Longest Increasing Subsequence

The longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.

For example, length of LIS for { 3,2,6,4,5,1 } is 2 and LIS is {2,4,5}.




Method 1) Brute Force Approch

for (i=N; i>0 ; --i){
   Find a SubSequence of length i
   if any increasing subsequence
       Break;
}

But finding all subsequence of length i will give Time Complexity in Exponential.
as all subsequences will be N! / i! * (N-i)!

Lets think directly in DP way...

Method 2) Dynamic Programming way.

Take Example D := { 3, 2 , 6 , 4 , 5 , 1 }    We are defining vector L of vector <Numbers> such that
L[i] = A vector, Longest Increasing sequence of D that ends with D[i].

L[0] = 3.
L[1] = 2 ( because 3 > 2, and we need to find increasing sub sequence )
L[2] = 2,6
L[3] = 2,4
L[4] = 2,4,5
L[5] = 1

say L[2] = 2,6 is '2' + '6'  where 2 is so far found max LIS, and '6' is current element

So formula
L[i] is  maximum of all L[j] where j<i and D[j]<D[i] + 'D[i]'

L[i] = MAX(L[j], where j<i, D[j]<D[i]) + D[i]

So now We have working code to print all Increasing SubSequence
Edit it your way to find larget increasing subsequence

Code :

#include <iostream>
using namespace std;
#include <iostream>
#include <vector>


void prt(vector<int>& arr, string msg = "") {
cout << msg << " ";
for  (auto i: arr) {
cout << i << " ";
}
cout << endl;
}


void calc_LIS(vector<int>& D) {
vector< vector<int> > L(D.size());  // The longest increasing subsequence ends with D[i]

   L[0].push_back(D[0]);

for (int i=1; i<D.size(); i++) {
for(int j=0; j<i; j++) {
if ( (D[j] < D[i]) && ( L[i].size() < L[j].size() ) ) {
L[i] = L[j];  
}         
}
      L[i].push_back(D[i]);
}

for (auto x: L) {
prt(x);
}

}

int main() {
int a[] = {3, 2, 6, 4, 5, 1};
vector<int> arr(a, a + sizeof(a)/sizeof(a[0]));






//prt(arr, "Data In:");
calc_LIS(arr);

return 0;
}

Working Code at : http://ideone.com/lOrktE