LIS O(nlogn) - Codeforces

madcannibal's blog

LIS O(nlogn)

By madcannibal, history, 11 years ago, In English

i wish that anyone explains to me in simple way the algorithm of the Longest Increasing Subsequence in O(nlogn) time , cuz i actually read all online articles about it and found no one explain it well , thanks in advance :)

-1

madcannibal
11 years ago
35

Comments (16)

Show archived | Write comment?

2360

11 years ago, hide # |

i learned LIS from here. it's pretty good :)

→ Reply

Apiram

6 years ago, hide # ^ |

-7

Is this possible to do binarysearch in an unsorted array?

→ Reply

holden

11 years ago, hide # |

the trick for the nlogn solution is to maintain, in addition to your dp table, an auxiliary table A so that A[ i ] holds the information "what is the minimum number in the array, such that an increasing subsequence of length i terminates at", it is easy to see that this array will be strictly increasing. Therefore, we can simply binary search the solution of every subproblem and update our array.

→ Reply

misof

11 years ago, hide # |

+11

It is also worth mentioning that you can do LIS in O(n log n) time simply by improving the O(n^2) DP. Everything you need is coordinate compression and a Fenwick tree / interval tree. The code is longer than with the bsearching algorithm, but for many people the idea is simpler because it uses tools they already know.

→ Reply

shenxy13

6 years ago, hide # ^ |

← Rev. 2 →

I think it can be done without coordinate compression by processing in increasing order of value and using a segment tree in O(n log n).

Edit: sorry for necroposting, didn't realise this blog was from 5 years ago.

→ Reply

SPyofgame

6 years ago, hide # |

← Rev. 3 →

Yes there is

Set implementation

#define itr iterator
...
typedef set<int> si;
...
si S;
for (int x : a) {
    S.insert(x);
    si::itr it = S.upper_bound(x);
    if(it != S.end()) S.erase(it);
}
/// LIS length = S.size();

→ Reply

Elegy_No_Programmer

6 years ago, hide # ^ |

But how to retain the LIS? This is fine for calculating its length. But what about the LIS itself?

→ Reply

TwentyOneHundredOrBust

6 years ago, hide # ^ |

It's not possible with the set method, just use segment tree.

→ Reply

Elegy_No_Programmer

6 years ago, hide # ^ |

Could you tell me how to retain LIS(sequence itself not the length) using segment trees?

→ Reply

TwentyOneHundredOrBust

6 years ago, hide # ^ |

Do the DP and find longest sequence with end value less than current value at index below us, augment the segment tree to return both max and index, then just step back the indices.

→ Reply

RezwanArefin01

6 years ago, hide # ^ |

It is possible with set. At each $$$i$$$, you should store the previous element's index in the LIS that contains $$$a_i$$$. For details see this comment.

→ Reply

SPyofgame

6 years ago, hide # ^ |

Here it is

vector with binary search and finding LIS offline

const int INF = 1e9;
int lis(const vector<int> &a)
{
    int n = a.size();
    vector<int> f(n), b(n);

    /// F[i] is binary search position
    /// B[i] is unordered LIS
    int res = 0;
    for (int i = 0; i < n; ++i) {
        b[i] = +INF;
        f[i] = lower_bound(b.begin(), b.begin() + res + 1, a[i]) - b.begin();
        res = max(res, f[i] + 1);
        b[f[i]] = a[i];          
    }
    /// res is the size of LIS

    vector<int> LIS; /// Finding real LIS
    while (n--)
    {
       if (f[n] == res)
       {
            LIS.emplace_back(a[n]);
            --res;
       }
    }

    reverse(all(LIS);
    return LIS.size();
}

→ Reply

TimeTraveler

6 years ago, hide # ^ |

It does not work if there are duplicates in a. e.g a = [1, 2, 3, 4, 1] It will give 3 but correct ans is 4

→ Reply

SPyofgame

6 years ago, hide # ^ |

Yes it is, sorry for not writting the notice

Nondecreasing Longest Subsequence - With unique

#include <iostream>
#include <vector>
#include <set>

using namespace std;

int main()
{
    int n;
    cin >> n;
    
    vector<int> a(n);
    for (int &x : a) cin >> x;
    
    set<int> S;
    set<int>::iterator it;
    for (int x : a)
    {
        S.insert(x);
        it = S.upper_bound(x);
        if (it != S.end()) S.erase(x);
    }
    
    cout << S.size();
    return 0;
}

Nondecreasing Longest Subsequence - With duplicates

#include <iostream>
#include <vector>
#include <set>

using namespace std;

int main()
{
    int n;
    cin >> n;
    
    vector<int> a(n);
    for (int &x : a) cin >> x;
    
    multiset<int> S;
    multiset<int>::iterator it;
    for (int x : a)
    {
        S.insert(x);
        it = S.upper_bound(x);
        if (it != S.end()) S.erase(it);
    }
    
    cout << S.size();
    return 0;
}

Strict Increasing Longest Subsequence

#include <iostream>
#include <vector>
#include <set>

using namespace std;

int main()
{
    int n;
    cin >> n;
    
    vector<int> a(n);
    for (int &x : a) cin >> x;
    
    set<int> S;
    set<int>::iterator it;
    for (int x : a)
    {
        S.insert(x);
        it = S.upper_bound(x);
        if (it != S.end()) S.erase(it);
    }
    
    cout << S.size();
    return 0;
}

→ Reply

TimeTraveler

6 years ago, hide # ^ |

Found this Implementation with vector

→ Reply

calmachint

6 years ago, hide # |

Well yeah,there are a lot of articles and finally I found a video which explains the NlogN algorithm with an example step by step and the actual idea/logic behind it.You can refer it : https://youtu.be/nf3YG4CnTbg

→ Reply