There is a chain of pairs is given. In each pair, there are two integers and the first integer is always smaller, and the second one is greater, the same rule can also be applied for the chain construction. A pair (x, y) can be added after a pair (p, q), only if q < x.
To solve this problem, at first, we have to sort given pairs in increasing order of the first element. After that, we will compare the second element of a pair, with the first element of the next pair.
Input and Output
Input: A chain of number pairs. {(5, 24), (15, 25), (27, 40), (50, 60)} Output: Largest length of the chain as given criteria. Here the length is 3.
Algorithm
maxChainLength(arr, n)
Each element of the chain will contain two elements a and b
Input − The array of pairs, number of items in the array.
Output − Maximum length.
Begin define maxChainLen array of size n, and fill with 1 max := 0 for i := 1 to n, do for j := 0 to i-1, do if arr[i].a > arr[j].b and maxChainLen[i] < maxChainLen[j] + 1 maxChainLen[i] := maxChainLen[j] + 1 done done max := maximum length in maxChainLen array return max End
Example
#include<iostream> #include<algorithm> using namespace std; struct numPair { //define pair as structure int a; int b; }; int maxChainLength(numPair arr[], int n) { int max = 0; int *maxChainLen = new int[n]; //create array of size n for (int i = 0; i < n; i++ ) //Initialize Max Chain length values for all indexes maxChainLen[i] = 1; for (int i = 1; i < n; i++ ) for (int j = 0; j < i; j++ ) if ( arr[i].a > arr[j].b && maxChainLen[i] < maxChainLen[j] + 1) maxChainLen[i] = maxChainLen[j] + 1; // maxChainLen[i] now holds the max chain length ending with pair i for (int i = 0; i < n; i++ ) if ( max < maxChainLen[i] ) max = maxChainLen[i]; //find maximum among all chain length values delete[] maxChainLen; //deallocate memory return max; } int main() { struct numPair arr[] = {{5, 24},{15, 25},{27, 40},{50, 60}}; int n = 4; cout << "Length of maximum size chain is " << maxChainLength(arr, n); }
Output
Length of maximum size chain is 3