Given an array arr[] consisting of N integers, the task is to print all adjacent integer pairs from the array which appears more than once in the given array. If the array contains more than one such pair, print all pairs in sorted order.
Examples:
Input: arr[] = {1, 2, 5, 1, 2}
Output:
1 2
Explanation:
1 2 is the only repeating integer pair in the array.Input: arr[] = {1, 2, 3, 4, 1, 2, 3, 4, 1, 2}
Output:
1 2
2 3
3 4
4 1
Explanation:
Since the array has more than one repeating pair, all the pairs are printed in the sorted order.
Approach: The simplest approach is to traverse the array and store every adjacent pair in a Map. Print all such pairs having frequency greater than 1. Follow the steps below to solve the problem:
- Create a Map M to store all adjacent pairs in an array.
- Traverse the given array and store every adjacent pair in the Map M.
- After the above step, traverse the map, and if the frequency of any pair is at least one, then insert it into a vector V.
- Sort the vector v in ascending order and print all the pairs stored in it.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to print adjacent pairs // in sorted order void repeated_pairs( int arr[], int N)
{ // Store the frequency of all
// the adjacent pairs
map<pair< int , int >, int > m;
// Stores the resultant pairs
vector<pair< int , int > > v;
int i, j;
// Stores the count of all
// adjacent pairs
for (i = 0; i < N - 1; i++) {
pair< int , int > p
= { arr[i], arr[i + 1] };
// Increment the count of pair
m[p]++;
}
// Store pairs that appears more
// than once
for ( auto i = m.begin();
i != m.end(); i++) {
// If frequency is at least 1
if (i->second > 1) {
pair< int , int > p = i->first;
// Insert pair into vector
v.push_back(p);
}
}
// Sort the vector
sort(v.begin(), v.end());
// Print the pairs
for (i = 0; i < v.size(); i++) {
pair< int , int > p = v[i];
// Print the pair
cout << p.first << " "
<< p.second << endl;
}
} // Driver Code int main()
{ // Given arr[]
int arr[] = { 1, 2, 3, 4, 1,
2, 3, 4, 1, 2 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
repeated_pairs(arr, N);
return 0;
} |
// Java code of above approach import java.util.*;
import java.lang.*;
class pair
{ int first, second;
pair( int f, int s)
{
this .first = f;
this .second = s;
}
@Override
public boolean equals(Object obj)
{
// if both the object references are
// referring to the same object.
if ( this == obj)
return true ;
if (obj == null || obj.getClass() != this .getClass())
return false ;
// type casting of the argument.
pair p = (pair) obj;
return (p.first == this .first && p.second == this .second);
}
@Override
public int hashCode()
{
return this .first + this .second/ 2 ;
}
} class GFG {
// Function to print adjacent pairs
// in sorted order
static void repeated_pairs( int arr[], int N)
{
// Store the frequency of all
// the adjacent pairs
Map<pair, Integer> m= new HashMap<>();
// Stores the resultant pairs
ArrayList<pair> v = new ArrayList<>();
int i, j;
// Stores the count of all
// adjacent pairs
for (i = 0 ; i < N - 1 ; i++)
{
pair p = new pair(arr[i], arr[i + 1 ]);
// Increment the count of pair
m.put(p,m.getOrDefault(p, 0 ) + 1 );
}
// Store pairs that appears more
// than once
for (Map.Entry<pair,Integer> k: m.entrySet())
{
// If frequency is at least
if (k.getValue() > 1 )
{
// Insert pair into vector
v.add(k.getKey());
}
}
// Sort the vector
Collections.sort(v, (a, b)->a.first-b.first);
// Print the pairs
for (pair k:v)
{
// Print the pair
System.out.println(k.first + " " + k.second);
}
}
// Driver code
public static void main(String[] args)
{
// Given arr[]
int arr[] = { 1 , 2 , 3 , 4 , 1 ,
2 , 3 , 4 , 1 , 2 };
int N = arr.length;
// Function call
repeated_pairs(arr, N);
}
} // This code is contributed by offbeat |
# Python3 program for the above approach # Function to print adjacent pairs # in sorted order def repeated_pairs(arr, N):
# Store the frequency of all
# the adjacent pairs
m = {}
# Stores the resultant pairs
v = []
# Stores the count of all
# adjacent pairs
for i in range (N - 1 ):
p = (arr[i], arr[i + 1 ])
# Increment the count of pair
m[p] = m.get(p, 0 ) + 1
# Store pairs that appears more
# than once
for i in m:
# If frequency is at least 1
if (m[i] > 1 ):
p = i
# Insert pair into vector
v.append(p)
# Sort the vector
v = sorted (v)
# Print the pairs
for i in range ( len (v)):
p = v[i]
# Print the pair
print (p[ 0 ], p[ 1 ])
# Driver Code if __name__ = = '__main__' :
# Given arr[]
arr = [ 1 , 2 , 3 , 4 , 1 ,
2 , 3 , 4 , 1 , 2 ]
N = len (arr)
# Function call
repeated_pairs(arr, N)
# This code is contributed by mohit kumar 29 |
// C# code of above approach using System;
using System.Collections.Generic;
using System.Linq;
class pair
{ public int first, second;
public pair( int f, int s)
{
this .first = f;
this .second = s;
}
public override bool Equals(Object obj)
{
// if both the object references are
// referring to the same object.
if ( this == obj)
return true ;
if (obj == null || obj.GetType() != this .GetType())
return false ;
// type casting of the argument.
pair p = (pair)obj;
return (p.first == this .first && p.second == this .second);
}
public override int GetHashCode()
{
return this .first + this .second / 2;
}
} class GFG
{ // Function to print adjacent pairs
// in sorted order
static void repeated_pairs( int [] arr, int N)
{
// Store the frequency of all
// the adjacent pairs
Dictionary<pair, int > m = new Dictionary<pair, int >();
// Stores the resultant pairs
List<pair> v = new List<pair>();
// Stores the count of all
// adjacent pairs
for ( int i = 0; i < N - 1; i++)
{
pair p = new pair(arr[i], arr[i + 1]);
if (!m.ContainsKey(p))
{
m[p] = 0;
}
// Increment the count of pair
m[p]++;
}
// Store pairs that appears more
// than once
foreach (KeyValuePair<pair, int > k in m)
{
// If frequency is at least
if (k.Value > 1)
{
// Insert pair into vector
v.Add(k.Key);
}
}
// Sort the vector
v.Sort((a, b) => a.first - b.first);
// print the pair
foreach (pair k in v)
{
// print pair
Console.WriteLine(k.first + " " + k.second);
}
}
// Driver code
public static void Main( string [] args)
{
// Given arr[]
int [] arr = { 1, 2, 3, 4, 1, 2, 3, 4, 1, 2 };
int N = arr.Length
// Function call
repeated_pairs(arr, N);
}
} // This code is contributed by Aditya Sharma |
// Function to print adjacent pairs // in sorted order function repeated_pairs(arr, N) {
// Store the frequency of all
// the adjacent pairs
const m = {};
// Stores the resultant pairs
const v = [];
// Stores the count of all
// adjacent pairs
for (let i = 0; i < N - 1; i++) {
const p = [arr[i], arr[i + 1]];
// Increment the count of pair
m[p] = (m[p] || 0) + 1;
}
// Store pairs that appears more
// than once
for (let i in m) {
// If frequency is at least 1
if (m[i] > 1) {
const p = i.split( ',' );
// Insert pair into vector
v.push([parseInt(p[0]), parseInt(p[1])]);
}
}
// Sort the vector
v.sort();
// Print the pairs
for (let i = 0; i < v.length; i++) {
const p = v[i];
// Print the pair
console.log(p[0], p[1]);
}
} // Driver Code const arr = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2]; const N = arr.length; // Function call repeated_pairs(arr, N); // This code is contributed by Aditya Sharma |
1 2 2 3 3 4 4 1
Time Complexity: O(N*log N)
Auxiliary Space: O(N)