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. Source: Amazon Interview | Set 2

For example, if the given pairs are {{5, 24}, {39, 60}, {15, 28}, {27, 40}, {50, 90} }, then the longest chain that can be formed is of length 3, and the chain is {{5, 24}, {27, 40}, {50, 90}}

This problem is a variation of standard Longest Increasing Subsequence problem. Following is a simple two step process.

1) Sort given pairs in increasing order of first (or smaller) element.

2) Now run a modified LIS process where we compare the second element of already finalized LIS with the first element of new LIS being constructed.

The following code is a slight modification of method 2 of this post.

## C

#include<stdio.h> #include<stdlib.h> // Structure for a pair struct pair { int a; int b; }; // This function assumes that arr[] is sorted in increasing order // according the first (or smaller) values in pairs. int maxChainLength( struct pair arr[], int n) { int i, j, max = 0; int *mcl = (int*) malloc ( sizeof( int ) * n ); /* Initialize MCL (max chain length) values for all indexes */ for ( i = 0; i < n; i++ ) mcl[i] = 1; /* Compute optimized chain length values in bottom up manner */ for ( i = 1; i < n; i++ ) for ( j = 0; j < i; j++ ) if ( arr[i].a > arr[j].b && mcl[i] < mcl[j] + 1) mcl[i] = mcl[j] + 1; // mcl[i] now stores the maximum chain length ending with pair i /* Pick maximum of all MCL values */ for ( i = 0; i < n; i++ ) if ( max < mcl[i] ) max = mcl[i]; /* Free memory to avoid memory leak */ free( mcl ); return max; } /* Driver program to test above function */ int main() { struct pair arr[] = { {5, 24}, {15, 25}, {27, 40}, {50, 60} }; int n = sizeof(arr)/sizeof(arr[0]); printf("Length of maximum size chain is %d\n", maxChainLength( arr, n )); return 0; }

## Java

class Pair{ int a; int b; public Pair(int a, int b) { this.a = a; this.b = b; } // This function assumes that arr[] is sorted in increasing order // according the first (or smaller) values in pairs. static int maxChainLength(Pair arr[], int n) { int i, j, max = 0; int mcl[] = new int[n]; /* Initialize MCL (max chain length) values for all indexes */ for ( i = 0; i < n; i++ ) mcl[i] = 1; /* Compute optimized chain length values in bottom up manner */ for ( i = 1; i < n; i++ ) for ( j = 0; j < i; j++ ) if ( arr[i].a > arr[j].b && mcl[i] < mcl[j] + 1) mcl[i] = mcl[j] + 1; // mcl[i] now stores the maximum chain length ending with pair i /* Pick maximum of all MCL values */ for ( i = 0; i < n; i++ ) if ( max < mcl[i] ) max = mcl[i]; return max; } /* Driver program to test above function */ public static void main(String[] args) { Pair arr[] = new Pair[] {new Pair(5,24), new Pair(15, 25), new Pair (27, 40), new Pair(50, 60)}; System.out.println("Length of maximum size chain is " + maxChainLength(arr, arr.length)); } }

## Python3

class Pair(object): def __init__(self, a, b): self.a = a self.b = b # This function assumes that arr[] is sorted in increasing # order according the first (or smaller) values in pairs. def maxChainLength(arr, n): max = 0 # Initialize MCL(max chain length) values for all indices mcl = [1 for i in range(n)] # Compute optimized chain length values in bottom up manner for i in range(1, n): for j in range(0, i): if (arr[i].a > arr[j].b and mcl[i] < mcl[j] + 1): mcl[i] = mcl[j] + 1 # mcl[i] now stores the maximum # chain length ending with pair i # Pick maximum of all MCL values for i in range(n): if (max < mcl[i]): max = mcl[i] return max # Driver program to test above function arr = [Pair(5, 24), Pair(15, 25), Pair(27, 40), Pair(50, 60)] print('Length of maximum size chain is', maxChainLength(arr, len(arr))) # This code is contributed by Soumen Ghosh

Output:

Length of maximum size chain is 3

Time Complexity: O(n^2) where n is the number of pairs.

### Asked in: Amazon

The given problem is also a variation of Activity Selection problem and can be solved in (nLogn) time. To solve it as a activity selection problem, consider the first element of a pair as start time in activity selection problem, and the second element of pair as end time. Thanks to Palash for suggesting this approach.

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