We have discussed Overlapping Subproblems and Optimal Substructure properties.
Let us discuss 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}.
More 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}
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.
Thus, we see the LIS problem satisfies the optimal substructure property as the main problem can be solved using solutions to subproblems.
Following is a simple recursive implementation of the LIS problem. It follows the recursive structure discussed above.
C/C++
/* A Naive C/C++ recursive implementation of LIS problem */#include<stdio.h> #include<stdlib.h> /* 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 */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[0], arr[1] ... 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[0]); printf("Length of lis is %dn", 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[0], arr[1] ... 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) + "
"); } } /*This code is contributed by Rajat Mishra*/ |
Python
# 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 maximum global 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[0], arr[1]..arr[n-2] IF arr[n-1] is maller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it""" for i in xrange(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 # lenght 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 function arr = [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[0], arr[1] ... 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) + "
"); } } |
Length of lis is 5
Overlapping Subproblems:
Considering the above implementation, following is 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 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. Following is a tabulated implementation for the LIS problem.
C++
/* Dynamic Programming C++ implementation of LIS problem */#include<bits/stdc++.h> 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[0] = 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[0]); printf("Length of lis is %d
", 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 ) + "
" ); } } /*This code is contributed by Rajat Mishra*/ |
Python
# 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 = [1]*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 function arr = [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 ) + "
" ); } // This code is contributed by Ryuga } |
Output:
Length of lis is 5
Note that the time complexity of the above Dynamic Programming (DP) solution is O(n^2) and there is a O(nLogn) 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)
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