Greedy vs DP How to Know Which to Apply in DSA Java - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the problem has optimal substructure property.

DSA Java

if (problem.has[1]()) {
    System.out.println("Optimal substructure exists");
}

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

AOptimalSubstructure

BGreedyChoiceProperty

COverlappingSubproblems

DMemoization

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing greedy choice property with optimal substructure.

Assuming memoization is a property of the problem.

The optimal substructure property means the optimal solution can be built from optimal solutions of subproblems.

2fill in blank

medium

Complete the code to check if the problem has greedy choice property.

DSA Java

if (problem.has[1]()) {
    System.out.println("Greedy choice property exists");
}

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

ABacktracking

BGreedyChoiceProperty

COverlappingSubproblems

DOptimalSubstructure

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing greedy choice property with overlapping subproblems.

Thinking backtracking is a property of the problem.

The greedy choice property means a global optimal solution can be arrived at by choosing a local optimum at each step.

3fill in blank

hard

Fix the error in the code that decides whether to use greedy or DP.

DSA Java

if (problem.hasOptimalSubstructure() && problem.has[1]()) {
    System.out.println("Use Greedy Algorithm");
} else if (problem.hasOptimalSubstructure() && problem.hasOverlappingSubproblems()) {
    System.out.println("Use Dynamic Programming");
}

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

ADivideAndConquer

BMemoization

CGreedyChoiceProperty

DBacktracking

Attempts:

3 left

💡 Hint

Common Mistakes

Using memoization as a property for greedy choice.

Confusing divide and conquer with greedy choice property.

Greedy algorithms require both optimal substructure and greedy choice property.

4fill in blank

hard

Fill both blanks to complete the method that decides algorithm type based on problem properties.

DSA Java

public String chooseAlgorithm(Problem problem) {
    if (problem.hasOptimalSubstructure() && problem.has[1]()) {
        return "Greedy";
    } else if (problem.hasOptimalSubstructure() && problem.has[2]()) {
        return "Dynamic Programming";
    } else {
        return "No suitable algorithm";
    }
}

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

AGreedyChoiceProperty

BMemoization

COverlappingSubproblems

DBacktracking

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing memoization with overlapping subproblems.

Using backtracking as a property for either algorithm.

Greedy algorithms need greedy choice property; dynamic programming needs overlapping subproblems.

5fill in blank

hard

Fill all three blanks to complete the code that explains when to use greedy or DP.

DSA Java

public void explainChoice(Problem problem) {
    if (problem.hasOptimalSubstructure() && problem.has[1]()) {
        System.out.println("Use Greedy because it has [2] and [3].");
    } else if (problem.hasOptimalSubstructure() && problem.hasOverlappingSubproblems()) {
        System.out.println("Use Dynamic Programming because it has overlapping subproblems and optimal substructure.");
    } else {
        System.out.println("Neither greedy nor DP is suitable.");
    }
}

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

AGreedyChoiceProperty

Bgreedy choice property

Coptimal substructure

Dmemoization

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up property names and explanation phrases.

Using memoization instead of optimal substructure in explanation.

Greedy algorithms require both greedy choice property and optimal substructure.

Practice

1. Which of the following best describes when to use a greedy algorithm instead of dynamic programming?

easy

A. When the problem needs to store all intermediate results

B. When the problem requires checking all possible combinations

C. When the problem has overlapping subproblems but no optimal substructure

D. When the problem has the greedy choice property and optimal substructure

Solution

1. Step 1: Understand greedy choice property

   A greedy algorithm works when choosing the best option at each step leads to a global optimum.

2. Step 2: Recognize optimal substructure

   Optimal substructure means the optimal solution can be built from optimal solutions of subproblems.

3. Final Answer:

   When the problem has the greedy choice property and optimal substructure -> Option D

4. Quick Check:

   Greedy = greedy choice + optimal substructure [OK]

Hint: Use greedy if local best leads to global best [OK]

Common Mistakes:

• Confusing overlapping subproblems with greedy applicability
• Using greedy when problem needs full exploration
• Ignoring optimal substructure requirement

2. Which Java syntax correctly declares a 2D array for dynamic programming table initialization?

easy

A. int[][] dp = new int[5];

B. int dp[][] = new int(5,5);

C. int[][] dp = new int[5][5];

D. int dp[5][5] = new int[][];

Solution

1. Step 1: Recall Java 2D array syntax

   Java uses new int[rows][columns] to declare 2D arrays.

2. Step 2: Check each option

   int[][] dp = new int[5][5]; uses correct syntax. Options A, B, C have syntax errors or incomplete declarations.

3. Final Answer:

   int[][] dp = new int[5][5]; -> Option C

4. Quick Check:

   Correct 2D array declaration = int[][] dp = new int[5][5]; [OK]

Hint: Use new int[rows][cols] for 2D arrays in Java [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Missing size in second dimension
• Incorrect array initialization syntax

3. What is the output of the following Java code snippet?

int[] coins = {1, 2, 5};
int amount = 11;
int count = 0;
for (int i = coins.length - 1; i >= 0; i--) {
    count += amount / coins[i];
    amount %= coins[i];
}
System.out.println(count);

medium

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Trace the loop from largest coin to smallest

   Start with amount=11, coins = [1,2,5]. i=2 (coin=5): count += 11/5=2, amount=11%5=1.

2. Step 2: Continue with next coins

   i=1 (coin=2): count += 1/2=0, amount=1%2=1; i=0 (coin=1): count += 1/1=1, amount=1%1=0.

3. Step 3: Calculate total count

   Total count = 2 + 0 + 1 = 3.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Greedy coin count = 3 [OK]

Hint: Divide and mod with largest coin first [OK]

Common Mistakes:

• Adding counts incorrectly
• Mixing up division and modulo
• Ignoring order of coins

4. The following code attempts to solve the coin change problem using greedy approach. What is the bug?

int[] coins = {1, 3, 4};
int amount = 6;
int count = 0;
for (int i = coins.length - 1; i >= 0; i--) {
    count += amount / coins[i];
    amount %= coins[i];
}
System.out.println(count);

medium

A. Division by zero error

B. Greedy fails because coin set lacks greedy choice property

C. Array index out of bounds error

D. Incorrect initialization of count variable

Solution

1. Step 1: Understand greedy choice property failure

   Coins {1,3,4} do not guarantee greedy choice property for amount 6.

2. Step 2: Trace greedy approach output

   Greedy picks 4 once (count=1), remainder 2; then picks two 1s (count=3). But optimal is two 3s (count=2).

3. Final Answer:

   Greedy fails because coin set lacks greedy choice property -> Option B

4. Quick Check:

   Greedy fails if no greedy choice property [OK]

Hint: Check if greedy choice property holds before using greedy [OK]

Common Mistakes:

• Assuming greedy always works for coin change
• Ignoring optimal substructure requirement
• Confusing runtime errors with logic errors

5. Given a set of tasks with start and end times, which approach is best to find the maximum number of non-overlapping tasks?

class Task {
    int start, end;
    Task(int s, int e) { start = s; end = e; }
}

Choose the correct approach and why.

hard

A. Use greedy by sorting tasks by earliest finish time to select tasks

B. Use dynamic programming to check all subsets of tasks

C. Use greedy by sorting tasks by longest duration first

D. Use dynamic programming to store maximum tasks ending at each time

Solution

1. Step 1: Identify problem type

   This is the activity selection problem, classic greedy example.

2. Step 2: Choose greedy approach

   Sorting by earliest finish time ensures maximum tasks without overlap.

3. Step 3: Compare options

   Use greedy by sorting tasks by earliest finish time to select tasks matches known greedy solution; DP is possible but less efficient here.

4. Final Answer:

   Use greedy by sorting tasks by earliest finish time to select tasks -> Option A

5. Quick Check:

   Activity selection = greedy by finish time [OK]

Hint: Sort tasks by finish time for max non-overlapping tasks [OK]

Common Mistakes:

• Sorting by start time or duration instead of finish time
• Using DP unnecessarily increasing complexity
• Ignoring greedy choice property in scheduling