Overlapping Subproblems Property in Dynamic Programming | DP-1

• Difficulty Level : Easy
• Last Updated : 31 Dec, 2021

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 suggests that the given problem can be solved using Dynamic programming.

In this post, we will discuss the first property (Overlapping Subproblems) in detail. The second property of Dynamic programming is discussed in the 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 the 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 be 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 an example of following recursive program for Fibonacci Numbers, there are many subproblems that are solved again and again.

C

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

Java

 /*package whatever //do not write package name here *//* a simple recursive program for Fibonacci numbers */static int fib(int n){    if (n <= 1)        return n;     return fib(n - 1) + fib(n - 2);} // This code is contributed by umadevi9616

Python

 #  a simple recursive program for Fibonacci numbersdef fib(n):    if n <= 1:        return n     return fib(n - 1) + fib(n - 2)

C#

 /* a simple recursive program for Fibonacci numbers */static int fib(int n){    if (n <= 1)        return n;     return fib(n - 1) + fib(n - 2);}  // This code contributed by umadevi9616

Javascript



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 fib(3) is being called 2 times. If we would have stored the value of fib(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 looks into a lookup table before computing solutions. We initialize a lookup array with all initial values as NIL. Whenever we need the 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 the lookup table so that it can be reused later.

Following is the memoized version for the nth Fibonacci Number.

C++

 /* C++ program for Memoized versionfor nth Fibonacci number */#include using namespace std;#define NIL -1#define MAX 100 int lookup[MAX]; /* Function to initialize NILvalues 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];} // Driver codeint main(){    int n = 40;    _initialize();    cout << "Fibonacci number is " << fib(n);    return 0;} // This is code is contributed by rathbhupendra

C

 /* C program for Memoized version for nth Fibonacci number */#include #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

 # a program for Memoized version of nth Fibonacci number # function to calculate nth Fibonacci number  def fib(n, lookup):     # base case    if 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)

C#

 // C# program for Memoized versionof nth Fibonacci numberusing System; class GFG {     static int MAX = 100;    static int NIL = -1;    static int[] lookup = new int[MAX];     /* Function to initialize NIL    values in lookup table */    static void initialize()    {        for (int i = 0; i < MAX; i++)            lookup[i] = NIL;    }     /* function for nth Fibonacci number */    static 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];    }     // Driver code    public static void Main()    {         int n = 40;        initialize();        Console.Write("Fibonacci number is"                      + " " + fib(n));    }} // This Code is Contributed by Sam007

Javascript


Output
Fibonacci number is 102334155

b) Tabulation (Bottom Up): The tabulated program for a given problem builds a table in bottom-up fashion and returns the last entry from the 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 program for Tabulated version */#include int fib(int n){    int f[n + 1];    int i;    f = 0;    f = 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;        f = 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) versiondef fib(n):     # array declaration    f =  * (n + 1)     # base case assignment    f = 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)

C#

 // C# program for Tabulated versionusing System; class GFG {    static int fib(int n)    {        int[] f = new int[n + 1];        f = 0;        f = 1;        for (int i = 2; i <= n; i++)            f[i] = f[i - 1] + f[i - 2];        return f[n];    }     public static void Main()    {         int n = 9;        Console.Write("Fibonacci number is"                      + " " + fib(n));    }} // This Code is Contributed by Sam007



PHP


Output
Fibonacci number is 34

Both Tabulated and Memoized store the solutions of subproblems. In Memoized version, the table is filled on demand while in the 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 the 40th Fibonacci number:
Recursive solution
Memoized solution
Tabulated solution
Time taken by the Recursion method is much more than the two Dynamic Programming techniques mentioned above – Memorization and Tabulation!

Also, see method 2 of the 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 the 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 Memorization and Tabulation?