Count number of common elements between a sorted array and a reverse sorted array
Last Updated :
22 Mar, 2023
Given two arrays consisting of N distinct integers such that the array A[] and B[] are sorted in ascending and descending order respectively, the task is to find the number of values common in both arrays.
Examples:
Input: A[] = {1, 10, 100}, B[] = {200, 20, 2}
Output: 0
Input: A[] = {2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999}, B[] = {109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1}
Output: 4
Naive Approach:- Check for all elements in array A that is present in array B or not if Yes increase the count of pair.
Implementation:-
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// elements common in both the arrays
int countEqual(int A[], int B[], int N)
{
//variable to store answer
int ans=0;
//first loop for array A
for(int i=0;i<N;i++)
{
//This loop to find array A element in B
for(int j=0;j<N;j++)
{
//if found then increase count and exit the loop
if(A[i]==B[j])
{
ans++;
break;
}
}
}
return ans;
}
// Driver Code
int main()
{
int A[] = { 2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999 };
int B[] = { 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 };
int N = sizeof(A) / sizeof(int);
cout << countEqual(A, B, N);
return 0;
}
//This code contributed by shubhamrajput6156
import java.io.*;
class GFG
{
// Java program for the above approach
// Function to count the number of
// elements common in both the arrays
public static int countEqual(int[] A, int[] B, int N)
{
// variable to store answer
int ans = 0;
// first loop for array A
for (int i = 0;i < N;i++)
{
// This loop to find array A element in B
for (int j = 0;j < N;j++)
{
// if found then increase count and exit the loop
if (A[i] == B[j])
{
ans++;
break;
}
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] A = {2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999};
int[] B = {109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1};
int N = A.length;
System.out.print(countEqual(A, B, N));
}
}
// This code contributed by bhardwajji
// JS code to implement the approach
// JavaScript code for the above approach
function countEqual(A, B, N)
{
// variable to store answer
let ans = 0;
// first loop for array A
for (let i = 0; i < N; i++)
{
// This loop to find array A element in B
for (let j = 0; j < N; j++)
{
// if found then increase count and exit the
// loop
if (A[i] == B[j]) {
ans++;
break;
}
}
}
return ans;
}
// Driver Code
let A = [ 2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999 ];
let B = [ 109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1 ];
let N = A.length;
console.log(countEqual(A, B, N));
// This code is contributed by phasing17
// C# program for the above approach
using System;
class GFG
{
// Function to count the number of
// elements common in both the arrays
public static int countEqual(int[] A, int[] B, int N)
{
// variable to store answer
int ans = 0;
// first loop for array A
for (int i = 0;i < N;i++)
{
// This loop to find array A element in B
for (int j = 0;j < N;j++)
{
// if found then increase count and exit the loop
if (A[i] == B[j])
{
ans++;
break;
}
}
}
return ans;
}
// Driver Code
public static void Main()
{
int[] A = {2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999};
int[] B = {109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1};
int N = A.Length;
Console.WriteLine(countEqual(A, B, N));
}
}
// This code is contributed by Pushpesh Raj.
# python program for the above approach
# Function to count the number of
# elements common in both the arrays
def countEqual(A, B, N):
# variable to store answer
ans = 0
# first loop for array A
for i in range(N):
# This loop to find array A element in B
for j in range(N):
# if found then increase count and exit the loop
if A[i] == B[j]:
ans += 1
break
return ans
# driver code
A = [2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999]
B = [109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1]
N = len(A)
print(countEqual(A, B, N))
Time Complexity:- O(N^2)
Auxiliary Space:- O(1)
Approach: The given problem can be solved by using the Two Pointer Approach. Follow the steps below to solve the problem:
- Initialize two variables, say first as 0 and second as (N - 1) that is used to traverse the array A[] and B[] from the front and back respectively.
- Initialize a variable, say count as 0 that stores the count of numbers common in the array A[] and B[].
- Iterate a loop until first < N and second >= 0 and perform the following steps:
- If the value of A[first] is equal to B[second], then increment the values of count and first and decrement the value of the second.
- If the value of A[first] is less than B[second], then increment the value of first.
- If the value of A[first] is greater than B[second], then decrement the value of the second.
- After completing the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// elements common in both the arrays
int countEqual(int A[], int B[], int N)
{
// Used to traverse array A[] and
// B[] from the front and the back
int first = 0;
int second = N - 1;
// Stores the count of numbers
// common in both array
int count = 0;
while (first < N && second >= 0) {
// If A[first] is less than
// B[second]
if (A[first] < B[second]) {
// Increment the value
// of first
first++;
}
// IF B[second] is less
// than A[first]
else if (B[second] < A[first]) {
// Decrement the value
// of second
second--;
}
// A[first] is equal to
// B[second]
else {
// Increment the value
// of count
count++;
// Increment the value
// of first
first++;
// Decrement the value
// of second
second--;
}
}
// Return the value of count
return count;
}
// Driver Code
int main()
{
int A[] = { 2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999 };
int B[] = { 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 };
int N = sizeof(A) / sizeof(int);
cout << countEqual(A, B, N);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
// Function to count the number of
// elements common in both the arrays
static int countEqual(int A[], int B[], int N)
{
// Used to traverse array A[] and
// B[] from the front and the back
int first = 0;
int second = N - 1;
// Stores the count of numbers
// common in both array
int count = 0;
while (first < N && second >= 0) {
// If A[first] is less than
// B[second]
if (A[first] < B[second]) {
// Increment the value
// of first
first++;
}
// IF B[second] is less
// than A[first]
else if (B[second] < A[first]) {
// Decrement the value
// of second
second--;
}
// A[first] is equal to
// B[second]
else {
// Increment the value
// of count
count++;
// Increment the value
// of first
first++;
// Decrement the value
// of second
second--;
}
}
// Return the value of count
return count;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999 };
int B[] = { 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 };
int N = A.length;
System.out.println(countEqual(A, B, N));
}
}
// This code is contributed by susmitakundugoaldanga.
# Python program for the above approach
# Function to count the number of
# elements common in both the arrays
def countEqual(A, B, N) :
# Used to traverse array A[] and
# B[] from the front and the back
first = 0
second = N - 1
# Stores the count of numbers
# common in both array
count = 0
while (first < N and second >= 0) :
# If A[first] is less than
# B[second]
if (A[first] < B[second]) :
# Increment the value
# of first
first += 1
# IF B[second] is less
# than A[first]
elif (B[second] < A[first]) :
# Decrement the value
# of second
second -= 1
# A[first] is equal to
# B[second]
else :
# Increment the value
# of count
count += 1
# Increment the value
# of first
first += 1
# Decrement the value
# of second
second -= 1
# Return the value of count
return count
# Driver Code
A= [ 2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999 ]
B = [ 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 ]
N = len(A)
print(countEqual(A, B, N))
# This code is contributed by sanjou_62.
// C# program for the above approach
using System;
class GFG{
// Function to count the number of
// elements common in both the arrays
static int countEqual(int[] A, int[] B, int N)
{
// Used to traverse array A[] and
// B[] from the front and the back
int first = 0;
int second = N - 1;
// Stores the count of numbers
// common in both array
int count = 0;
while (first < N && second >= 0)
{
// If A[first] is less than
// B[second]
if (A[first] < B[second])
{
// Increment the value
// of first
first++;
}
// IF B[second] is less
// than A[first]
else if (B[second] < A[first])
{
// Decrement the value
// of second
second--;
}
// A[first] is equal to
// B[second]
else
{
// Increment the value
// of count
count++;
// Increment the value
// of first
first++;
// Decrement the value
// of second
second--;
}
}
// Return the value of count
return count;
}
// Driver code
static void Main()
{
int[] A = { 2, 4, 5, 8, 12, 13,
17, 18, 20, 22, 309, 999 };
int[] B = { 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 };
int N = A.Length;
Console.WriteLine(countEqual(A, B, N));
}
}
// This code is contributed by abhinavjain194
<script>
// Javascript program for the above approach
// Function to count the number of
// elements common in both the arrays
function countEqual(A, B, N)
{
// Used to traverse array A[] and
// B[] from the front and the back
let first = 0;
let second = N - 1;
// Stores the count of numbers
// common in both array
let count = 0;
while (first < N && second >= 0) {
// If A[first] is less than
// B[second]
if (A[first] < B[second]) {
// Increment the value
// of first
first++;
}
// IF B[second] is less
// than A[first]
else if (B[second] < A[first]) {
// Decrement the value
// of second
second--;
}
// A[first] is equal to
// B[second]
else {
// Increment the value
// of count
count++;
// Increment the value
// of first
first++;
// Decrement the value
// of second
second--;
}
}
// Return the value of count
return count;
}
// Driver Code
let A = [2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999];
let B = [109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1];
let N = A.length;
document.write(countEqual(A, B, N));
// This code is contributed _saurabh_jaiswal
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
Another Approach : We will use Binary search to check if the element of array B[] is present in the array A[] or not because array A[] is already sorted in increasing order. So , we can use binary search for finding elements.
Below is the implementation of the above approach :
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
//Function to check if x is present in the array or not
bool binarysearch(int arr[], int N, int x)
{
int l = 0, r = N - 1;
while (l <= r) {
int mid = (l + r) / 2;
// Checking if the middle element is equal to x
if (arr[mid] == x) {
return true;
}
else if (arr[mid] < x) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
// return true , if element x is present in the array
// else false
return false;
}
// Function to count the number of
// elements common in both the arrays
int countEqual(int A[], int B[], int N, int M)
{ int count = 0;
// Iterate each element of array B
for (int i = 0; i < M; i++)
{
// Checking if the element of array B is present in
// array A using the binary search
if (binarysearch(A, N, B[i]))
{
count++;
}
}
// Return count of common element
return count;
}
// Driver Code
int main()
{
int A[] = { 2, 4, 5, 8, 12, 13, 17,
18, 20, 22, 309, 999 };
int B[] = { 109, 99, 68, 54, 22
, 19,17, 13, 11, 5, 3, 1 };
int N = sizeof(A) / sizeof(int);
int M = sizeof(B) / sizeof(int);
//Function call
cout << countEqual(A, B, N, M)<<endl;
return 0;
}
// This code is contributed by nikhilsainiofficial546
// Java program for the above approach
import java.util.Arrays;
class Main {
// Function to check if x is present in the array or not
static boolean binarySearch(int[] arr, int N, int x) {
int l = 0, r = N - 1;
while (l <= r) {
int mid = (l + r) / 2;
// Checking if the middle element is equal to x
if (arr[mid] == x) {
return true;
} else if (arr[mid] < x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
// return true , if element x is present in the array
// else false
return false;
}
// Function to count the number of elements common in both the arrays
static int countEqual(int[] A, int[] B, int N, int M) {
int count = 0;
// Sort array A
Arrays.sort(A);
// Iterate each element of array B
for (int i = 0; i < M; i++) {
// Checking if the element of array B is present in array A using the binary search
if (binarySearch(A, N, B[i])) {
count++;
}
}
// Return count of common element
return count;
}
// Driver Code
public static void main(String[] args) {
int[] A = { 2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999 };
int[] B = { 109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1 };
int N = A.length;
int M = B.length;
//Function call
System.out.println(countEqual(A, B, N, M));
}
}
#Python program for the above approach
# Function to check if x is present in the array or not
def binarySearch(arr, N, x):
l = 0
r = N - 1
while l <= r:
mid = (l + r) // 2
# Checking if the middle element is equal to x
if arr[mid] == x:
return True
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
# return true , if element x is present in the array
# else false
return False
# Function to count the number of elements common in both the arrays
def countEqual(A, B, N, M):
count = 0
# Sort array A
A.sort()
# Iterate each element of array B
for i in range(M):
# Checking if the element of array B is present in array A using the binary search
if binarySearch(A, N, B[i]):
count += 1
# Return count of common element
return count
# Driver Code
A = [ 2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999 ]
B = [ 109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1 ]
N = len(A)
M = len(B)
#Function call
print(countEqual(A, B, N, M))
// C# program for the above approach
using System;
class Program
{
// Function to check if x is present in the array or not
static bool BinarySearch(int[] arr, int N, int x)
{
int l = 0, r = N - 1;
while (l <= r) {
int mid = (l + r) / 2;
// Checking if the middle element is equal to x
if (arr[mid] == x) {
return true;
}
else if (arr[mid] < x) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
// return true , if element x is present in the
// array else false
return false;
}
// Function to count the number of
// elements common in both the arrays
static int CountEqual(int[] A, int[] B, int N, int M)
{
int count = 0;
// Iterate each element of array B
for (int i = 0; i < M; i++) {
// Checking if the element of array B is present
// in array A using the binary search
if (BinarySearch(A, N, B[i])) {
count++;
}
}
// Return count of common element
return count;
}
// Driver Code
static void Main()
{
int[] A = { 2, 4, 5, 8, 12, 13,
17, 18, 20, 22, 309, 999 };
int[] B = { 109, 99, 68, 54, 22, 19,
17, 13, 11, 5, 3, 1 };
int N = A.Length;
int M = B.Length;
// Function call
Console.WriteLine(CountEqual(A, B, N, M));
Console.ReadLine();
}
}
// JavaScript program to implement the approach
// Function to check if x is present in the array or not
function binarysearch(arr, N, x) {
let l = 0, r = N - 1;
while (l <= r) {
let mid = Math.floor((l + r) / 2);
// Checking if the middle element is equal to x
if (arr[mid] === x) {
return true;
}
else if (arr[mid] < x) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
// return true , if element x is present in the array
// else false
return false;
}
// Function to count the number of
// elements common in both the arrays
function countEqual(A, B, N, M) {
let count = 0;
// Iterate each element of array B
for (let i = 0; i < M; i++) {
// Checking if the element of array B is present in
// array A using the binary search
if (binarysearch(A, N, B[i])) {
count++;
}
}
// Return count of common element
return count;
}
// Driver Code
(() => {
const A = [2, 4, 5, 8, 12, 13, 17, 18, 20, 22, 309, 999];
const B = [109, 99, 68, 54, 22, 19, 17, 13, 11, 5, 3, 1];
const N = A.length;
const M = B.length;
// Function call
console.log(countEqual(A, B, N, M));
})();
// This code is contributed by phasing17
Time Complexity: O(M*log(N))
Auxiliary Space: O(1)
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