Overlapping Subproblems and Optimal Substructure in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Fibonacci Recursive Calls

What is the output of the following C code that prints Fibonacci numbers using recursion with overlapping subproblems?

DSA C

#include <stdio.h>

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

int main() {
    for (int i = 0; i < 5; i++) {
        printf("%d ", fib(i));
    }
    return 0;
}

A 0 1 1 2 3

B 1 1 2 3 5

C 0 1 2 3 5

D 1 2 3 5 8

💡 Hint

Remember Fibonacci sequence starts with 0 and 1 for fib(0) and fib(1).

The Fibonacci sequence starts with 0 and 1. The recursive function returns fib(0)=0 and fib(1)=1, then sums previous two results. So the first five numbers are 0,1,1,2,3.

🧠 Conceptual

intermediate

Identifying Overlapping Subproblems

Which of the following problems best demonstrates the concept of overlapping subproblems in dynamic programming?

A Sorting an array using quicksort

B Finding the shortest path in a graph using Dijkstra's algorithm

C Calculating factorial of a number using recursion

D Computing Fibonacci numbers using naive recursion

💡 Hint

Overlapping subproblems means the same smaller problem is solved multiple times.

Naive recursive Fibonacci calculation solves the same fib(n) multiple times, showing overlapping subproblems. Factorial and quicksort do not reuse subproblems, and Dijkstra's algorithm is a greedy approach.

🔧 Debug

advanced

Debugging Memoization Implementation

What is the output of this C code with memoization for Fibonacci, and why?

DSA C

#include <stdio.h>

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() {
    for (int i = 0; i < 7; i++) {
        printf("%d ", fib(i));
    }
    return 0;
}

A 0 1 1 2 3 5 8

B 0 1 2 3 5 8 13

C 0 1 1 2 3 5 0

D 0 1 1 2 3 5 21

💡 Hint

Memoization stores results to avoid repeated calculations.

Memoization caches fib(n) results. Since fib(6) = fib(5) + fib(4) = 5 + 3 = 8, the output is the first 7 Fibonacci numbers starting from 0.

🚀 Application

advanced

Optimal Substructure in Coin Change Problem

Given unlimited coins of denominations 1, 3, and 4, what is the minimum number of coins needed to make amount 6 using optimal substructure?

A 3

B 2

C 4

D 5

💡 Hint

Try breaking amount 6 into smaller amounts and combine optimal solutions.

Minimum coins for 6 can be found by checking 6-1=5, 6-3=3, 6-4=2. Minimum coins for 5 is 2 (4+1), for 3 is 1 (3), for 2 is 2 (1+1). So min is 1 + min(coins for 5,3,2) = 1 + 1 = 2. The minimum is 2 coins (e.g., 3+3).

🧠 Conceptual

expert

Recognizing Problems Without Optimal Substructure

Which problem does NOT exhibit optimal substructure, meaning its optimal solution cannot be constructed from optimal solutions of subproblems?

A Shortest path in a weighted graph without negative cycles

B Matrix chain multiplication

C Longest simple path in a graph

D Knapsack problem with fractional items

💡 Hint

Optimal substructure means combining optimal subproblem solutions gives global optimum.

Longest simple path problem does not have optimal substructure because subpaths of the longest path are not necessarily longest paths themselves. Others have optimal substructure.

Practice

1. Which of the following best describes the concept of overlapping subproblems in dynamic programming?

easy

A. The problem requires sorting elements before solving.

B. The problem has a unique solution without repeating any subproblems.

C. The problem can be broken down into smaller subproblems which are solved multiple times.

D. The problem can be solved only using recursion without optimization.

Solution

1. Step 1: Understand overlapping subproblems

   Overlapping subproblems means the same smaller problems appear multiple times during the solution process.

2. Step 2: Match with options

   The problem can be broken down into smaller subproblems which are solved multiple times. correctly states that subproblems are solved multiple times, which is the key idea behind overlapping subproblems.

3. Final Answer:

   The problem can be broken down into smaller subproblems which are solved multiple times. -> Option C

4. Quick Check:

   Overlapping subproblems = repeated smaller problems [OK]

Hint: Look for repeated smaller problems solved multiple times [OK]

Common Mistakes:

• Confusing overlapping subproblems with unique subproblems
• Thinking sorting is required for overlapping subproblems
• Assuming recursion alone solves overlapping subproblems

2. Which of the following is the correct syntax to declare a memoization array for storing results of subproblems in C?

easy

A. int memo[] = 0;

B. int memo = new int[1000];

C. int memo(1000);

D. int memo[1000] = {0};

Solution

1. Step 1: Understand C array declaration

   In C, arrays are declared with fixed size and can be initialized with {0} to set all elements to zero.

2. Step 2: Check options for valid C syntax

   int memo[1000] = {0}; uses correct C syntax: int memo[1000] = {0}; initializes an array of 1000 integers to zero.

3. Final Answer:

   int memo[1000] = {0}; -> Option D

4. Quick Check:

   C array declaration with initialization = int memo[1000] = {0}; [OK]

Hint: Use fixed size array with braces to initialize in C [OK]

Common Mistakes:

• Using C++ style vector or new keyword in C
• Declaring arrays without size
• Incorrect initialization syntax

3. Consider the following C code for Fibonacci with memoization:

int memo[100] = {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;
}

What is the output when this code runs?

medium

A. 0

B. 5

C. 3

D. 8

Solution

1. Step 1: Trace fib(5) calls with memoization

   fib(5) calls fib(4) + fib(3). fib(4) calls fib(3) + fib(2), and so on. Base cases fib(0)=0, fib(1)=1.

2. Step 2: Calculate fib(5)

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

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Fibonacci(5) = 5 [OK]

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

Common Mistakes:

• Confusing index with Fibonacci value
• Ignoring memoization effect
• Miscomputing base cases

4. The following code attempts to compute the nth Fibonacci number using memoization but has a bug:

int memo[100] = {0};

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

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

What is the bug in this code?

medium

A. fib(0) returns 0 but memo[0] is never set, causing repeated calls.

B. The memo array is not initialized to zero.

C. The base case for fib(1) is missing.

D. The memoization check uses wrong index.

Solution

1. Step 1: Analyze base cases and memo usage

   fib(0) returns 0 but does not store it in memo[0]. So memo[0] remains 0, which is the default value.

2. Step 2: Understand effect on memoization

   Because the base case results are not stored in memo[n], fib(0) (and fib(1)) will be called repeatedly in the recursion tree for larger n, instead of returning a cached value.

3. Final Answer:

   fib(0) returns 0 but memo[0] is never set, causing repeated calls. -> Option A

4. Quick Check:

   Base case not stored in memo causes repeated calls [OK]

Hint: Store base case results in memo to avoid repeated calls [OK]

Common Mistakes:

• Assuming memo array auto-initializes for base cases
• Missing base case for fib(1)
• Incorrect memo check condition

5. You want to find the minimum cost to reach the top of stairs where each step has a cost. You can climb 1 or 2 steps at a time. The cost array is cost[] = {10, 15, 20}. Using dynamic programming with overlapping subproblems and optimal substructure, what is the minimum cost to reach the top (beyond last step)?

hard

A. 15

B. 20

C. 25

D. 10

Solution

1. Step 1: Define subproblems and recurrence

   Minimum cost to reach step i: dp[i] = cost[i] + min(dp[i-1], dp[i-2]). dp[0]=10, dp[1]=15.

2. Step 2: Calculate minimum cost step by step

   dp[2] = 20 + min(10,15) = 30. To reach top (beyond last step), min(dp[1], dp[2]) = min(15,30) = 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Minimum cost to top = 15 [OK]

Hint: Use min cost of last two steps for final answer [OK]

Common Mistakes:

• Adding all costs instead of choosing min path
• Ignoring that you can start from step 0 or 1
• Confusing cost to step vs cost beyond last step