Overlapping Subproblems and Optimal Substructure in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the nth Fibonacci number using simple recursion.

DSA Java

public int fib(int n) {
    if (n <= 1) {
        return n;
    }
    return fib(n - 1) + fib([1]);
}

Drag options to blanks, or click blank then click option'

An + 1

Bn - 2

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n or n+1 instead of n-2 causes incorrect recursion and wrong results.

The Fibonacci number is calculated by adding fib(n-1) and fib(n-2). So the blank should be n - 2.

2fill in blank

medium

Complete the code to use memoization for Fibonacci to avoid repeated calculations.

DSA Java

public int fibMemo(int n, int[] memo) {
    if (n <= 1) {
        return n;
    }
    if (memo[n] != 0) {
        return memo[n];
    }
    memo[n] = fibMemo(n - 1, memo) + fibMemo([1], memo);
    return memo[n];
}

Drag options to blanks, or click blank then click option'

A1

Bn + 1

Cn

Dn - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong index like n or n+1 causes wrong memoization and incorrect results.

Memoization stores results of fib(n-1) and fib(n-2). The second call must be fibMemo(n - 2, memo).

3fill in blank

hard

Fix the error in the bottom-up Fibonacci implementation to correctly fill the dp array.

DSA Java

public int fibBottomUp(int n) {
    if (n <= 1) {
        return n;
    }
    int[] dp = new int[n + 1];
    dp[0] = 0;
    dp[1] = 1;
    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[[1]];
    }
    return dp[n];
}

Drag options to blanks, or click blank then click option'

Ai

Bi + 1

Ci - 2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i or i+1 causes index errors or wrong sums.

The Fibonacci number at position i is the sum of dp[i-1] and dp[i-2]. So the blank must be i - 2.

4fill in blank

hard

Fill both blanks to complete the recursive function with memoization for the coin change problem.

DSA Java

public int coinChange(int[] coins, int amount, int[] memo) {
    if (amount == 0) {
        return 0;
    }
    if (amount < 0) {
        return -1;
    }
    if (memo[amount] != 0) {
        return memo[amount];
    }
    int min = Integer.MAX_VALUE;
    for (int coin : coins) {
        int res = coinChange(coins, amount - coin, memo);
        if (res != -1 && res [1] min) {
            min = res + 1;
        }
    }
    memo[amount] = (min == Integer.MAX_VALUE) ? -1 : min;
    return memo[amount];
}

Drag options to blanks, or click blank then click option'

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or == causes wrong min updates and incorrect results.

We want the minimum number of coins, so we update min if res is less than min. The blank should be <.

5fill in blank

hard

Fill all three blanks to complete the dynamic programming solution for the longest common subsequence problem.

DSA Java

public int lcs(String s1, String s2) {
    int m = s1.length();
    int n = s2.length();
    int[][] dp = new int[m + 1][n + 1];
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                dp[i][j] = dp[i - 1][j - 1] + [1];
            } else {
                dp[i][j] = Math.[2](dp[i - 1][j], dp[i][[3]]);
            }
        }
    }
    return dp[m][n];
}

Drag options to blanks, or click blank then click option'

A1

Bmax

Cj - 1

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max causes wrong LCS length.

Using wrong indices causes index errors.

When characters match, add 1 to dp[i-1][j-1]. Otherwise, take the max of dp[i-1][j] and dp[i][j-1]. So blanks are 1, max, and j - 1.

Practice

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

easy

A. The problem has no repeated subproblems and each subproblem is unique.

B. The problem can be broken into smaller subproblems that are solved multiple times.

C. The solution requires sorting the input data first.

D. The problem can be solved using a greedy approach without recursion.

Solution

1. Step 1: Understand overlapping subproblems

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

2. Step 2: Match with options

   The problem can be broken into smaller subproblems that are solved multiple times. correctly states that subproblems repeat and are solved multiple times.

3. Final Answer:

   The problem can be broken into smaller subproblems that are solved multiple times. -> Option B

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
• Mixing greedy approach with dynamic programming

2. Which of the following Java method signatures correctly represents a memoized Fibonacci function?

easy

A. public int fib(int n, int[] memo)

B. public void fib(int n)

C. public int fib(String n, int[] memo)

D. public int fib(int n, String[] memo)

Solution

1. Step 1: Identify correct parameter types

   Memoization requires storing results, so an int array for memo is needed. The input n should be int.

2. Step 2: Check method signatures

   public int fib(int n, int[] memo) uses int n and int[] memo, which is correct for memoized Fibonacci.

3. Final Answer:

   public int fib(int n, int[] memo) -> Option A

4. Quick Check:

   Memo array type matches input type [OK]

Hint: Memo array must match return type and input must be int [OK]

Common Mistakes:

• Using void return type for Fibonacci function
• Using String type for numeric input or memo array
• Not passing memo array as parameter

3. What is the output of the following Java code snippet implementing a simple memoized Fibonacci?

public class FibMemo {
    static int[] memo = new int[6];
    public static 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];
    }
    public static void main(String[] args) {
        System.out.println(fib(5));
    }
}

medium

A. 8

B. 0

C. 5

D. StackOverflowError

Solution

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

   fib(5) calls fib(4) and 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 C

4. Quick Check:

   Fibonacci(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
• Expecting sum of all calls instead of fib(5)

4. Identify the error in this Java code snippet for computing Fibonacci with memoization:

public class Fib {
    static int[] memo = new int[5];
    public static 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];
    }
    public static void main(String[] args) {
        System.out.println(fib(5));
    }
}

medium

A. No error, prints correct Fibonacci number

B. StackOverflowError due to infinite recursion

C. Compilation error due to missing return

D. ArrayIndexOutOfBoundsException due to memo size

Solution

1. Step 1: Check memo array size and access

   Memo array size is 5, so valid indices are 0 to 4. fib(5) tries to access memo[5], causing error.

2. Step 2: Identify exception type

   Accessing memo[5] causes ArrayIndexOutOfBoundsException at runtime.

3. Final Answer:

   ArrayIndexOutOfBoundsException due to memo size -> Option D

4. Quick Check:

   Array size < access index = error [OK]

Hint: Check array size matches max index accessed [OK]

Common Mistakes:

• Assuming memo size is enough without checking
• Thinking recursion causes stack overflow here
• Ignoring runtime exceptions

5. Given the problem of finding the minimum cost path in a grid from top-left to bottom-right, which property justifies using dynamic programming to solve it efficiently?

hard

A. The problem has overlapping subproblems and optimal substructure.

B. The problem can be solved by sorting all paths.

C. The problem requires only greedy choices without recursion.

D. The problem has no repeated subproblems and no optimal substructure.

Solution

1. Step 1: Understand minimum cost path problem

   The problem breaks into smaller subproblems (cost to reach each cell) that overlap.

2. Step 2: Check properties for dynamic programming

   Optimal substructure means optimal solution uses optimal solutions of subproblems. Overlapping subproblems means subproblems repeat.

3. Step 3: Match with options

   The problem has overlapping subproblems and optimal substructure. correctly states both properties, justifying dynamic programming.

4. Final Answer:

   The problem has overlapping subproblems and optimal substructure. -> Option A

5. Quick Check:

   Overlapping subproblems + optimal substructure = DP [OK]

Hint: Look for repeated subproblems and optimal smaller solutions [OK]

Common Mistakes:

• Thinking sorting solves the problem
• Assuming greedy approach always works
• Ignoring overlapping subproblems