Memoization Top Down DP in DSA Python - Time & Space Complexity

Time Complexity: Memoization Top Down DP

O(n)

Understanding Time Complexity

We want to understand how using memoization in top-down dynamic programming affects the time it takes to solve problems.

Specifically, how does remembering past results change the number of steps as input grows?

Scenario Under Consideration

Analyze the time complexity of the following Fibonacci function using memoization.

    def fib(n, memo=None):
        if memo is None:
            memo = {}
        if n in memo:
            return memo[n]
        if n <= 1:
            return n
        memo[n] = fib(n-1, memo) + fib(n-2, memo)
        return memo[n]
    

This code calculates the nth Fibonacci number by storing results to avoid repeated work.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to fib with smaller n values.
• How many times: Each unique n from 0 to n is computed once due to memoization.

How Execution Grows With Input

Each number from 0 up to n is calculated once and stored, so the steps grow roughly in a straight line with n.

Input Size (n)	Approx. Operations
10	About 10 calculations
100	About 100 calculations
1000	About 1000 calculations

Pattern observation: The work grows directly with n, not faster.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute grows in a straight line as the input number increases.

Common Mistake

[X] Wrong: "Memoization still does all recursive calls like the simple recursive method, so it's slow."

[OK] Correct: Memoization saves results, so each unique call happens once, avoiding repeated work and making it much faster.

Interview Connect

Understanding memoization shows you can improve slow recursive solutions by saving work, a key skill in many coding problems.

Self-Check

"What if we removed memoization and just used plain recursion? How would the time complexity change?"

Practice

1. What is the main purpose of memoization in top-down dynamic programming?

easy

A. To store results of subproblems to avoid repeated calculations

B. To sort the input data before processing

C. To convert recursive code into iterative code

D. To increase the size of the input for better accuracy

Solution

1. Step 1: Understand memoization concept

   Memoization saves the results of subproblems so they are not recalculated multiple times.

2. Step 2: Compare options

   Only To store results of subproblems to avoid repeated calculations describes this purpose correctly. Other options describe unrelated actions.

3. Final Answer:

   To store results of subproblems to avoid repeated calculations -> Option A

4. Quick Check:

   Memoization = Store subproblem results [OK]

Hint: Memoization means saving answers to reuse later [OK]

Common Mistakes:

• Confusing memoization with sorting or iteration
• Thinking memoization increases input size
• Believing memoization converts recursion to iteration

2. Which of the following is the correct way to define a memo dictionary in Python for top-down DP?

easy

A. memo = []

B. memo = {}

C. memo = set()

D. memo = ()

Solution

1. Step 1: Identify data structure for memo

   Memoization requires a dictionary to store key-value pairs of subproblem inputs and results.

2. Step 2: Match with options

   Only memo = {} uses a dictionary ({}). Lists, sets, and tuples do not support key-value storage.

3. Final Answer:

   memo = {} -> Option B

4. Quick Check:

   Memo dictionary = {} [OK]

Hint: Use {} for memo dictionary in Python [OK]

Common Mistakes:

• Using list or set instead of dictionary
• Confusing tuple with dictionary
• Not initializing memo before recursion

3. What will be the output of the following code?

def fib(n, memo={}):
    if n in memo:
        return memo[n]
    if n <= 1:
        return n
    memo[n] = fib(n-1, memo) + fib(n-2, memo)
    return memo[n]

print(fib(5))

medium

A. 3

B. 8

C. 5

D. Error

Solution

1. Step 1: Understand Fibonacci calculation with memo

   fib(5) calculates Fibonacci number at position 5 using memo to save results.

2. Step 2: Calculate fib(5)

   fib(0)=0, fib(1)=1, fib(2)=1, fib(3)=2, fib(4)=3, fib(5)=5.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Fibonacci(5) = 5 using memoized recursion [OK]

Common Mistakes:

• Confusing Fibonacci sequence values
• Expecting 8 instead of 5 for fib(5)
• Thinking code causes error due to default dict

4. Identify the error in this memoized Fibonacci function:

def fib(n, memo={}):
    if n <= 1:
        return n
    if n in memo:
        return memo[n]
    memo[n] = fib(n-1, memo) + fib(n-2, memo)
    return memo[n]

medium

A. Missing base case for n=0

B. Function lacks return statement

C. Memo dictionary is not used correctly

D. Default mutable argument causes shared state across calls

Solution

1. Step 1: Analyze default argument usage

   Using memo={} as default argument creates shared dictionary across calls, causing unexpected behavior.

2. Step 2: Identify correct practice

   Default mutable arguments should be avoided; better to use None and initialize inside function.

3. Final Answer:

   Default mutable argument causes shared state across calls -> Option D

4. Quick Check:

   Mutable default argument = bug [OK]

Hint: Avoid mutable default args like {} in function definitions [OK]

Common Mistakes:

• Ignoring mutable default argument issue
• Thinking base case is missing
• Assuming memo dictionary usage is wrong

5. You want to find the number of ways to climb a staircase of n steps, taking 1 or 2 steps at a time. Which memoized top-down function correctly solves this problem?

hard

A. def climb(n, memo=None):
if memo is None:
memo = {}
if n < 0:
return 0
if n == 0:
return 1
if n in memo:
return memo[n]
memo[n] = climb(n-1, memo) + climb(n-2, memo)
return memo[n]

B. def climb(n):
if n <= 1:
return n
return climb(n-1) + climb(n-2)

C. def climb(n, memo={}):
if n == 0:
return 0
if n in memo:
return memo[n]
memo[n] = climb(n-1, memo) + climb(n-2, memo)
return memo[n]

D. def climb(n, memo=None):
if memo is None:
memo = []
if n <= 1:
return 1
if n in memo:
return memo[n]
memo[n] = climb(n-1, memo) + climb(n-2, memo)
return memo[n]

Solution

1. Step 1: Understand problem and base cases

   Number of ways to climb n steps with 1 or 2 steps is classic DP with base cases: ways(0)=1, ways(-1)=0.

2. Step 2: Check each option for correctness

   def climb(n, memo=None): if memo is None: memo = {} if n < 0: return 0 if n == 0: return 1 if n in memo: return memo[n] memo[n] = climb(n-1, memo) + climb(n-2, memo) return memo[n] correctly initializes memo, handles negative and zero cases, and uses memoization properly. Others have wrong base cases or memo usage.

3. Final Answer:

   Option A function correctly solves the problem -> Option A

4. Quick Check:

   Proper base cases + memo dict = correct solution [OK]

Hint: Base cases and memo dict initialization are key [OK]

Common Mistakes:

• Wrong base case returns
• Using mutable default memo argument
• Using list instead of dict for memo