Longest Increasing Subsequence (LIS) in $O(N \log N)$

Longest Increasing Subsequence (LIS) in $O(N \log N)$ - Preparation to IOI Part 3

1. Set len = 0.
2. For every position i from 1 to n:
   2.1. Set x = a[i].
   2.2. If len = 0 or tail[len] < x:
        2.2.1. Increase len by 1.
        2.2.2. Set tail[len] = x.
   2.3. Otherwise, find the first position with tail[position] >= x:
        2.3.1. Set l = 1 and r = len.
        2.3.2. While l < r:
               2.3.2.1. Set m = (l + r) / 2.
               2.3.2.2. If tail[m] < x, set l = m + 1.
               2.3.2.3. Otherwise, set r = m.
        2.3.3. Set tail[l] = x.
3. Output len.

#include <bits/stdc++.h>
using namespace std;

const int N = 200200;

int n;
int a[N];
int tail[N];
int len;

int get_pos(int x) {
    int l = 1;
    int r = len;

    while (l < r) {
        int m = (l + r) / 2;
        if (tail[m] < x) {
            l = m + 1;
        } else {
            r = m;
        }
    }

    return l;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    len = 0;
    for (int i = 1; i <= n; i++) {
        if (len == 0 || tail[len] < a[i]) {
            len++;
            tail[len] = a[i];
        } else {
            int p = get_pos(a[i]);
            tail[p] = a[i];
        }
    }

    cout << len << endl;
    return 0;
}urn 0;
}

Longest Increasing Subsequence (LIS) in $O(N \log N)$

Introduction

Why a smaller ending is better

Why binary search appears

Why this only finds the length

Main algorithm

Implementation

Complexity