Given an array of pairs of numbers of size N. In every pair, the first number is always smaller than the second number. A pair (c, d) can follow another pair (a, b) if b < c. The chain of pairs can be formed in this fashion. The task is to find the length of the longest chain which can be formed from a given set of pairs.
Examples:
Input: N = 5, arr={{5, 24}, {39, 60}, {15, 28}, {27, 40}, {50, 90} }
Output: 3
The longest chain that can be formed is of length 3, and the chain is {{5, 24}, {27, 40}, {50, 90}}.Input : N = 2, arr={{5, 10}, {1, 11}}
Output :1
Approach: A dynamic programming approach for the problem has been discussed here.
Idea is to solve the problem using the greedy approach which is the same as Activity Selection Problem.
- Sort all pairs in increasing order of second number of each pair.
- Select first no as the first pair of chain and set a variable s(say) with the second value of the first pair.
- Iterate from the second pair to last pair of the array and if the value of the first element of the current pair is greater then previously selected pair then select the current pair and update the value of maximum length and variable s.
- Return the value of Max length of chain.
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Comparator function which can compare// the second element of the pair used to// sort pairs in increasing order of second value.const bool comparator(const pair<int,int>& p1, const pair<int,int>& p2){ return (p1.second < p2.second);}// Function for finding max length chainint maxChainLen(vector<pair<int,int> >p, int n){ // Initialize length l = 1 int l = 1; // Sort all pair in increasing order // according to second no of pair sort(p.begin(),p.end(),comparator); // Pick up the first pair and assign the // value of second element of pair to a // temporary variable s int s = p[0].second; // Iterate from second pair (index of // the second pair is 1) to the last pair for (int i = 1; i < n; i++) { // If first of current pair is greater // than previously selected pair then // select current pair and update // value of l and s if (p[i].first > s) { l++; s = p[i].second; } } // Return maximum length return l;}// Driver Codeint main(){ // Declaration of vector of pairs vector<pair<int,int>> p = { { 5, 24 }, { 39, 60 }, { 15, 28 }, { 27, 40 }, { 50, 90 } }; int n = p.size(); // Function call cout << maxChainLen(p, n) << endl; return 0;} |
Java
// Java implementation of the above approach import java.util.*;// Structure for storing pairs // of first and second values.class GFG{ // Class for storing pairs // of first and second values. static class Pair { int first; int second; Pair(int first, int second) { this.first = first; this.second = second; }}; // Function for finding max length chain static int maxChainLen(Pair p[], int n) { // Initialize length l = 1 int l = 1; // Sort all pair in increasing order // according to second no of pair Arrays.sort(p, new Comparator<Pair>() { public int compare(Pair a, Pair b) { return a.second - b.second; } }); // Pick up the first pair and assign the // value of second element of pair to a // temporary variable s int s = p[0].second; // Iterate from second pair (index of // the second pair is 1) to the last pair for(int i = 1; i < n; i++) { // If first of current pair is greater // than previously selected pair then // select current pair and update // value of l and s if (p[i].first > s) { l++; s = p[i].second; } } // Return maximum length return l; } // Driver Code public static void main(String args[]) { // Declaration of array of structure Pair p[] = new Pair[5]; p[0] = new Pair(5, 24); p[1] = new Pair(39, 60); p[2] = new Pair(15, 28); p[3] = new Pair(27, 40); p[4] = new Pair(50, 90); int n = p.length; // Function call System.out.println(maxChainLen(p, n));} }// This code is contributed by adityapande88 |
C#
// C# implementation of the above approach using System;using System.Linq;// Structure for storing pairs // of first and second values.class GFG{ // Class for storing pairs // of first and second values. class Pair : IComparable<Pair>{ public int first, second; public Pair(int first, int second) { this.first = first; this.second = second; } public int CompareTo(Pair p) { return this.second-p.second; }} // Function for finding max length chain static int maxChainLen(Pair []p, int n) { // Initialize length l = 1 int l = 1; // Sort all pair in increasing order // according to second no of pair Array.Sort(p); // Pick up the first pair and assign the // value of second element of pair to a // temporary variable s int s = p[0].second; // Iterate from second pair (index of // the second pair is 1) to the last pair for(int i = 1; i < n; i++) { // If first of current pair is greater // than previously selected pair then // select current pair and update // value of l and s if (p[i].first > s) { l++; s = p[i].second; } } // Return maximum length return l; } // Driver Code public static void Main(String []args) { // Declaration of array of structure Pair []p = new Pair[5]; p[0] = new Pair(5, 24); p[1] = new Pair(39, 60); p[2] = new Pair(15, 28); p[3] = new Pair(27, 40); p[4] = new Pair(50, 90); int n = p.Length; // Function call Console.WriteLine(maxChainLen(p, n));} }// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the above approach# Function for finding max length chaindef maxChainLen(p, n): # Initialize length l = 1 l = 1 # Sort all pair in increasing order # according to second no of pair p.sort(key=lambda x:x[1]) # Pick up the first pair and assign the # value of second element of pair to a # temporary variable s s = p[0][1] # Iterate from second pair (index of # the second pair is 1) to the last pair for i in range(n): # If first of current pair is greater # than previously selected pair then # select current pair and update # value of l and s if (p[i][0] > s) : l+=1 s = p[i][1] # Return maximum length return l# Driver Codeif __name__ == '__main__': # Declaration of vector of pairs p = [(5, 24) , (39, 60) , (15, 28) , (27, 40) , (50, 90)] n = len(p) # Function call print(maxChainLen(p, n)) |
Javascript
<script>// JavaScript implementation of the above approach// Function for finding max length chainfunction maxChainLen(p, n){ // Initialize length l = 1 let l = 1 // Sort all pair in increasing order // according to second no of pair p.sort((x,y)=>x[1]-y[1]) // Pick up the first pair and assign the // value of second element of pair to a // temporary variable s let s = p[0][1] // Iterate from second pair (index of // the second pair is 1) to the last pair for(let i=0;i<n;i++){ // If first of current pair is greater // than previously selected pair then // select current pair and update // value of l and s if (p[i][0] > s){ l+=1 s = p[i][1] } } // Return maximum length return l}// Driver Code// Declaration of vector of pairslet p = [[5, 24] , [39, 60] , [15, 28] , [27, 40] , [50, 90]]let n = p.length// Function calldocument.write(maxChainLen(p, n),"</br>")// This code is contributed by shinjanpatra.</script> |
Output:
3
Time complexity : O(N*log(N))
Auxiliary Space: O(1)
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