Climbing Stairs

This problem can be solved using a simple dynamic programming approach, as it exhibits both overlapping subproblems and optimal substructure.

Here’s how you can approach this:

  1. Recognize the Pattern: The number of ways to climb n stairs is the sum of ways to climb n-1 stairs and n-2 stairs. It’s because you can reach the nth step by either taking a single step from n-1 or taking a double step from n-2.

  2. Use a Bottom-Up Approach: Start from the known base cases and build up the solution.

Here’s the code:

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class Solution:
    def climbStairs(self, n: int) -> int:
        if n == 1:
            return 1
        if n == 2:
            return 2

        ways = [0] * (n + 1)
        ways[1] = 1
        ways[2] = 2

        for i in range(3, n + 1):
            ways[i] = ways[i - 1] + ways[i - 2]

        return ways[n]

Explanation:

  • ways[i] represents the number of ways to climb i stairs.
  • The base cases are ways[1] = 1 and ways[2] = 2, representing the number of ways to climb 1 and 2 stairs, respectively.
  • For each subsequent step, you add the previous two ways.

The time complexity of this solution is (O(n)), and the space complexity is also (O(n)), where (n) is the given input. You can further reduce the space complexity to (O(1)) by using only two variables to keep track of the previous two steps instead of a full array.

  1. Base case : n = 1, 1

  2. f(n) = f(n-1) + f(n-2) if n > 1 = 1 for n = 1

  3. Recursion

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# @param {Integer} n
# @return {Integer}

def wrapper(n, memo)
    if memo.key?(n)
        return memo[n]
    end
  if  n == 1 || n == 0
       return 1
   end

   memo[n] = wrapper(n-1, memo) + wrapper(n-2, memo)
   memo[n]
end

def climb_stairs(n)
    memo = {}
    wrapper(n, memo)
end

Identifying Problem Isomorphism

“Climbing Stairs” is approximately isomorphic to “Fibonacci Number”.

In “Climbing Stairs”, you are asked to find the number of distinct ways you can climb to the top of a staircase that has ’n’ steps, given that you can only take 1 or 2 steps at a time.

In “Fibonacci Number”, you are asked to calculate the ’n’th Fibonacci number. The Fibonacci sequence is one where each number is the sum of the two preceding ones, usually starting with 0 and 1.

The reason is that both problems share a similar underlying structure in terms of calculating the ’n’th term based on the two preceding terms. In “Climbing Stairs”, the number of ways to reach the ’n’th step is the sum of the ways to reach the (’n-1’)th step and the (’n-2’)th step. Similarly, the ’n’th Fibonacci number is the sum of the (’n-1’)th and the (’n-2’)th Fibonacci numbers.

“Fibonacci Number” is simpler than “Climbing Stairs”, mainly because the Fibonacci sequence has no constraints. In contrast, “Climbing Stairs” involves a specific problem context, i.e., climbing steps one or two at a time. If we were to arrange them in terms of complexity:

  • “Fibonacci Number”
  • “Climbing Stairs”

“Climbing Stairs” is a basic dynamic programming problem that many users find accessible even without much prior experience with dynamic programming. To build up your problem-solving skills with some simpler problems first, consider:

  1. Two Sum (LeetCode 1): This problem involves finding two numbers in an array that add up to a specific target. It’s a simple introduction to using hash maps for fast lookups.

  2. Reverse Integer (LeetCode 7): This problem requires you to reverse the digits of an integer. It’s a straightforward problem that helps build basic programming skills.

  3. Palindrome Number (LeetCode 9): Here, you need to determine if an integer is a palindrome. It’s another simple problem that can help you get comfortable with coding.

  4. Roman to Integer (LeetCode 13): This problem involves converting a Roman numeral to an integer. It helps build skills in string manipulation.

  5. Valid Parentheses (LeetCode 20): This problem requires you to determine if a string of parentheses is valid. It introduces the concept of using a stack for matching pairs.

  6. Remove Duplicates from Sorted Array (LeetCode 26): This problem involves removing duplicates from a sorted array. It helps build array manipulation skills.

  7. Implement strStr() (LeetCode 28): This problem asks you to implement a basic string searching function. It helps build skills in string manipulation.

  8. Search Insert Position (LeetCode 35): Here, you need to find the index where a target value should be inserted in a sorted array. It’s a good introduction to binary search.

  9. Count and Say (LeetCode 38): This problem involves generating the nth term in a specific sequence. It can help build skills in recursion, which is related to the concept of dynamic programming used in “Climbing Stairs”.

  10. Maximum Subarray (LeetCode 53): This problem asks you to find the contiguous subarray with the maximum sum. It’s a simple dynamic programming problem that can serve as a stepping stone to “Climbing Stairs”.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

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class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        memo ={}
        memo[1] = 1
        memo[2] = 2
        
        def climb(n):
            if n in memo: # if the recurssion already done before first take a look-up in the look-up table
                return memo[n]
            else:   # Store the recurssion function in the look-up table and reuturn the stored look-up table function
                memo[n] =  climb(n-1) + climb(n-2)
                return memo[n]
        
        return climb(n)

  1. Base case : n = 1, 1

  2. f(n) = f(n-1) + f(n-2) if n > 1 = 1 for n = 1

  3. Recursion

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# @param {Integer} n
# @return {Integer}

def wrapper(n, memo)
    if memo.key?(n)
        return memo[n]
    end
  if  n == 1 || n == 0
       return 1
   end
    
   memo[n] = wrapper(n-1, memo) + wrapper(n-2, memo)
   memo[n]
end

def climb_stairs(n)
    memo = {}
    wrapper(n, memo)
end

Problem Classification

  1. Dynamic Programming: The problem uses the concept of “memoization” which is a common technique in dynamic programming. Dynamic programming involves breaking down a larger problem into simpler subproblems and storing the solutions of each subproblem to avoid solving the same subproblem multiple times.

  2. Recursion: The solution involves a function calling itself which is the essence of recursion. Recursion is often used when a problem can be broken down into smaller, simpler subproblems.

  3. Combinatorics: The problem is about counting the number of distinct ways one can climb ’n’ steps, which falls under the branch of mathematics known as combinatorics.

  4. Sequence Generation (Fibonacci Sequence): The number of ways to climb ’n’ steps follows the Fibonacci sequence, where each number (after the first two) is the sum of the two preceding ones. Therefore, this problem can also be seen as generating a sequence based on a specific rule or pattern.

Language Agnostic Coding Drills

The code is a Python implementation of a function to find the number of distinct ways one can climb a staircase with ’n’ steps, where each time you can either climb 1 or 2 steps. The function employs a recursive approach with memoization to improve its efficiency.

Here’s a breakdown of this code into smaller units of learning:

  1. Variables and Data Types: The concept of variables and their types is critical here. The code involves integers (n), dictionaries (memo), and functions (climb).

  2. Functions and/or Methods: The concept of functions is key in this code. We define an inner function (climb) within our main function (climbStairs). This inner function is recursive and calls itself.

  3. Recursion: Recursion is a process in which a function calls itself as a subroutine. This allows the function to be repeated several times, as it can call itself during its execution.

  4. Conditional Statements (If and Else): Conditional statements are used to check whether a specific condition is met. In this case, we check whether ’n’ is already a key in the dictionary ‘memo’.

  5. Dictionary (Hash Map): The code uses a dictionary (also known as a hash map) to store previously computed results (memoization). This avoids redundant calculations and improves efficiency.

  6. Arithmetic Operations: Basic arithmetic operations are used. The addition operation is used to calculate the number of ways to reach the nth step.

Problem-Solving Approach:

  1. Understand the problem: The problem is to find the number of distinct ways one can climb a staircase of ’n’ steps, given one can climb either 1 or 2 steps at a time.

  2. Define the problem requirements: Given an integer ’n’ (number of steps), find the total number of unique ways to reach the top.

  3. Break down the problem: The problem can be broken down into subproblems:

    • The number of ways to reach the nth step is the sum of ways to reach (n-1)th step and (n-2)th step.

  4. Develop a solution: Pseudocode for the solution can be written as follows:

    • Define a dictionary ‘memo’ to store the number of ways to reach each step.
    • Define a recursive function ‘climb’ that takes ’n’ as an argument. This function checks if ’n’ is already in ‘memo’. If it is, it returns the value. If not, it calculates the value by summing ‘climb(n-1)’ and ‘climb(n-2)’, stores this value in ‘memo’, and then returns it.
    • Call the function ‘climb’ with ’n’ and return its value.

  5. Test the solution: The solution should be tested with various inputs to ensure that it handles all scenarios.

Targeted Drills in Python

  1. Variables and Data Types:
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# Integer
step_count = 5
print(step_count)

# Dictionary
memo = {1: 1, 2: 2}
print(memo)
  1. Functions and/or Methods:
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# Function definition and call
def say_hello(name):
    return "Hello, " + name

print(say_hello("World"))
  1. Recursion:
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# Recursive function to compute factorial of a number
def factorial(n):
    if n == 0:
        return 1
    else:
        return n * factorial(n-1)

print(factorial(5))
  1. Conditional Statements (If and Else):
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# Checking if a key is in dictionary
memo = {1: 1, 2: 2}
key = 1

if key in memo:
    print(f"{key} is in memo")
else:
    print(f"{key} is not in memo")
  1. Dictionary (Hash Map):
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# Storing and retrieving values from dictionary
memo = {}
memo[1] = 1
memo[2] = 2

print(memo[1])
print(memo[2])
  1. Arithmetic Operations:
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# Addition
sum = 5 + 5
print(sum)

# Subtraction
diff = 10 - 5
print(diff)

Problem Specific Drills:

  1. Recursive function to compute Fibonacci sequence: The problem of climbing stairs is similar to computing the Fibonacci sequence. Here’s a drill to compute the nth Fibonacci number using recursion:
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def fibonacci(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fibonacci(n-1) + fibonacci(n-2)

print(fibonacci(5))
  1. Memoization with Fibonacci sequence: The function can be made more efficient by storing previously computed results. Here’s a drill for that:
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memo = {0: 0, 1: 1}

def fibonacci(n):
    if n not in memo:
        memo[n] = fibonacci(n-1) + fibonacci(n-2)
    return memo[n]

print(fibonacci(5))

Each of these drills targets a specific concept required for the final solution. They can be combined and modified to form the final solution.