# Print Maximum Length Chain of Pairs

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 â€“

## C++

 `/* Dynamic Programming solution to construct` `   ``Maximum Length Chain of Pairs */` `#include ` `using` `namespace` `std;`   `struct` `Pair` `{` `    ``int` `a;` `    ``int` `b;` `};`   `// comparator function for sort function` `int` `compare(Pair x, Pair y)` `{` `    ``return` `x.a < y.a;` `}`   `// Function to construct Maximum Length Chain` `// of Pairs` `void` `maxChainLength(vector 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 > 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 maxChain;` `    ``for` `(vector x : L)` `        ``if` `(x.size() > maxChain.size())` `            ``maxChain = x;`   `    ``for` `(Pair pair : maxChain)` `        ``cout << ``"("` `<< pair.a << ``", "` `             ``<< pair.b << ``") "``;` `}`   `// Driver Function` `int` `main()` `{` `    ``Pair a[] = {{5, 29}, {39, 40}, {15, 28},` `                ``{27, 40}, {50, 90}};` `    ``int` `n = ``sizeof``(a)/``sizeof``(a[0]);`   `    ``vector arr(a, a + n);`   `    ``maxChainLength(arr);`   `    ``return` `0;` `}`

## Java

 `// Java program to implement the approach` `import` `java.util.ArrayList;` `import` `java.util.Collections;` `import` `java.util.List;`   `// User Defined Pair Class` `class` `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 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> L = ``new` `ArrayList<>();`   `    ``// L[0] is equal to arr[0]` `    ``List 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 maxChain = ``new` `ArrayList<>();` `    ``for` `(List 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 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 Pairs` `class` `Pair:`   `    ``def` `__init__(``self``, a, b):` `        ``self``.a ``=` `a` `        ``self``.b ``=` `b`   `    ``def` `__lt__(``self``, other):` `        ``return` `self``.a < other.a`   `def` `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 Code` `if` `__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 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> L = ``new` `List>();`   `    ``// L[0] is equal to arr[0]` `    ``L.Add(``new` `List { arr[0] });` `    ``for` `(``int` `i = 0; i < arr.Count - 1; i++)` `      ``L.Add(``new` `List());`   `    ``// 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 maxChain = ``new` `List();` `    ``foreach` `(List 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 arr = ``new` `List(a);`   `    ``MaxChainLength(arr);` `  ``}` `}`

## Javascript

 ``

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). If you like GeeksforGeeks 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 GeeksforGeeks 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.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next