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


View More Similar Problems

Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →

Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →


This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →