What is Dynamic Programming?

Introduction

0/1 Knapsack

Solution: 0/1 Knapsack

Equal Subset Sum Partition

Solution: Equal Subset Sum Partition

Subset Sum

Solution: Subset Sum

Minimum Subset Sum Difference

Solution: Minimum Subset Sum Difference

Count of Subset Sum

Solution: Count of Subset Sum

Target Sum

Solution: Target Sum

Pattern 1: 0/1 Knapsack

Unbounded Knapsack

Solution: Unbounded Knapsack

Rod Cutting

Solution: Rod Cutting

Coin Change

Solution: Coin Change

Minimum Coin Change

Solution: Minimum Coin Change 

Maximum Ribbon Cut

Solution: Maximum Ribbon Cut

Pattern 2: Unbounded Knapsack

Fibonacci numbers

Solution: Fibonacci numbers

Staircase

Solution: Staircase

Number factors

Solution: Number factors

Minimum jumps to reach the end

Solution: Minimum jumps to reach the end

Minimum jumps with fee

Solution: Minimum jumps with fee

House thief

Solution: House thief

Pattern 3: Fibonacci Numbers

Longest Palindromic Subsequence

Solution: Longest Palindromic Subsequence

Longest Palindromic Substring

Solution: Longest Palindromic Substring

Count of Palindromic Substrings

Solution: Count of Palindromic Substrings

Minimum Deletions in a String to make it a Palindrome

Solution: Minimum Deletions in a String to make it a Palindrome

Palindromic Partitioning

Solution: Palindromic Partitioning

Pattern 4: Palindromic Subsequence

Longest Common Substring

Solution: Longest Common Substring

Longest Common Subsequence

Solution: Longest Common Subsequence

Minimum Deletions & Insertions to Transform a String into another

Solution: Minimum Deletions & Insertions to Transform a String into another

Longest Increasing Subsequence

Solution: Longest Increasing Subsequence

Maximum Sum Increasing Subsequence

Solution: Maximum Sum Increasing Subsequence

Shortest Common Super-sequence

Solution: Shortest Common Super-sequence

Minimum Deletions to Make a Sequence Sorted

Solution: Minimum Deletions to Make a Sequence Sorted

Longest Repeating Subsequence

Solution: Longest Repeating Subsequence

Subsequence Pattern Matching

Solution: Subsequence Pattern Matching

Longest Bitonic Subsequence

Solution: Longest Bitonic Subsequence

Longest Alternating Subsequence

Solution: Longest Alternating Subsequence

Edit Distance

Solution: Edit Distance

Strings Interleaving

Solution: Strings Interleaving

Pattern 5: Longest Common Substring

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Appendix

Grokking Dynamic Programming Patterns for Coding Interviews

## Problem Statement
Write a function to calculate the nth Fibonacci number.

Fibonacci numbers are a series of numbers in which each number is the sum of the two preceding numbers. First few Fibonacci numbers are: 0, 1, 1, 2, 3, 5, 8, ...

Mathematically we can define the Fibonacci numbers as:

```js
    Fib(n) = Fib(n-1) + Fib(n-2), for n > 1

    Given that: Fib(0) = 0, and Fib(1) = 1
```

**Constraints:**

- `0 <= n <= 30`

## Basic Solution
A basic solution could be to have a recursive implementation of the mathematical formula discussed above:

The time complexity of the above algorithm is exponential $O(2^n)$ as we are making two recursive calls in the same function. The space complexity is $O(n)$ which is used to store the recursion stack.

Let's visually draw the recursion for `CalculateFibonacci(4)` to see the overlapping subproblems:



We can clearly see the overlapping subproblem pattern: `fib(2)` has been called twice and `fib(1)` has been called thrice. We can optimize this using memoization to store the results for subproblems.

## Top-down Dynamic Programming with Memoization
We can use an array to store the already solved subproblems. Here is the code:


## Bottom-up Dynamic Programming

Let's try to populate our `dp[]` array from the above solution, working in a bottom-up fashion. Since every Fibonacci number is the sum of the previous two numbers, we can use this fact to populate our array.

Here is the code for the bottom-up dynamic programming approach:

The above solution has time and space complexity of $O(n)$.

## Memory optimization
We can optimize the space used in our previous solution. We don't need to store all the Fibonacci numbers up to 'n', as we only need two previous numbers to calculate the next Fibonacci number. We can use this fact to further improve our solution:

The above solution has a time complexity of $O(n)$ but a constant space complexity $O(1)$.
