Dynamic Programming | Set 20 (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. 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.

#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; }

Output:

Length of maximum size chain is 3

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

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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