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

Time Complexity: Memoization Top Down DP

O(n)

Understanding Time Complexity

We analyze how memoization in top-down dynamic programming reduces repeated work.

We want to know how the number of operations changes as input size grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int fib(int n, int memo[]) {
    if (n <= 1) return n;
    if (memo[n] != -1) return memo[n];
    memo[n] = fib(n-1, memo) + fib(n-2, memo);
    return memo[n];
}

This code calculates the nth Fibonacci number using recursion with memoization to store results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to fib with memo checks.
• How many times: Each unique fib(n) is computed once; repeated calls return stored results instantly.

How Execution Grows With Input

Each number from 0 to n is computed once, so operations grow roughly in a straight line with n.

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

Pattern observation: Doubling n roughly doubles the work, showing linear growth.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows in direct proportion to the input size n.

Common Mistake

[X] Wrong: "Memoization still does all recursive calls like plain recursion."

[OK] Correct: Memoization stores results, so each unique call runs once, avoiding repeated work.

Interview Connect

Understanding memoization's time savings shows you can optimize slow recursive solutions effectively.

Self-Check

"What if we removed memoization? How would the time complexity change?"

Practice

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

easy

A. To convert recursive code into iterative code

B. To sort the input data before processing

C. To store results of subproblems to avoid repeated calculations

D. To increase the number of recursive calls

Solution

1. Step 1: Understand memoization concept

   Memoization saves answers of smaller problems so they don't get solved again.

2. Step 2: Identify its role in top-down DP

   It stores subproblem results during recursion to speed up the process.

3. Final Answer:

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

4. Quick Check:

   Memoization = Store subproblem results [OK]

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

Common Mistakes:

• Confusing memoization with sorting
• Thinking memoization converts recursion to iteration
• Believing memoization increases recursive calls

2. Which of the following is the correct way to declare a memoization array for Fibonacci numbers in C?

easy

A. int memo[] = new int[100];

B. int memo[100] = {0};

C. int memo = new int[100];

D. int memo(100);

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared with fixed size and optional initialization like int memo[100] = {0};

2. Step 2: Check each option

   int memo[100] = {0}; uses correct C syntax. Options A and C use Java/C++ style. int memo(100); is invalid in C.

3. Final Answer:

   int memo[100] = {0}; -> Option B

4. Quick Check:

   C array declaration = int memo[100] = {0}; [OK]

Hint: C arrays use fixed size with square brackets [OK]

Common Mistakes:

• Using Java or C++ style array declarations in C
• Forgetting to initialize memo array to zero
• Using parentheses instead of brackets for arrays

3. What will be the output of the following C code snippet using memoization for Fibonacci?

int memo[10] = {0};
int fib(int n) {
  if (n <= 1) return n;
  if (memo[n] != 0) return memo[n];
  memo[n] = fib(n-1) + fib(n-2);
  return memo[n];
}

int main() {
  printf("%d", fib(5));
  return 0;
}

medium

A. 0

B. 8

C. 3

D. 5

Solution

1. Step 1: Trace fib(5) calls and memo usage

   fib(5) calls fib(4) + fib(3), and so on. Memo stores results to avoid repeats.

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 D

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing fib(5) with fib(6)
• Ignoring memoization effect
• Assuming memo array starts with all ones

4. Identify the error in this memoized Fibonacci function in C:

int memo[10];
int fib(int n) {
  if (n <= 1) return n;
  if (memo[n] != -1) return memo[n];
  memo[n] = fib(n-1) + fib(n-2);
  return memo[n];
}

int main() {
  printf("%d", fib(5));
  return 0;
}

medium

A. memo array is not initialized to -1

B. fib function should return void

C. Base case should be n == 0 only

D. memo array size is too small

Solution

1. Step 1: Check memo array initialization

   Memo is checked for -1 but never initialized, so contains garbage values.

2. Step 2: Understand impact on memoization

   Without initialization, memo[n] != -1 may be true incorrectly, causing wrong returns.

3. Final Answer:

   memo array is not initialized to -1 -> Option A

4. Quick Check:

   Initialize memo to -1 before use [OK]

Hint: Always initialize memo array before use [OK]

Common Mistakes:

• Forgetting to initialize memo array
• Changing base case incorrectly
• Assuming default zero initialization in C

5. Given the following top-down memoized code to count ways to climb stairs (1 or 2 steps), what is the output of countWays(4)? Assume memo array is initialized with -1.

int memo[100];
int countWays(int n) {
  if (n == 0) return 1;
  if (n < 0) return 0;
  if (memo[n] != -1) return memo[n];
  memo[n] = countWays(n-1) + countWays(n-2);
  return memo[n];
}

int main() {
  for(int i=0; i<100; i++) memo[i] = -1;
  printf("%d", countWays(4));
  return 0;
}

hard

A. 5

B. 7

C. 8

D. 4

Solution

1. Step 1: Understand problem and base cases

   Ways to climb 0 steps = 1, negative steps = 0, ways(n) = ways(n-1)+ways(n-2).

2. Step 2: Calculate countWays(4)

   countWays(1)=1, countWays(2)=2, countWays(3)=3, countWays(4)=5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Ways to climb 4 steps = 5 [OK]

Hint: Count ways follow Fibonacci pattern starting with 1,1 [OK]

Common Mistakes:

• Confusing base cases for counting ways
• Not initializing memo array
• Mixing up countWays with Fibonacci values