Longest Increasing Subsequence | DP-3

• Difficulty Level : Medium
• Last Updated : 17 Jan, 2022

We have already discussed Overlapping Subproblems and Optimal Substructure properties.
Now, let us discuss the Longest Increasing Subsequence (LIS) problem as an example problem that can be solved using Dynamic Programming.

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}

Method 1: Recursion.
Optimal Substructure: Let arr[0..n-1] be the input array and L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS.

Then, L(i) can be recursively written as:

L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or
L(i) = 1, if no such j exists.

To find the LIS for a given array, we need to return max(L(i)) where 0 < i < n.
Formally, the length of the longest increasing subsequence ending at index i, will be 1 greater than the maximum of lengths of all longest increasing subsequences ending at indices before i, where arr[j] < arr[i] (j < i).
Thus, we see the LIS problem satisfies the optimal substructure property as the main problem can be solved using solutions to subproblems.

The recursive tree given below will make the approach clearer:

Input  : arr[] = {3, 10, 2, 11}
f(i): Denotes LIS of subarray ending at index 'i'

(LIS(1)=1)

f(4)  {f(4) = 1 + max(f(1), f(2), f(3))}
/    |    \
f(1)  f(2)  f(3) {f(3) = 1, f(2) and f(1) are > f(3)}
|      |  \
f(1)  f(2)  f(1) {f(2) = 1 + max(f(1)}
|
f(1) {f(1) = 1}

Below is the implementation of the recursive approach:

C++

 /* A Naive C++ recursive implementationof LIS problem */#include using namespace std;  /* To make use of recursive calls, thisfunction must return two things:1) Length of LIS ending with element arr[n-1].    We use max_ending_here for this purpose2) Overall maximum as the LIS may end with    an element before arr[n-1] max_ref is    used this purpose.The value of LIS of full array of size nis stored in *max_ref which is our final result*/int _lis(int arr[], int n, int* max_ref){    /* Base case */    if (n == 1)        return 1;      // 'max_ending_here' is length of LIS    // ending with arr[n-1]    int res, max_ending_here = 1;      /* Recursively get all LIS ending with arr,    arr ... arr[n-2]. If arr[i-1] is smaller    than arr[n-1], and max ending with arr[n-1]    needs to be updated, then update it */    for (int i = 1; i < n; i++) {        res = _lis(arr, i, max_ref);        if (arr[i - 1] < arr[n - 1]            && res + 1 > max_ending_here)            max_ending_here = res + 1;    }      // Compare max_ending_here with the overall    // max. And update the overall max if needed    if (*max_ref < max_ending_here)        *max_ref = max_ending_here;      // Return length of LIS ending with arr[n-1]    return max_ending_here;}  // The wrapper function for _lis()int lis(int arr[], int n){    // The max variable holds the result    int max = 1;      // The function _lis() stores its result in max    _lis(arr, n, &max);      // returns max    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);    cout <<"Length of lis is "<< lis(arr, n);    return 0;}// This code is contributed by shivanisinghss2110

C

 /* A Naive C recursive implementationof LIS problem */#include #include   /* To make use of recursive calls, thisfunction must return two things:1) Length of LIS ending with element arr[n-1].    We use max_ending_here for this purpose2) Overall maximum as the LIS may end with    an element before arr[n-1] max_ref is    used this purpose.The value of LIS of full array of size nis stored in *max_ref which is our final result*/int _lis(int arr[], int n, int* max_ref){    /* Base case */    if (n == 1)        return 1;      // 'max_ending_here' is length of LIS    // ending with arr[n-1]    int res, max_ending_here = 1;      /* Recursively get all LIS ending with arr,    arr ... arr[n-2]. If arr[i-1] is smaller    than arr[n-1], and max ending with arr[n-1]    needs to be updated, then update it */    for (int i = 1; i < n; i++) {        res = _lis(arr, i, max_ref);        if (arr[i - 1] < arr[n - 1]            && res + 1 > max_ending_here)            max_ending_here = res + 1;    }      // Compare max_ending_here with the overall    // max. And update the overall max if needed    if (*max_ref < max_ending_here)        *max_ref = max_ending_here;      // Return length of LIS ending with arr[n-1]    return max_ending_here;}  // The wrapper function for _lis()int lis(int arr[], int n){    // The max variable holds the result    int max = 1;      // The function _lis() stores its result in max    _lis(arr, n, &max);      // returns max    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);    printf("Length of lis is %d", lis(arr, n));    return 0;}

Java

 /* A Naive Java Program for LIS Implementation */class LIS {    static int max_ref; // stores the LIS      /* To make use of recursive calls, this function must    return two things: 1) Length of LIS ending with element    arr[n-1]. We use max_ending_here for this purpose 2)    Overall maximum as the LIS may end with an element       before arr[n-1] max_ref is used this purpose.    The value of LIS of full array of size n is stored in    *max_ref which is our final result */    static int _lis(int arr[], int n)    {        // base case        if (n == 1)            return 1;          // 'max_ending_here' is length of LIS ending with        // arr[n-1]        int res, max_ending_here = 1;          /* Recursively get all LIS ending with arr,           arr ... arr[n-2]. If   arr[i-1] is smaller           than arr[n-1], and max ending with arr[n-1] needs           to be updated, then update it */        for (int i = 1; i < n; i++) {            res = _lis(arr, i);            if (arr[i - 1] < arr[n - 1]                && res + 1 > max_ending_here)                max_ending_here = res + 1;        }          // Compare max_ending_here with the overall max. And        // update the overall max if needed        if (max_ref < max_ending_here)            max_ref = max_ending_here;          // Return length of LIS ending with arr[n-1]        return max_ending_here;    }      // The wrapper function for _lis()    static int lis(int arr[], int n)    {        // The max variable holds the result        max_ref = 1;          // The function _lis() stores its result in max        _lis(arr, n);          // returns max        return max_ref;    }      // driver program to test above functions    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)                           + "\n");    }}/*This code is contributed by Rajat Mishra*/

Python3

 # A naive Python implementation of LIS problem  """ To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in *max_ref which is our final result """  # global variable to store the maximumglobal maximum    def _lis(arr, n):      # to allow the access of global variable    global maximum      # Base Case    if n == 1:        return 1      # maxEndingHere is the length of LIS ending with arr[n-1]    maxEndingHere = 1      """Recursively get all LIS ending with arr, arr..arr[n-2]       IF arr[n-1] is smaller than arr[n-1], and max ending with       arr[n-1] needs to be updated, then update it"""    for i in range(1, n):        res = _lis(arr, i)        if arr[i-1] < arr[n-1] and res+1 > maxEndingHere:            maxEndingHere = res + 1      # Compare maxEndingHere with overall maximum. And    # update the overall maximum if needed    maximum = max(maximum, maxEndingHere)      return maxEndingHere    def lis(arr):      # to allow the access of global variable    global maximum      # length of arr    n = len(arr)      # maximum variable holds the result    maximum = 1      # The function _lis() stores its result in maximum    _lis(arr, n)      return maximum    # Driver program to test the above functionarr = [10, 22, 9, 33, 21, 50, 41, 60]n = len(arr)print ("Length of lis is ", lis(arr))  # This code is contributed by NIKHIL KUMAR SINGH

C#

 using System;  /* A Naive C# Program for LIS Implementation */class LIS {    static int max_ref; // stores the LIS      /* To make use of recursive calls, this function must    return two things: 1) Length of LIS ending with element    arr[n-1]. We use max_ending_here for this purpose 2)    Overall maximum as the LIS may end with an element       before arr[n-1] max_ref is used this purpose.    The value of LIS of full array of size n is stored in    *max_ref which is our final result */    static int _lis(int[] arr, int n)    {        // base case        if (n == 1)            return 1;          // 'max_ending_here' is length of LIS ending with        // arr[n-1]        int res, max_ending_here = 1;          /* Recursively get all LIS ending with arr,           arr ... arr[n-2]. If   arr[i-1] is smaller           than arr[n-1], and max ending with arr[n-1] needs           to be updated, then update it */        for (int i = 1; i < n; i++) {            res = _lis(arr, i);            if (arr[i - 1] < arr[n - 1]                && res + 1 > max_ending_here)                max_ending_here = res + 1;        }          // Compare max_ending_here with the overall max. And        // update the overall max if needed        if (max_ref < max_ending_here)            max_ref = max_ending_here;          // Return length of LIS ending with arr[n-1]        return max_ending_here;    }      // The wrapper function for _lis()    static int lis(int[] arr, int n)    {        // The max variable holds the result        max_ref = 1;          // The function _lis() stores its result in max        _lis(arr, n);          // returns max        return max_ref;    }      // driver program to test above functions    public static void Main()    {        int[] arr = { 10, 22, 9, 33, 21, 50, 41, 60 };        int n = arr.Length;        Console.Write("Length of lis is " + lis(arr, n)                      + "\n");    }}

Javascript


Output
Length of lis is 5

Output:

Length of lis is 5

Complexity Analysis:

• Time Complexity: The time complexity of this recursive approach is exponential as there is a case of overlapping subproblems as explained in the recursive tree diagram above.
• Auxiliary Space: O(1). No external space used for storing values apart from the internal stack space.

Method 2: Dynamic Programming.
We can see that there are many subproblems in the above recursive solution which 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.

The simulation of approach will make things clear:

Input  : arr[] = {3, 10, 2, 11}
LIS[] = {1, 1, 1, 1} (initially)

Iteration-wise simulation :

1. arr > arr {LIS = max(LIS , LIS+1)=2}
2. arr < arr {No change}
3. arr < arr {No change}
4. arr > arr {LIS = max(LIS , LIS+1)=2}
5. arr > arr {LIS = max(LIS , LIS+1)=3}
6. arr > arr {LIS = max(LIS , LIS+1)=3}

We can avoid recomputation of subproblems by using tabulation as shown in the below code:

Below is the implementation of the above approach:

C++

 /* Dynamic Programming C++ implementation   of LIS problem */#include 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[n];      lis = 1;      /* Compute optimized LIS values in       bottom up manner */    for (int i = 1; i < n; i++) {        lis[i] = 1;        for (int j = 0; j < i; j++)            if (arr[i] > arr[j] && lis[i] < lis[j] + 1)                lis[i] = lis[j] + 1;    }      // Return maximum value in lis[]    return *max_element(lis, lis + n);}  /* Driver program to test above function */int main(){    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };    int n = sizeof(arr) / sizeof(arr);    printf("Length of lis is %d\n", lis(arr, n));      return 0;}

Java

 /* Dynamic Programming Java implementation   of LIS problem */  class LIS {    /* lis() returns the length of the longest       increasing subsequence in arr[] of size n */    static int lis(int arr[], int n)    {        int lis[] = new int[n];        int i, j, max = 0;          /* 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];          return max;    }      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)                           + "\n");    }}/*This code is contributed by Rajat Mishra*/

Python3

 # Dynamic programming Python implementation# of LIS problem  # lis returns length of the longest# increasing subsequence in arr of size n    def lis(arr):    n = len(arr)      # Declare the list (array) for LIS and    # initialize LIS values for all indexes    lis = *n      # Compute optimized LIS values in bottom up manner    for i in range(1, n):        for j in range(0, i):            if arr[i] > arr[j] and lis[i] < lis[j] + 1:                lis[i] = lis[j]+1      # Initialize maximum to 0 to get    # the maximum of all LIS    maximum = 0      # Pick maximum of all LIS values    for i in range(n):        maximum = max(maximum, lis[i])      return maximum# end of lis function    # Driver program to test above functionarr = [10, 22, 9, 33, 21, 50, 41, 60]print ("Length of lis is", lis(arr))# This code is contributed by Nikhil Kumar Singh

C#

 /* Dynamic Programming C# implementation of LIS problem */  using System;class LIS {    /* lis() returns the length of the longest increasing    subsequence in arr[] of size n */    static int lis(int[] arr, int n)    {        int[] lis = new int[n];        int i, j, max = 0;          /* 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];          return max;    }      public static 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)                          + "\n");    }      // This code is contributed by Ryuga}

Javascript


Output
Length of lis is 5

Complexity Analysis:

• Time Complexity: O(n2).
As nested loop is used.
• Auxiliary Space: O(n).
Use of any array to store LIS values at each index.

Note: The time complexity of the above Dynamic Programming (DP) solution is O(n^2) and there is a O(N log N) solution for the LIS problem. We have not discussed the O(N log N) solution here as the purpose of this post is to explain Dynamic Programming with a simple example. See below post for O(N log N) solution.
Longest Increasing Subsequence Size (N log N)

Method 3: Dynamic Programming

If we closely observe the problem then we can convert this problem to longest Common Subsequence Problem. Firstly we will create another array of unique elements of original array and sort it. Now the longest increasing subsequence of our array must be present as a subsequence in our sorted array. That’s why our problem is now reduced to finding the common subsequence between the two arrays.

Eg. arr =[50,3,10,7,40,80]
// Sorted array
arr1 = [3,7,10,40,50,80]
// LIS is longest common subsequence between the two arrays
ans = 4
The longest increasing subsequence is {3, 7, 40, 80}

C++

 // Dynamic Programming Approach of Finding LIS by reducing// the problem to longest common Subsequence#include using namespace std;  /* lis() returns the length of the longest  increasing subsequence in arr[] of size n */int lis(int arr[], int n){    vector b;    set s;        // setting iterator for set    set::iterator it;        // storing unique elements    for (int i = 0; i < n; i++) {        s.insert(arr[i]);    }        // creating sorted vector    for (it = s.begin(); it != s.end(); it++) {        b.push_back(*it);    }    int m = b.size(), i, j;    int dp[m + 1][n + 1];        // storing -1 in dp multidimentional array    for (i = 0; i < m + 1; i++) {        for (j = 0; j < n + 1; j++) {            dp[i][j] = -1;        }    }    // Finding Longest common Subsequence of the two    // arrays    for (i = 0; i < m + 1; i++) {        for (j = 0; j < n + 1; j++) {            if (i == 0 || j == 0) {                dp[i][j] = 0;            }            else if (arr[i - 1] == b[j - 1]) {                dp[i][j] = 1 + dp[i - 1][j - 1];            }            else {                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);            }        }    }    return dp[m][n];}  /* Driver program to test above function */int main(){    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };    int n = sizeof(arr) / sizeof(arr);    printf("Length of lis is %d\n", lis(arr, n));}  /* This code is contributed by Arun Bang */

Java

 import static java.lang.Math.max;  import java.util.SortedSet;import java.util.TreeSet;  // Dynamic Programming Approach of Finding LIS by reducing// the problem to longest common Subsequencepublic class Main {      /* lis() returns the length of the longest    increasing subsequence in arr[] of size n */    static int lis(int arr[], int n)    {        SortedSet hs = new TreeSet();        // Storing and Sorting unique elements.        for (int i = 0; i < n; i++)            hs.add(arr[i]);        int lis[] = new int[hs.size()];        int k = 0;        // Storing all the unique values in a sorted manner.        for (int val : hs) {            lis[k] = val;            k++;        }        int m = k, i, j;        int dp[][] = new int[m + 1][n + 1];          // Storing -1 in dp multidimensional array.        for (i = 0; i < m + 1; i++) {            for (j = 0; j < n + 1; j++) {                dp[i][j] = -1;            }        }          // Finding the Longest Common Subsequence of the two        // arrays        for (i = 0; i < m + 1; i++) {            for (j = 0; j < n + 1; j++) {                if (i == 0 || j == 0) {                    dp[i][j] = 0;                }                else if (arr[i - 1] == lis[j - 1]) {                    dp[i][j] = 1 + dp[i - 1][j - 1];                }                else {                    dp[i][j]                        = max(dp[i - 1][j], dp[i][j - 1]);                }            }        }        return dp[m][n];    }      // Driver Program for the above test function.    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)                           + "\n");    }}  // This Code is Contributed by Omkar Subhash Ghongade

Python3

 # Dynamic Programming Approach of Finding LIS by reducing the problem to longest common Subsequence    def lis(a):    n = len(a)    # Creating the sorted list    b = sorted(list(set(a)))    m = len(b)      # Creating dp table for storing the answers of sub problems    dp = [[-1 for i in range(m+1)] for j in range(n+1)]      # Finding Longest common Subsequence of the two arrays    for i in range(n+1):          for j in range(m+1):            if i == 0 or j == 0:                dp[i][j] = 0            elif a[i-1] == b[j-1]:                dp[i][j] = 1+dp[i-1][j-1]            else:                dp[i][j] = max(dp[i-1][j], dp[i][j-1])    return dp[-1][-1]    # Driver program to test above functionarr = [10, 22, 9, 33, 21, 50, 41, 60]print("Length of lis is ", lis(arr))# This code is Contributed by Dheeraj Khatri
Output
Length of lis is  5

Complexity Analysis : O(n*n)

As nested loop is used

Space Complexity : O(n*n)

As a matrix is used for storing the values.