Dynamic programming is an approach to solve a larger problem with the help of the results of smaller subproblems. It is a technique used to avoid computing multiple time the same subproblem in a recursive algorithm. I find a lot of students asking me question around, how do I know this problem is a dynamic programming problem? There is a definite way to arrive at the conclusion if a problem is a dynamic programming problem or not?

The first thing I would recommend you to read before going down is this beautiful explanation of dynamic programming to 4 years old.

The first thing you will notice about dynamic programming problems (not all problems) is they are optimization problem. Either it will be finding minimum or maximum of some entity. For example, find minimum edit between two strings or find longest common subsequence etc. However, problems like Fibonacci series are not exactly like an optimization problem, these are more like Combinatorial problems. Still, this can be a good hint that a problem can be a DP problem.

Second, you will notice that the problem can be divided into a pattern like an `fn(n) = C + fn(n-k)` where k can be anything between 1 and n.

This property is called optimum subproblem structure, where an optimum solution to the subproblem leads to the optimum solution to the larger problem.

Once you get the equation, it is very easy to come with a recursive solution to the problem. I would advise you to write the recursive solution and try to calculate the complexity of the solution. It will exponential in big-O notation.

`fn(j)`

relies on `fn(j − 1)`

. Thus the full recursion tree generally has polynomial depth and an exponential number of nodes.However, it turns out that most of these nodes are repeats, that there are not too many distinct subproblems among them. Efficiency is therefore obtained by explicitly enumerating the distinct subproblems and solving them in the right order.

Reference

This will lead us to the third property, which is overlapping subproblems. Once, you draw the execution tree of the recursive solution of the problem, it will appear that a lot of problems are being solved again and again at different levels of recursion.

The intuition behind dynamic programming is that we trade space for time, i.e. to say that instead of calculating all the subproblems taking a lot of time but no space, we take up space to store the results of all the subproblems to save time later. The typical characteristics of a dynamic programming problem are optimization problems, optimal substructure property, overlapping subproblems, trade space for time, implementation via bottom-up/memoization.

## Dynamic programming in action

Enough of theory, let’s take an example and see how dynamic programming works on real problems. I will take a very commonly used but most effective problem to explain DP in action. Problem is known as the Fibonacci series. Fibonacci series is a series of integers where each integer is the sum of previous two integers. For example, 1,1,2,3,5,8,13,17 is a Fibonacci series of eight integers. Now, the question is given a number n, output the integer which will be at the n^{th} integer in Fibonacci series. For example for n = 4, the output should be 3 and for n=6, it should 8.

First hint: It is a combinatorial problem, so maybe a DP problem. Second, it is already given in the problem that current integer depends on the sum of previous two integers, that means f(n) = f(n-1) + f(n-2). This implies that the solution to subproblems will lead to a solution to the bigger problem which is optimal substructure property.

Next step is to implement the recursive function.

public int fibonacci (int n) { if (n < 2) //base case return 1; return fibonacci(n-1) + fibonacci(n-2); }

Great, next step is to draw the execution tree of this function. It looks like below for n = 6. It is apparent how many times the same problem is solved at different levels.

So, now we know three things about the Fibonacci problem: It is combinatorial problem, there is optimal substructure and there are overlapping subproblems. As in dynamic programming, we side with more space than time, we will try to use extra space to avoid recalculating subproblems.

The first way is to use a case, which stores the value of fab(n) if it is already calculated. This is called memoization or top-down approach.

Map<Integer, Integer> cache = new HashMap<>(); public int fibonacci(int n){ if (n == 0) return 0; if (n == 1) return 1; if(cache.containsKey(n)) return cache.get(n); cache.put(n, fibonacci(n - 1) + fibonacci(n - 2)); return cache.get(n); }

Another approach is bottom up, where the smaller problems are solved in an order which helps us with solving bigger problems. Here also, we use memoization but in a different way. We store the solution of smaller subproblems and directly use this to build the solution.

int[] fib = new int[n]; fib[0] = fib[1] = 1; public int fibonacci(int n){ for(int i=2; i<=n; i++){ fib[n] = fib[n-1] + fib[n-2]; } return fib[n]; }

Above solution requires extra O(n) space, however, the time complexity is also reduced to O(n) with each subproblem solved only once.

Follow longest increasing subsequence problem, how we have applied the same pattern while we solved the problem.

Final thoughts

Where to apply dynamic programming : If you solution is based on optimal substructure and overlapping sub problem then in that case using the earlier calculated value will be useful so you do not have to recompute it. It is bottom up approach. Suppose you need to calculate fib(n) in that case all you need to do is add the previous calculated value of fib(n-1) and fib(n-2)

Recursion : Basically subdividing you problem into smaller part to solve it with ease but keep it in mind it does not avoid re computation if we have same value calculated previously in other recursion call.

Memoization : Basically storing the old calculated recursion value in table is known as memoization which will avoid re-computation if its already been calculated by some previous call so any value will be calculated once. So before calculating we check whether this value has already been calculated or not if already calculated then we return the same from table instead of recomputing. It is also top down approach.

Reference: Answer by Endeavour

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