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Dynamic Programming

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Dynamic Programming Examples

Question : Calculate the nth fibonacci number.

Lets explore the steps to coming up with DP solution :

1) Think of a recursive approach to solving the problem.
This part is simple.
We already know
fibo(n) = fibo(n - 1) + fibo(n - 2)
and we satisfy the condition of Yi < X as
n - 1 < n and n - 2 < n.

2) Write a recursive code for the approach you just thought of.

        int fibo(int n) {
            if (n <= 1) return n;
            return fibo(n - 1) + fibo(n - 2);

Try to think of the time complexity of the above function.
We analyzed it previously in one of the lessons on recursion. You can check it out again.
It is essentially exponential in terms of n.

3) Save the results you get for every function run so that if solve(A1, A2, A3, ... ) is called again, you do not recompute the whole thing.

Ok. So, we try to save the value we calculate somewhere so that we dont compute it again. This is also called memoization.

Lets declare a global variable then.

        int memo[100] = {0};
        int fibo(int n) {
            if (n <= 1) return n;
            // If we have processed this function before, return the result from the last time. 
            if (memo[n] != 0) return memo[n]; 
            // Otherwise calculate the result and remember it. 
            memo[n] = fibo(n - 1) + fibo(n - 2);
            return memo[n];

4) Analyze the space and time requirements, and improve it if possible.

Lets look at the space complexity first. We have an array of size n allocated for storing the results which has space complexity of O(n).
We also have the stack memory overhead of recursion which is also O(n). So, overall space complexity is O(n).

Lets now look at the time complexity.
Lets look at fibo(n).

**Note: ** When fibo(n - 1) is called, it makes a call to fibo(n - 2). So when the call comes back to the original call from fibo(n), fibo(n-2) would already be calculated. Hence the call to fibo(n - 2) will be O(1).

            Hence, T(n) = T(n - 1) + c where c is a constant. 
                    = T(n - 2) + 2c
                    = T(n - 3) + 3c
                    = T(n - k) + kc
                    = T(0) + n * c = 1 + n * c = O(n). 

And voila indeed! Thanks to DP, we reduced a exponential problem to a linear problem.

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Dynamic Programming Problems

Greedy or dp
Problem Score Companies Time Status
Tushar's Birthday Bombs 200
Jump Game Array 225 41:16
Min Jumps Array 300 71:44
Tree dp
Problem Score Companies Time Status
Max edge queries! 200 56:34
Max Sum Path in Binary Tree 400 55:07
Suffix / prefix dp
Derived dp
Problem Score Companies Time Status
Chain of Pairs 200 42:09
Max Sum Without Adjacent Elements 225 58:10
Merge elements 300 57:43
Problem Score Companies Time Status
Flip Array 200
Tushar's Birthday Party 200 70:50
0-1 Knapsack 200 47:57
Equal Average Partition 350 71:49
Problem Score Companies Time Status
Best Time to Buy and Sell Stocks II 225 40:18
Dp optimized backtrack
Problem Score Companies Time Status
Word Break II 350
Multiply dp
Problem Score Companies Time Status
Unique Binary Search Trees II 400 36:06
Count Permutations of BST 400
Breaking words
Problem Score Companies Time Status
Palindrome Partitioning II 400 62:02
Word Break 400

Additional Practice

Problem Score Companies Time Status
Potions 200 56:52
Dice Throw 400 49:10
Double Increasing Series 200 45:07
Dice Rolls 300 27:32
Palindromic Substrings 200
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