Greedy vs DP How to Know Which to Apply in DSA Typescript - Why the Distinction Matters

The Big Idea

Can you tell when a quick choice is enough or when you need to think deeper to get the best answer?

The Scenario

Imagine you have a big jar of coins and you want to give exact change using the fewest coins possible. You try picking the biggest coin first every time, but sometimes this doesn't work and you end up with more coins than needed.

The Problem

Picking coins greedily (always the biggest coin first) can be fast but sometimes leads to wrong answers. Trying every possible combination manually is slow and confusing. Without a clear method, you waste time and make mistakes.

The Solution

Greedy and Dynamic Programming (DP) are smart ways to solve such problems. Greedy works when local best choices lead to global best. DP tries all options but remembers results to avoid repeating work. Knowing when to use each saves time and gets correct answers.

Before vs After

✗ Before

function getChange(coins: number[], amount: number): number {
  let count = 0;
  while (amount > 0) {
    let coin = coins.reduce((max, c) => c <= amount ? Math.max(max, c) : max, 0);
    if (coin === 0) return -1;
    amount -= coin;
    count++;
  }
  return count;
}

✓ After

function getChangeDP(coins: number[], amount: number): number {
  const dp = Array(amount + 1).fill(Infinity);
  dp[0] = 0;
  for (let i = 1; i <= amount; i++) {
    for (const coin of coins) {
      if (coin <= i) dp[i] = Math.min(dp[i], dp[i - coin] + 1);
    }
  }
  return dp[amount] === Infinity ? -1 : dp[amount];
}

What It Enables

It enables you to solve complex optimization problems efficiently and correctly by choosing the right approach.

Real Life Example

Choosing the best route for delivery trucks: Greedy might pick the closest next stop, but DP considers all routes to find the shortest total path.

Key Takeaways

Greedy is fast but only works if local choices lead to global best.

Dynamic Programming tries all options but remembers results to avoid repeated work.

Knowing when to use each saves time and ensures correct solutions.

Practice

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

easy

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

B. When the problem requires checking all possible combinations

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

D. When the problem needs to store all intermediate results

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: Check 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 A

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 syntax correctly declares a memoization object in TypeScript for dynamic programming?

easy

A. var memo: number[] = [];

B. let memo = new Map();

C. const memo: Record = {};

D. const memo = {} as number[];

Solution

1. Step 1: Understand memo type for DP

   Memoization stores results with keys, so Record is suitable.

2. Step 2: Check syntax correctness

   const memo: Record = {}; correctly declares an empty object with number keys and values.

3. Final Answer:

   const memo: Record = {}; -> Option C

4. Quick Check:

   Memo object uses Record type [OK]

Hint: Use Record for memo objects [OK]

Common Mistakes:

• Using var instead of const or let
• Incorrect type assertion for memo
• Using Map without type parameters

3. Consider this TypeScript code snippet for a problem that can be solved by greedy or DP:

function maxSum(arr: number[]): number {
  let sum = 0;
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] > 0) sum += arr[i];
  }
  return sum;
}
console.log(maxSum([-1, 2, 3, -5, 4]));

What is the output?

medium

A. 9

B. 4

C. 0

D. Error

Solution

1. Step 1: Trace the loop and sum positive numbers

   Positive numbers are 2, 3, and 4. Sum = 2 + 3 + 4 = 9.

2. Step 2: Confirm return value

   The function returns sum = 9, so console.log prints 9.

3. Final Answer:

   9 -> Option A

4. Quick Check:

   Sum positives = 9 [OK]

Hint: Sum only positive numbers for max sum greedy [OK]

Common Mistakes:

• Adding negative numbers mistakenly
• Returning zero instead of sum
• Confusing output with array length

4. The following TypeScript function is intended to solve the Fibonacci sequence using DP with memoization. What is the bug?

function fib(n: number, memo: Record = {}): number {
  if (n <= 1) return n;
  if (memo[n]) return memo[n];
  memo[n] = fib(n - 1, memo) + fib(n - 2, memo);
  return memo[n];
}
console.log(fib(5));

medium

A. The recursive calls should be fib(n + 1) and fib(n + 2)

B. The base case should be 'if (n < 0) return 0;'

C. The function should not use default parameter for memo

D. The memo check should be 'if (memo[n] !== undefined)' to handle zero

Solution

1. Step 1: Analyze memo check condition

   Checking 'if (memo[n])' fails when memo[n] is 0, which is falsy.

2. Step 2: Correct memo check

   Use 'if (memo[n] !== undefined)' to correctly detect stored values including zero.

3. Final Answer:

   The memo check should be 'if (memo[n] !== undefined)' to handle zero -> Option D

4. Quick Check:

   Memo check must handle zero values [OK]

Hint: Check memo with !== undefined, not truthy check [OK]

Common Mistakes:

• Using truthy check for memo causing wrong cache hits
• Wrong base case for Fibonacci
• Incorrect recursive call directions

5. You have a set of tasks with durations and deadlines. You want to maximize the number of tasks completed before their deadlines. Which approach is best and why?

interface Task {
  duration: number;
  deadline: number;
}
function scheduleTasks(tasks: Task[]): number {
  // Implementation needed
}

hard

A. Use dynamic programming to try all subsets and pick max count

B. Use a greedy algorithm sorting tasks by earliest deadline to maximize count

C. Use greedy sorting by longest duration first to minimize total time

D. Use dynamic programming to minimize total duration ignoring deadlines

Solution

1. Step 1: Understand problem goal

   Maximize number of tasks completed before deadlines means scheduling tasks efficiently.

2. Step 2: Identify best approach

   Greedy algorithm sorting by earliest deadline ensures maximum tasks fit before deadlines.

3. Final Answer:

   Use a greedy algorithm sorting tasks by earliest deadline to maximize count -> Option B

4. Quick Check:

   Earliest deadline first greedy maximizes tasks [OK]

Hint: Sort tasks by earliest deadline for max count [OK]

Common Mistakes:

• Trying all subsets with DP causing inefficiency
• Sorting by longest duration wastes time
• Ignoring deadlines in scheduling