You are given n pairs of numbers. 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. Chain of pairs can be formed in this fashion. Find the longest chain which can be formed from a given set of pairs. Examples:
Input: (5, 24), (39, 60), (15, 28), (27, 40), (50, 90)
Output: (5, 24), (27, 40), (50, 90)
Input: (11, 20), {10, 40), (45, 60), (39, 40)
Output: (11, 20), (39, 40), (45, 60)
In previous post, we have discussed about Maximum Length Chain of Pairs problem. However, the post only covered code related to finding the length of maximum size chain, but not to the construction of maximum size chain. In this post, we will discuss how to construct Maximum Length Chain of Pairs itself. The idea is to first sort given pairs in increasing order of their first element. Let arr[0..n-1] be the input array of pairs after sorting. We define vector L such that L[i] is itself is a vector that stores Maximum Length Chain of Pairs of arr[0..i] that ends with arr[i]. Therefore for an index i, L[i] can be recursively written as –
L[0] = {arr[0]}
L[i] = {Max(L[j])} + arr[i] where j < i and arr[j].b < arr[i].a
= arr[i], if there is no such j
For example, for (5, 24), (39, 60), (15, 28), (27, 40), (50, 90)
L[0]: (5, 24) L[1]: (5, 24) (39, 60) L[2]: (15, 28) L[3]: (5, 24) (27, 40) L[4]: (5, 24) (27, 40) (50, 90)
Please note sorting of pairs is done as we need to find the maximum pair length and ordering doesn’t matter here. If we don’t sort, we will get pairs in increasing order but they won’t be maximum possible pairs. Below is implementation of above idea –
C++
/* Dynamic Programming solution to construct Maximum Length Chain of Pairs */#include <bits/stdc++.h>using namespace std;struct Pair{ int a; int b;};// comparator function for sort functionint compare(Pair x, Pair y){ return x.a < y.a;}// Function to construct Maximum Length Chain// of Pairsvoid maxChainLength(vector<Pair> arr){ // Sort by start time sort(arr.begin(), arr.end(), compare); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. vector<vector<Pair> > L(arr.size()); // L[0] is equal to arr[0] L[0].push_back(arr[0]); // start from index 1 for (int i = 1; i < arr.size(); i++) { // for every j less than i for (int j = 0; j < i; j++) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if ((arr[j].b < arr[i].a) && (L[j].size() > L[i].size())) L[i] = L[j]; } L[i].push_back(arr[i]); } // print max length vector vector<Pair> maxChain; for (vector<Pair> x : L) if (x.size() > maxChain.size()) maxChain = x; for (Pair pair : maxChain) cout << "(" << pair.a << ", " << pair.b << ") ";}// Driver Functionint main(){ Pair a[] = {{5, 29}, {39, 40}, {15, 28}, {27, 40}, {50, 90}}; int n = sizeof(a)/sizeof(a[0]); vector<Pair> arr(a, a + n); maxChainLength(arr); return 0;} |
Java
// Java program to implement the approachimport java.util.ArrayList;import java.util.Collections;import java.util.List;// User Defined Pair Classclass Pair { int a; int b;}class GFG { // Custom comparison function public static int compare(Pair x, Pair y) { return x.a - (y.a); } public static void maxChainLength(List<Pair> arr) { // Sort by start time Collections.sort(arr, Main::compare); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. List<List<Pair>> L = new ArrayList<>(); // L[0] is equal to arr[0] List<Pair> l0 = new ArrayList<>(); l0.add(arr.get(0)); L.add(l0); for (int i = 0; i < arr.size() - 1; i++) { L.add(new ArrayList<>()); } // start from index 1 for (int i = 1; i < arr.size(); i++) { // for every j less than i for (int j = 0; j < i; j++) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if (arr.get(j).b < arr.get(i).a && L.get(j).size() > L.get(i).size()) L.set(i, L.get(j)); } L.get(i).add(arr.get(i)); } // print max length vector List<Pair> maxChain = new ArrayList<>(); for (List<Pair> x : L) if (x.size() > maxChain.size()) maxChain = x; for (Pair pair : maxChain) System.out.println("(" + pair.a + ", " + pair.b + ") "); } // Driver Code public static void main(String[] args) { Pair[] a = {new Pair() {{a = 5; b = 29;}}, new Pair() {{a = 39; b = 40;}}, new Pair() {{a = 15; b = 28;}}, new Pair() {{a = 27; b = 40;}}, new Pair() {{a = 50; b = 90;}}}; int n = a.length; List<Pair> arr = new ArrayList<>(); for (Pair anA : a) { arr.add(anA); } // Function call maxChainLength(arr); }}// This code is contributed by phasing17 |
Python3
# Dynamic Programming solution to construct# Maximum Length Chain of Pairsclass Pair: def __init__(self, a, b): self.a = a self.b = b def __lt__(self, other): return self.a < other.adef maxChainLength(arr): # Function to construct # Maximum Length Chain of Pairs # Sort by start time arr.sort() # L[i] stores maximum length of chain of # arr[0..i] that ends with arr[i]. L = [[] for x in range(len(arr))] # L[0] is equal to arr[0] L[0].append(arr[0]) # start from index 1 for i in range(1, len(arr)): # for every j less than i for j in range(i): # L[i] = {Max(L[j])} + arr[i] # where j < i and arr[j].b < arr[i].a if (arr[j].b < arr[i].a and len(L[j]) > len(L[i])): L[i] = L[j] L[i].append(arr[i]) # print max length vector maxChain = [] for x in L: if len(x) > len(maxChain): maxChain = x for pair in maxChain: print("({a},{b})".format(a = pair.a, b = pair.b), end = " ") print()# Driver Codeif __name__ == "__main__": arr = [Pair(5, 29), Pair(39, 40), Pair(15, 28), Pair(27, 40), Pair(50, 90)] n = len(arr) maxChainLength(arr)# This code is contributed # by vibhu4agarwal |
C#
using System;using System.Collections.Generic;public class Pair{ public int a; public int b;}public class Program{ public static int Compare(Pair x, Pair y) { return x.a - (y.a); } public static void MaxChainLength(List<Pair> arr) { // Sort by start time arr.Sort(Compare); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. List<List<Pair>> L = new List<List<Pair>>(); // L[0] is equal to arr[0] L.Add(new List<Pair> { arr[0] }); for (int i = 0; i < arr.Count - 1; i++) L.Add(new List<Pair>()); // start from index 1 for (int i = 1; i < arr.Count; i++) { // for every j less than i for (int j = 0; j < i; j++) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if (arr[j].b < arr[i].a && L[j].Count > L[i].Count) L[i] = L[j]; } L[i].Add(arr[i]); } // print max length vector List<Pair> maxChain = new List<Pair>(); foreach (List<Pair> x in L) if (x.Count > maxChain.Count) maxChain = x; foreach (Pair pair in maxChain) Console.WriteLine("(" + pair.a + ", " + pair.b + ") "); } public static void Main() { Pair[] a = { new Pair() { a = 5, b = 29 }, new Pair() { a = 39, b = 40 }, new Pair() { a = 15, b = 28 }, new Pair() { a = 27, b = 40 }, new Pair() { a = 50, b = 90 } }; int n = a.Length; List<Pair> arr = new List<Pair>(a); MaxChainLength(arr); }} |
Javascript
<script>// Dynamic Programming solution to construct// Maximum Length Chain of Pairsclass Pair{ constructor(a, b){ this.a = a this.b = b }}function maxChainLength(arr){ // Function to construct // Maximum Length Chain of Pairs // Sort by start time arr.sort((c,d) => c.a - d.a) // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. let L = new Array(arr.length).fill(0).map(()=>new Array()) // L[0] is equal to arr[0] L[0].push(arr[0]) // start from index 1 for (let i=1;i<arr.length;i++){ // for every j less than i for(let j=0;j<i;j++){ // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if (arr[j].b < arr[i].a && L[j].length > L[i].length) L[i] = L[j] } L[i].push(arr[i]) } // print max length vector let maxChain = [] for(let x of L){ if(x.length > maxChain.length) maxChain = x } for(let pair of maxChain) document.write(`(${pair.a}, ${pair.b}) `) document.write("</br>")}// driver codelet arr = [new Pair(5, 29), new Pair(39, 40), new Pair(15, 28), new Pair(27, 40), new Pair(50, 90)]let n = arr.lengthmaxChainLength(arr)/// This code is contributed by shinjanpatra</script> |
Output:
(5, 29) (39, 40) (50, 90)
Time complexity of above Dynamic Programming solution is O(n2) where n is the number of pairs. Auxiliary space used by the program is O(n2). This article is contributed by Aditya Goel. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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