Greedy vs DP How to Know Which to Apply in DSA C++ - Visual Comparison

Concept Flow - Greedy vs DP How to Know Which to Apply

Start Problem

↓

Check Problem Structure

↓

Is problem optimal substructure?

No→Greedy Not Suitable

Yes↓

Is problem overlapping subproblems?

No→Use Greedy

Yes↓

Use Dynamic Programming

↓

Implement Solution

↓

End

Start by checking if the problem has optimal substructure. If yes, check for overlapping subproblems. If overlapping exists, use DP; otherwise, greedy.

Execution Sample

DSA C++

bool isGreedyApplicable = !hasOverlappingSubproblems && hasOptimalSubstructure;
bool isDPApplicable = hasOverlappingSubproblems && hasOptimalSubstructure;

Determine applicability of greedy or DP based on problem properties.

Execution Table

Step	Check	Condition	Result	Decision
1	Check optimal substructure	Yes	True	Continue
2	Check overlapping subproblems	No	False	Apply Greedy
3	End	-	-	Stop

💡 Overlapping subproblems is False, so greedy approach is chosen.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
hasOptimalSubstructure	Unknown	True	True	True
hasOverlappingSubproblems	Unknown	Unknown	False	False
isGreedyApplicable	False	False	True	True
isDPApplicable	False	False	False	False

Key Moments - 3 Insights

Why do we check for optimal substructure first?

Because both greedy and DP require optimal substructure. If it's missing, neither applies well (see execution_table step 1).

What does overlapping subproblems mean and why does it matter?

It means the problem repeats the same smaller problems. If yes, DP is better; if no, greedy can work (see execution_table step 2).

Can greedy always replace DP if optimal substructure exists?

No, greedy only works if there are no overlapping subproblems. Otherwise, DP is needed to store repeated results.

Visual Quiz - 3 Questions

Look at the execution_table, at which step do we decide to apply greedy?

A Step 1

B Step 3

C Step 2

D Before Step 1

💡 Hint

Check the 'Decision' column in execution_table row 2.

According to variable_tracker, what is the value of isDPApplicable after step 2?

A False

B True

C Unknown

D Not set

💡 Hint

Look at the 'isDPApplicable' row under 'After Step 2' in variable_tracker.

If overlapping subproblems was True, what would be the decision at step 2?

A Apply Greedy

B Apply Dynamic Programming

C Stop

D Check optimal substructure again

💡 Hint

Refer to concept_flow where overlapping subproblems True leads to DP.

Concept Snapshot

Greedy vs DP Decision:
- Check if problem has optimal substructure.
- If no, neither applies well.
- If yes, check overlapping subproblems.
- Overlapping? Use DP.
- No overlapping? Use Greedy.
- Greedy is simpler; DP handles repeated subproblems.

Full Transcript

To decide between greedy and dynamic programming, first check if the problem has optimal substructure, meaning the best solution can be built from best solutions of subproblems. If not, neither approach fits well. If yes, check if the problem has overlapping subproblems, meaning the same subproblems appear multiple times. If overlapping exists, dynamic programming is suitable because it stores and reuses results. If not, greedy algorithms can be used as they make locally optimal choices leading to a global optimum. This decision flow helps choose the right approach efficiently.

Practice

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

easy

A. When the problem has no clear optimal substructure

B. When the problem has optimal substructure and no overlapping subproblems

C. When the problem requires checking all possible solutions

D. When the problem has overlapping subproblems and optimal substructure

Solution

1. Step 1: Understand greedy algorithm conditions

   Greedy algorithms work well when the problem has an optimal substructure and no overlapping subproblems, meaning making the best local choice leads to a global optimum.

2. Step 2: Compare with dynamic programming conditions

   Dynamic programming is used when there are overlapping subproblems and optimal substructure, requiring storing intermediate results to avoid recomputation.

3. Final Answer:

   When the problem has optimal substructure and no overlapping subproblems -> Option B

4. Quick Check:

   Greedy = optimal substructure + no overlap [OK]

Hint: Use greedy if no overlapping subproblems exist [OK]

Common Mistakes:

• Confusing overlapping subproblems as a greedy condition
• Using DP when greedy suffices
• Ignoring optimal substructure requirement

2. Which of the following C++ code snippets correctly initializes a DP array for a problem with 5 states?

easy

A. int dp[5];

B. int dp(5, 0);

C. vector<int> dp(5, 0);

D. int dp = new int[5];

Solution

1. Step 1: Check syntax for array initialization

   int dp[5]; leaves elements uninitialized. int dp(5, 0); is invalid syntax for array initialization. int dp = new int[5]; uses incorrect syntax and memory management in C++.

2. Step 2: Identify correct vector initialization

   vector<int> dp(5, 0); correctly creates a vector of size 5 with all elements initialized to 0, which is the preferred way to initialize DP arrays in modern C++.

3. Final Answer:

   vector<int> dp(5, 0); -> Option C

4. Quick Check:

   Use vector<int> dp(size, 0) for DP arrays [OK]

Hint: Use vector with size and initial value for DP arrays [OK]

Common Mistakes:

• Using incorrect syntax for array initialization
• Not initializing all elements to zero
• Using raw pointers without delete

3. Consider the following C++ code snippet for a problem that can be solved by either greedy or DP. What is the output?

int main() {
    int arr[] = {1, 2, 3, 4, 5};
    int n = 5;
    int sum = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            sum += arr[i];
        }
    }
    std::cout << sum << std::endl;
    return 0;
}

medium

A. 6

B. 9

C. 15

D. 10

Solution

1. Step 1: Identify even numbers in the array

   The array is {1, 2, 3, 4, 5}. Even numbers are 2 and 4.

2. Step 2: Calculate sum of even numbers

   Sum = 2 + 4 = 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Sum of even numbers = 6 [OK]

Hint: Sum only even numbers using modulo check [OK]

Common Mistakes:

• Summing all numbers instead of only even
• Off-by-one error in loop
• Misunderstanding modulo operator

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

int coinChange(int amount) {
    int coins[] = {25, 10, 5, 1};
    int count = 0;
    for (int i = 0; i < 4; i++) {
        while (amount >= coins[i]) {
            amount -= coins[i];
            count++;
        }
    }
    return count;
}

medium

A. Greedy approach fails for some coin sets, not always optimal

B. The loop condition is incorrect and causes infinite loop

C. Coins array is not sorted in ascending order

D. The function does not return the count

Solution

1. Step 1: Understand greedy approach for coin change

   The greedy approach picks the largest coin repeatedly. It works only if coin denominations are canonical (like US coins).

2. Step 2: Identify problem with greedy for arbitrary coins

   For some coin sets, greedy does not give minimum coins. Dynamic programming is needed for guaranteed optimality.

3. Final Answer:

   Greedy approach fails for some coin sets, not always optimal -> Option A

4. Quick Check:

   Greedy coin change not always optimal [OK]

Hint: Greedy coin change fails if coins not canonical [OK]

Common Mistakes:

• Assuming greedy always gives minimum coins
• Ignoring coin set properties
• Confusing loop correctness with algorithm correctness

5. You have a set of tasks each with a deadline and profit. You want to schedule tasks to maximize profit without missing deadlines. Which approach is best and why?

struct Task {
    int deadline;
    int profit;
};

hard

A. Use greedy algorithm by sorting tasks by deadline only

B. Use dynamic programming to try all subsets of tasks and pick max profit

C. Use brute force to check all permutations of tasks

D. Use greedy algorithm by sorting tasks by profit and scheduling latest available slot

Solution

1. Step 1: Understand task scheduling problem

   This is a classic greedy problem where tasks are sorted by profit descending, then scheduled in the latest free slot before deadline.

2. Step 2: Compare approaches

   Dynamic programming or brute force is too slow. Sorting by deadline only does not maximize profit. Greedy by profit with latest slot is efficient and optimal.

3. Final Answer:

   Use greedy algorithm by sorting tasks by profit and scheduling latest available slot -> Option D

4. Quick Check:

   Greedy by profit + latest slot = max profit [OK]

Hint: Sort tasks by profit, schedule latest free slot [OK]

Common Mistakes:

• Sorting only by deadline misses max profit
• Trying DP or brute force unnecessarily
• Scheduling tasks without checking deadlines