C++ Program for 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, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}. 

Examples:

Input  : arr[] = {3, 10, 2, 1, 20}
Output : Length of LIS = 3
The longest increasing subsequence is 3, 10, 20

Input  : arr[] = {3, 2}
Output : Length of LIS = 1
The longest increasing subsequences are {3} and {2}

Input : arr[] = {50, 3, 10, 7, 40, 80}
Output : Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}

Overlapping Subproblems:
Considering the above implementation, the following is a recursion tree for an array of size 4. lis(n) gives us the length of LIS for arr[].  

              lis(4)
        /        |     
      lis(3)    lis(2)   lis(1)
     /           /
   lis(2) lis(1) lis(1)
   /
lis(1)

We can see that there are many subproblems that are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. Following is a tabulated implementation for the LIS problem. 

C++

/* Dynamic Programming C/C++ implementation of LIS problem */
#include <iostream>
using namespace std;

/* lis() returns the length of the longest increasing
  subsequence in arr[] of size n */
int lis(int arr[], int n)
{
    int *lis, i, j, max = 0;
    lis = (int*)malloc(sizeof(int) * n);

    /* Initialize LIS values for all indexes */
    for (i = 0; i < n; i++)
        lis[i] = 1;

    /* Compute optimized LIS values in bottom up manner */
    for (i = 1; i < n; i++)
        for (j = 0; j < i; j++)
            if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
                lis[i] = lis[j] + 1;

    /* Pick maximum of all LIS values */
    for (i = 0; i < n; i++)
        if (max < lis[i])
            max = lis[i];

    /* Free memory to avoid memory leak */
    free(lis);

    return max;
}

/* Driver program to test above function */
int main()
{
    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout <<"Length of lis is "<< lis(arr, n);
    return 0;
}


// this code is contributed by shivanisinghss2110

C

/* Dynamic Programming C/C++ implementation of LIS problem */
#include <stdio.h>
#include <stdlib.h>

/* lis() returns the length of the longest increasing
  subsequence in arr[] of size n */
int lis(int arr[], int n)
{
    int *lis, i, j, max = 0;
    lis = (int*)malloc(sizeof(int) * n);

    /* Initialize LIS values for all indexes */
    for (i = 0; i < n; i++)
        lis[i] = 1;

    /* Compute optimized LIS values in bottom up manner */
    for (i = 1; i < n; i++)
        for (j = 0; j < i; j++)
            if (arr[i] > arr[j] && lis[i] < lis[j] + 1)
                lis[i] = lis[j] + 1;

    /* Pick maximum of all LIS values */
    for (i = 0; i < n; i++)
        if (max < lis[i])
            max = lis[i];

    /* Free memory to avoid memory leak */
    free(lis);

    return max;
}

/* Driver program to test above function */
int main()
{
    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Length of lis is %d\n", lis(arr, n));
    return 0;
}

Java

/* Dynamic Programming Java implementation
of LIS problem */
import java.util.*;

class GFG
{

    /*
    * lis() returns the length of the longest
    * increasing subsequence in arr[] of size n
    */
    static int lis(int[] arr, int n)
    {
        int max = 0;
        int[] lst = new int[n];

        // initialize LIS values for all indexes
        Arrays.fill(lst, 1);

        /* Compute optimized LIS values
        in bottom up manner */
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                if (arr[i] > arr[j] &&
                    lst[i] < lst[j] + 1)
                    lst[i] = lst[j] + 1;
            }
        }

        /* Pick maximum of all LIS values */
        for (int i = 0; i < n; i++)
            if (max < lst[i])
                max = lst[i];

        return max;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
        int n = arr.length;
        System.out.println("Length of lis is " +
                                   lis(arr, n));
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Dynamic Programming python3
# implementation of LIS problem

# lis() returns the length of the
# longest increasing subsequence
# in arr[] of size n
def lis(arr, n):
    i, j, maxm = 0, 0, 0

    # initialize LIS values for all indexes
    lst = [1 for s in range(n)]

    for i in range(1, n):
        for j in range(0, i):
            if (arr[i] > arr[j] and
                lst[i] < lst[j] + 1):
                lst[i] = lst[j] + 1

    # Pick maximum of all LIS values
    for i in range(0, n):
        if maxm < lst[i]:
            maxm = lst[i]

    return maxm

# Driver Code
arr = [10, 22, 9, 33, 21, 50, 41, 60]
n = len(arr)
print("Length of lst is", lis(arr, n))

# This code is contributed
# by Mohit kumar 29

C#

/* Dynamic Programming C# implementation
of LIS problem */
using System;
public class GFG
{

  /*
    * lis() returns the length of the longest
    * increasing subsequence in arr[] of size n
    */
  static int lis(int[] arr, int n)
  {
    int max = 0;
    int[] lst = new int[n];

    // initialize LIS values for all indexes
    Array.Fill(lst, 1);

    /* Compute optimized LIS values
        in bottom up manner */
    for (int i = 1; i < n; i++)
    {
      for (int j = 0; j < i; j++)
      {
        if (arr[i] > arr[j] && lst[i] < lst[j] + 1)
        {
          lst[i] = lst[j] + 1;
        }
      }
    }

    /* Pick maximum of all LIS values */
    for (int i = 0; i < n; i++)
      if (max < lst[i])
        max = lst[i];
    return max;
  }

  // Driver code
  static public void Main ()
  {
    int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };
    int n = arr.Length;
    Console.WriteLine("Length of lis is " +  lis(arr, n));
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>
/* Dynamic Programming Javascript implementation
of LIS problem */

    /*
    * lis() returns the length of the longest
    * increasing subsequence in arr[] of size n
    */
    function lis(arr,n)
    {
        let max = 0;
        let lst = new Array(n);

        // initialize LIS values for all indexes
        for(let i=0;i<lst.length;i++)
        {
            lst[i]=1;
        }


        /* Compute optimized LIS values
        in bottom up manner */
        for (let i = 1; i < n; i++)
        {
            for (let j = 0; j < i; j++)
            {
                if (arr[i] > arr[j] &&
                    lst[i] < lst[j] + 1)
                    lst[i] = lst[j] + 1;
            }
        }

        /* Pick maximum of all LIS values */
        for (let i = 0; i < n; i++)
            if (max < lst[i])
                max = lst[i];

        return max;
    }

    // Driver Code
    let arr=[10, 22, 9, 33, 21, 50, 41, 60 ];
    let n = arr.length;
    document.write("Length of lis is " +
                                   lis(arr, n));


// This code is contributed by patel2127
</script>

Output:

Length of lis is 5

Complexity Analysis:

Time Complexity: O(n²). As nested loop is used.
Auxiliary Space: O(n). 

Please refer complete article on Dynamic Programming | Set 3 (Longest Increasing Subsequence) for more details!

Method 2 : Lower Bound based approach

Algorithm :

1. Iterate the array.

2. Declare a new array ans to add the newly constructed increasing subsequence.

2. For every index, if lower_bound is points to the ending of the array ans, push it into a vector ans.

3. Return the ans array size.

C++

#include <bits/stdc++.h>
using namespace std;
int longest_increasing_subsequence(vector<int>& arr)
{
    vector<int> ans;
    int n = arr.size();
    for (int i = 0; i < n; i++) {
        auto it
            = lower_bound(ans.begin(), ans.end(), arr[i]);
        if (it == ans.end()) {
            ans.push_back(arr[i]);
        }
        else {
            *it = arr[i];
        }
    }
    return ans.size();
}
int main()
{
    vector<int> a = { 10, 22, 9, 33, 21, 50, 41, 60 };
    int ans = longest_increasing_subsequence(a);
    cout << ans;
    return 0;
}

Java

// Java code for above approach
import java.util.*;

public class Main {
    public static int longestIncreasingSubsequence(List<Integer> arr) {
        List<Integer> ans = new ArrayList<>();
        int n = arr.size();
        for (int i = 0; i < n; i++) {
            int it = arr.get(i);
            int idx = Collections.binarySearch(ans, it);
            if (idx < 0) {
                idx = -(idx + 1);
                if (idx == ans.size()) {
                    ans.add(it);
                } else {
                    ans.set(idx, it);
                }
            }
        }
        return ans.size();
    }

    public static void main(String[] args) {
        List<Integer> a = Arrays.asList(10, 22, 9, 33, 21, 50, 41, 60);
        int ans = longestIncreasingSubsequence(a);
        System.out.println(ans);
    }
}

// This code is contributed by Aman Kumar.

C#

using System;
using System.Collections.Generic;

public class Program {
    public static int
    LongestIncreasingSubsequence(List<int> arr)
    {
        List<int> ans = new List<int>();
        List<int> prevIdx = new List<int>();
        int n = arr.Count;

        for (int i = 0; i < n; i++) {
            var it = ans.BinarySearch(arr[i]);
            if (it < 0) {
                it = ~it;

                if (it == ans.Count) {
                    ans.Add(arr[i]);
                    prevIdx.Add(ans.Count - 2);
                }
                else {
                    ans[it] = arr[i];
                    prevIdx.Add(it == 0 ? -1
                                        : prevIdx[it - 1]);
                }
            }
            else {
                prevIdx.Add(it == 0 ? -1 : prevIdx[it - 1]);
            }
        }

        int len = ans.Count;
        List<int> subseq = new List<int>();
        int idx = len - 1;

        while (idx >= 0) {
            subseq.Add(ans[idx]);
            idx = prevIdx[idx];
        }

        subseq.Reverse();
        return len;
    }

    public static void Main()
    {
        List<int> a = new List<int>{ 10, 22, 9,  33,
                                     21, 50, 41, 60 };
        int ans = LongestIncreasingSubsequence(a);
        Console.WriteLine(ans);
    }
}
// This code is contributed by user_dtewbxkn77n

Output

5

Time Complexity: O(nlogn) , n = size of array.

Space Complexity: O(n)