Maximum Length Chain of Pairs | Set-2

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.

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Structure for storing pairs
// of first and second values.
struct val {
    int first;
    int second;
};
  
// Comparator function which can compare
// the second element of structure used to
// sort pairs in increasing order of second value.
bool comparator(struct val p1, struct val p2)
{
    return (p1.second < p2.second);
}
  
// Function for finding max length chain
int maxChainLen(struct val p[], int n)
{
    // Initialize length l = 1
    int l = 1;
  
    // Sort all pair in increasing order
    // according to second no of pair
    sort(p, p + n, comparator);
  
    // Pick up the first pair and assign the
    // value of second element fo 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
int main()
{
  
    // Declaration of array of structure
    val p[] = { { 5, 24 }, { 39, 60 }, 
              { 15, 28 }, { 27, 40 }, { 50, 90 } };
  
    int n = sizeof(p) / sizeof(p[0]);
  
    // Fucntion call
    cout << maxChainLen(p, n) << endl;
  
    return 0;
}

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Output:

3

Time complexity : O(N*log(N))



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