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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print adjacent pairs// in sorted ordervoid 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 Codeint 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
// Java code of above approachimport 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
# Python3 program for the above approach# Function to print adjacent pairs# in sorted orderdef 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 Codeif __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#
// C# code of above approachusing 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 |
Javascript
// Function to print adjacent pairs// in sorted orderfunction 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 Codeconst arr = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2];const N = arr.length;// Function callrepeated_pairs(arr, N);// This code is contributed by Aditya Sharma |
Output:
1 2 2 3 3 4 4 1
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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