We have discussed Overlapping Subproblems and Optimal Substructure properties in Set 1 and Set 2 respectively.

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

**Optimal Substructure:**

Let arr[0..n-1] be the input array and L(i) be the length of the LIS till index i such that arr[i] is part of LIS and arr[i] is the last element in LIS, then L(i) can be recursively written as.

*L(i) = { 1 + Max ( L(j) ) } where j < i and arr[j] < arr[i] and if there is no such j then L(i) = 1*

To get LIS of a given array, we need to return max(L(i)) where 0 < i < n
So the LIS problem has optimal substructure property as the main problem can be solved using solutions to subproblems.
**Overlapping Subproblems:**

Following is simple recursive implementation of the LIS problem. The implementation simply follows the recursive structure mentioned above. The value of lis ending with every element is returned using max_ending_here. The overall lis is returned using pointer to a variable max.

## 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 %d\n", lis( arr, n )); return 0; }

## 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 , 41 , 60] n = len(arr) print "Length of LIS is ", lis(arr) # This code is contributed by NIKHIL KUMAR SINGH

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 tabluated implementation for the LIS problem.

## C/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; }

## 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

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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