bm, then add it to res[] and decrease its frequency by 1 in bm.
- Return res[]
C++
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<int> findCommon(vector<int>& a, vector<int>& b) {
// Store the frequency of elements in b[]
unordered_map<int, int> bm;
for (int x : b) {
bm[x]++;
}
// Traverse a[] and conisder an
// item as common only if its
// frequency is more than 0 in bm.
vector<int> res;
for (int x : a) {
if (bm[x] > 0) {
res.push_back(x);
bm[x]--; // Decrease freq
}
}
return res;
}
int main() {
vector<int> a = {1, 2, 1, 3, 1};
vector<int> b = {3, 1, 3, 4, 1};
vector<int> res = findCommon(a, b);
for (int x : res) {
cout << x << " ";
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#define MAX 1000
// Structure to represent a hashmap in C
typedef struct {
int key;
int count;
} HashMap;
void findCommon(int a[], int sizeA, int b[], int sizeB, int res[], int *res_size) {
HashMap bm[MAX] = {0}; // hashmap array with default 0 frequency
// Store the frequency of elements in b[]
for (int i = 0; i < sizeB; i++) {
bm[b[i]].key = b[i];
bm[b[i]].count++;
}
// Traverse a[] and consider an
// item as common only if its
// frequency is more than 0 in bm.
for (int i = 0; i < sizeA; i++) {
if (bm[a[i]].count > 0) {
res[(*res_size)++] = a[i];
bm[a[i]].count--; // Decrease frequency
}
}
}
int main() {
int a[] = {1, 2, 1, 3, 1};
int b[] = {3, 1, 3, 4, 1};
int res[5];
int res_size = 0;
findCommon(a, 5, b, 5, res, &res_size);
for (int i = 0; i < res_size; i++) {
printf("%d ", res[i]);
}
return 0;
}
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
class GfG {
static List<Integer> findCommon(int[] a, int[] b) {
List<Integer> res = new ArrayList<>();
// Store the frequency of elements in b[]
HashMap<Integer, Integer> bm = new HashMap<>();
for (int x : b) {
bm.put(x, bm.getOrDefault(x, 0) + 1);
}
// Traverse a[] and consider an
// item as common only if its
// frequency is more than 0 in bm.
for (int x : a) {
if (bm.getOrDefault(x, 0) > 0) {
res.add(x);
bm.put(x, bm.get(x) - 1); // Decrease frequency
}
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 2, 1, 3, 1};
int[] b = {3, 1, 3, 4, 1};
List<Integer> res = findCommon(a, b);
for (int x : res) {
System.out.print(x + " ");
}
}
}
def findCommon(a, b):
# Store the frequency of elements in b[]
bm = {}
for x in b:
bm[x] = bm.get(x, 0) + 1
# Traverse a[] and consider an
# item as common only if its
# frequency is more than 0 in bm.
res = []
for x in a:
if bm.get(x, 0) > 0:
res.append(x)
bm[x] -= 1 # Decrease frequency
return res
# Driver code
a = [1, 2, 1, 3, 1]
b = [3, 1, 3, 4, 1]
result = findCommon(a, b)
print(" ".join(map(str, result)))
using System;
using System.Collections.Generic;
class GfG
{
static List<int> findCommon(int[] a, int[] b) {
List<int> res = new List<int>();
// Store the frequency of elements in b[]
Dictionary<int, int> bm = new Dictionary<int, int>();
foreach (int x in b) {
if (bm.ContainsKey(x))
bm[x]++;
else
bm[x] = 1;
}
// Traverse a[] and consider an
// item as common only if its
// frequency is more than 0 in bm.
foreach (int x in a) {
if (bm.ContainsKey(x) && bm[x] > 0) {
res.Add(x);
bm[x]--; // Decrease frequency
}
}
return res;
}
static void Main() {
int[] a = {1, 2, 1, 3, 1};
int[] b = {3, 1, 3, 4, 1};
List<int> result = findCommon(a, b);
foreach (int x in result) {
Console.Write(x + " ");
}
}
}
function findCommon(a, b) {
const res = [];
// Store the frequency of elements in b[]
const bm = {};
for (let x of b) {
bm[x] = (bm[x] || 0) + 1;
}
// Traverse a[] and consider an
// item as common only if its
// frequency is more than 0 in bm.
for (let x of a) {
if (bm[x] > 0) {
res.push(x);
bm[x]--; // Decrease frequency
}
}
return res;
}
// Driver code
const a = [1, 2, 1, 3, 1];
const b = [3, 1, 3, 4, 1];
const result = findCommon(a, b);
console.log(result.join(" "));
We can do better if two arrays are sorted. Please refer Intersection of Two Sorted Arrays for details.
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