# Dynamic Programming | Set 1 (Overlapping Subproblems Property)

Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems and stores the results of subproblems to avoid computing the same results again. Following are the two main properties of a problem that suggest that the given problem can be solved using Dynamic programming.

In this post, we will discuss first property (Overlapping Subproblems) in detail. The second property of Dynamic programming is discussed in next post i.e. Set 2.

1) Overlapping Subproblems
2) Optimal Substructure

1) Overlapping Subproblems:
Like Divide and Conquer, Dynamic Programming combines solutions to sub-problems. Dynamic Programming is mainly used when solutions of same subproblems are needed again and again. In dynamic programming, computed solutions to subproblems are stored in a table so that these don’t have to recomputed. So Dynamic Programming is not useful when there are no common (overlapping) subproblems because there is no point storing the solutions if they are not needed again. For example, Binary Search doesn’t have common subproblems. If we take example of following recursive program for Fibonacci Numbers, there are many subproblems which are solved again and again.

/* simple recursive program for Fibonacci numbers */
int fib(int n)
{
if ( n <= 1 )
return n;
return fib(n-1) + fib(n-2);
}

Recursion tree for execution of fib(5)

fib(5)
/             \
fib(4)                fib(3)
/      \                /     \
fib(3)      fib(2)         fib(2)    fib(1)
/     \        /    \       /    \
fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
/    \
fib(1) fib(0)

We can see that the function f(3) is being called 2 times. If we would have stored the value of f(3), then instead of computing it again, we could have reused the old stored value. There are following two different ways to store the values so that these values can be reused:
a) Memoization (Top Down)
b) Tabulation (Bottom Up)

a) Memoization (Top Down): The memoized program for a problem is similar to the recursive version with a small modification that it looks into a lookup table before computing solutions. We initialize a lookup array with all initial values as NIL. Whenever we need solution to a subproblem, we first look into the lookup table. If the precomputed value is there then we return that value, otherwise we calculate the value and put the result in lookup table so that it can be reused later.

Following is the memoized version for nth Fibonacci Number.

## C/C++

/* C/C++ program for Memoized version for nth Fibonacci number */
#include<stdio.h>
#define NIL -1
#define MAX 100

int lookup[MAX];

/* Function to initialize NIL values in lookup table */
void _initialize()
{
int i;
for (i = 0; i < MAX; i++)
lookup[i] = NIL;
}

/* function for nth Fibonacci number */
int fib(int n)
{
if (lookup[n] == NIL)
{
if (n <= 1)
lookup[n] = n;
else
lookup[n] = fib(n-1) + fib(n-2);
}

return lookup[n];
}

int main ()
{
int n = 40;
_initialize();
printf("Fibonacci number is %d ", fib(n));
return 0;
}

## Java

/* Java program for Memoized version */
public class Fibonacci
{
final int MAX = 100;
final int NIL = -1;

int lookup[] = new int[MAX];

/* Function to initialize NIL values in lookup table */
void _initialize()
{
for (int i = 0; i < MAX; i++)
lookup[i] = NIL;
}

/* function for nth Fibonacci number */
int fib(int n)
{
if (lookup[n] == NIL)
{
if (n <= 1)
lookup[n] = n;
else
lookup[n] = fib(n-1) + fib(n-2);
}
return lookup[n];
}

public static void main(String[] args)
{
Fibonacci f = new Fibonacci();
int n = 40;
f._initialize();
System.out.println("Fibonacci number is" + " " + f.fib(n));
}

}
// This Code is Contributed by Saket Kumar

## Python

# Python program for Memoized version of nth Fibonacci number

# Function to calculate nth Fibonacci number
def fib(n, lookup):

# Base case
if n == 0 or n == 1 :
lookup[n] = n

# If the value is not calculated previously then calculate it
if lookup[n] is None:
lookup[n] = fib(n-1 , lookup)  + fib(n-2 , lookup)

# return the value corresponding to that value of n
return lookup[n]
# end of function

# Driver program to test the above function
def main():
n = 34
# Declaration of lookup table
# Handles till n = 100
lookup = [None]*(101)
print "Fibonacci Number is ", fib(n, lookup)

if __name__=="__main__":
main()

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

b) Tabulation (Bottom Up): The tabulated program for a given problem builds a table in bottom up fashion and returns the last entry from table. For example, for the same Fibonacci number, we first calculate fib(0) then fib(1) then fib(2) then fib(3) and so on. So literally, we are building the solutions of subproblems bottom-up.

Following is the tabulated version for nth Fibonacci Number.

## C/C++

/* C program for Tabulated version */
#include<stdio.h>
int fib(int n)
{
int f[n+1];
int i;
f[0] = 0;   f[1] = 1;
for (i = 2; i <= n; i++)
f[i] = f[i-1] + f[i-2];

return f[n];
}

int main ()
{
int n = 9;
printf("Fibonacci number is %d ", fib(n));
return 0;
}

## Java

/* Java program for Tabulated version */
public class Fibonacci
{
int fib(int n)
{
int f[] = new int[n+1];
f[0] = 0;
f[1] = 1;
for (int i = 2; i <= n; i++)
f[i] = f[i-1] + f[i-2];
return f[n];
}

public static void main(String[] args)
{
Fibonacci f = new Fibonacci();
int n = 9;
System.out.println("Fibonacci number is" + " " + f.fib(n));
}

}
// This Code is Contributed by Saket Kumar

## Python

# Python program Tabulated (bottom up) version
def fib(n):

# array declaration
f = [0]*(n+1)

# base case assignment
f[1] = 1

# calculating the fibonacci and storing the values
for i in xrange(2 , n+1):
f[i] = f[i-1] + f[i-2]
return f[n]

# Driver program to test the above function
def main():
n = 9
print "Fibonacci number is " , fib(n)

if __name__=="__main__":
main()

# This code is contributed by Nikhil Kumar Singh (nickzuck_007)

Output:

Fibonacci number is 34

Both Tabulated and Memoized store the solutions of subproblems. In Memoized version, table is filled on demand while in Tabulated version, starting from the first entry, all entries are filled one by one. Unlike the Tabulated version, all entries of the lookup table are not necessarily filled in Memoized version. For example, Memoized solution of the LCS problem doesn’t necessarily fill all entries.

To see the optimization achieved by Memoized and Tabulated solutions over the basic Recursive solution, see the time taken by following runs for calculating 40th Fibonacci number:

Time taken by Recursion method is much more than the two Dynamic Programming techniques mentioned above – Memoization and Tabulation!

Also see method 2 of Ugly Number post for one more simple example where we have overlapping subproblems and we store the results of subproblems.

We will be covering Optimal Substructure Property and some more example problems in future posts on Dynamic Programming.

Try following questions as an exercise of this post.
1) Write a Memoized solution for LCS problem. Note that the Tabular solution is given in the CLRS book.
2) How would you choose between Memoization and Tabulation?

References:

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