The Longest Increasing Subsequence

Problem Statement :

An Introduction to the Longest Increasing Subsequence Problem

The task is to find the length of the longest subsequence in a given array of integers such that all elements of the subsequence are sorted in strictly ascending order. This is called the Longest Increasing Subsequence (LIS) problem.

For example, the length of the LIS for [15,27,14,38,26,55,46,65,85] is 6 since the longest increasing subsequence is [15,27,38,55,65,85].

Here's a great YouTube video of a lecture from MIT's Open-CourseWare covering the topic.

This is one approach which solves this in quadratic time using dynamic programming. A more efficient algorithm which solves the problem in O(n log n) time is available here.

Given a sequence of integers, find the length of its longest strictly increasing subsequence.

Function Description

Complete the longestIncreasingSubsequence function in the editor below. It should return an integer that denotes the array's LIS.

longestIncreasingSubsequence has the following parameter(s):

arr: an unordered array of integers
Input Format

The first line contains a single integer n, the number of elements in arr.
Each of the next n lines contains an integer, arr[i]


1 <= n <= 10^6
1 <= arr[i] <= 10^5

Output Format

Print a single line containing a single integer denoting the length of the longest increasing subsequence.

Solution :


                            Solution in C :

In C++ :

#include <iostream>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <string>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <ctime>

using namespace std;

#define f first
#define s second
#define pb push_back
#define mp make_pair
#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)
#define N 1000100
#define all(a) a.begin(), a.end()
#define ll long long
#define pii pair <int, int>
#define sz(a) (int)a.size()
#define vi vector <int>
#define forn(i, n) for(int i = 0; i < n; i ++)

const int inf = (int)(1e9);
const int mod = inf + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;

int n, a[N];

int main () {

    #ifdef LOCAL
    freopen("", "r", stdin);
    freopen("a.out", "w", stdout);

    cin >> n;
    for(int i = 0; i < n; i++) scanf("%d", a + i);
    vector <int> d(N, 0);

    d[0] = -inf;
    for(int i = 1; i < N; i++) d[i] = inf;

    for(int i = 0; i < n; i++) {
        int j = upper_bound(all(d), a[i]) - d.begin();
        if(d[j - 1] < a[i] && a[i] < d[j]) d[j] = a[i];

    for(int i = N - 1;; i--) {
        if(d[i] != inf) {
            cout << i;
            return 0;

    return 0;

In Java :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner reader = new Scanner(;
        int n = reader.nextInt();
        int[] array = new int[n];
        for(int i = 0 ; i < n; i++ )
            array[i] = reader.nextInt();
    public static int getLengthOfLIS(int[] array)
        TreeSet<Integer> s = new TreeSet<Integer>();
        for( int i = 0 ; i < array.length ; i++)
           //if array[i] is newly added?
           if( s.add(array[i]) )
               //if array[i] is not the last element?
               if( array[i] != s.last() )
                    //remove it's next element; s.higher() gives the least element 
                   //which is greater than array[i]
        return s.size();

In C :

#include <stdio.h>

int n,a[1000000],b[1000000]; 

int getRight(int *Arr,int left,int right,int value)
    int mid;
        else left=mid;
    return right;
int CalcLIS()
    int i,res=1;
        else if(a[i]>b[res-1])
    return res;
int main()
    int i;
    return 0;

In Python3 :

import bisect

N = int(input())
A = [int(input()) for c in range(N)]
P = [0] * N
M = [0] * (N + 1)
L = 0

for i in range(N):
    lo, hi = 1, L
    while lo <= hi:
        mid = lo + (hi-lo) // 2
        if A[M[mid]] < A[i]:
            lo = mid+1
            hi = mid-1
    newL = lo
    P[i] = M[newL-1]
    M[newL] = i
    if newL > L:
        L = newL


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