DP vs Recursion vs Greedy Choosing the Right Tool in DSA Javascript - Key Differences

Recall & Review

beginner

What is recursion in problem solving?

Recursion is a method where a function calls itself to solve smaller parts of a problem until it reaches a simple case it can solve directly.

intermediate

What does DP (Dynamic Programming) do differently from simple recursion?

DP stores results of solved subproblems to avoid repeating work, making it faster than simple recursion when problems have overlapping subproblems.

beginner

When is a greedy approach suitable?

A greedy approach works well when choosing the best option at each step leads to an overall best solution, without needing to reconsider previous choices.

beginner

Why might recursion be inefficient for some problems?

Recursion can be inefficient because it may solve the same subproblems multiple times, leading to repeated work and slower performance.

intermediate

How do you decide between DP and greedy methods?

Use DP when subproblems overlap and you need to remember past results; use greedy when local best choices lead to a global best solution without backtracking.

Which approach stores results of subproblems to avoid repeated work?

A Brute Force

B Greedy

C Dynamic Programming

D Simple Recursion

Dynamic Programming saves results of subproblems to improve efficiency.

When is a greedy algorithm the best choice?

A When subproblems overlap

B When local choices lead to global optimum

C When recursion depth is too large

D When all solutions must be checked

Greedy algorithms work best when local best choices lead to the overall best solution.

What is a main drawback of simple recursion without optimization?

A Uses too much memory

B Cannot solve problems with base cases

C Always slower than greedy

D Recomputes the same subproblems multiple times

Simple recursion often repeats work by solving the same subproblems again.

Which method is best if you need to explore all possible solutions?

A Recursion

B Dynamic Programming

C Greedy

D Sorting

Recursion can explore all possibilities by calling itself on smaller parts.

What key feature helps DP improve over recursion?

A Memoization or tabulation

B Sorting inputs

C Choosing local best options

D Using loops instead of functions

DP uses memoization or tabulation to save results and avoid repeated work.

Explain in your own words when to use recursion, dynamic programming, and greedy algorithms.

Think about problem size, repeated work, and choice strategy.

Describe how dynamic programming improves efficiency compared to simple recursion.

Focus on saving and reusing answers.

Practice

1. Which approach is best suited when a problem can be broken down into overlapping subproblems and optimal substructure?

easy

A. Dynamic Programming

B. Greedy Algorithm

C. Recursion without memoization

D. Brute Force

Solution

1. Step 1: Understand overlapping subproblems and optimal substructure

   Dynamic Programming is designed for problems where subproblems repeat and solutions build on smaller solutions.

2. Step 2: Compare with other approaches

   Greedy algorithms do not always consider overlapping subproblems, recursion without memoization repeats work, and brute force is inefficient.

3. Final Answer:

   Dynamic Programming -> Option A

4. Quick Check:

   Overlapping subproblems + optimal substructure = Dynamic Programming [OK]

Hint: Look for repeated subproblems and optimal substructure [OK]

Common Mistakes:

• Confusing greedy with DP for overlapping subproblems
• Ignoring memoization in recursion
• Choosing brute force for efficiency

2. Which of the following is the correct syntax to define a recursive function in JavaScript?

easy

A. function factorial(n) { while(n > 0) return n * factorial(n - 1); }

B. function factorial(n) { if (n === 0) return 1; return n * factorial(n - 1); }

C. function factorial(n) { for(let i=0; i

D. function factorial(n) { if (n === 0) return 0; return n + factorial(n - 1); }

Solution

1. Step 1: Check base case correctness

   function factorial(n) { if (n === 0) return 1; return n * factorial(n - 1); } returns 1 when n is 0, which is correct for factorial.

2. Step 2: Check recursive call and return statement

   function factorial(n) { if (n === 0) return 1; return n * factorial(n - 1); } correctly multiplies n by factorial(n-1). Others have incorrect loops or wrong base cases.

3. Final Answer:

   function factorial(n) { if (n === 0) return 1; return n * factorial(n - 1); } -> Option B

4. Quick Check:

   Correct base case and recursive call = function factorial(n) { if (n === 0) return 1; return n * factorial(n - 1); } [OK]

Hint: Base case must stop recursion; recursive call must reduce problem size [OK]

Common Mistakes:

• Using loops inside recursion incorrectly
• Wrong base case value
• Returning addition instead of multiplication for factorial

3. What will be the output of the following JavaScript code?

function fib(n, memo = {}) {
  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. 8

B. 3

C. 5

D. 13

Solution

1. Step 1: Understand the Fibonacci function with memoization

   fib(5) calculates the 5th Fibonacci number using memo to avoid repeated work.

2. Step 2: Calculate fib(5) step-by-step

   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:

   fib(5) = 5 with memoization [OK]

Hint: Memoization caches results to speed recursion [OK]

Common Mistakes:

• Confusing Fibonacci index with value
• Ignoring memo object causing slow recursion
• Mixing up base cases

4. Find the error in this greedy algorithm code snippet for coin change problem:

function coinChange(coins, amount) {
  coins.sort();
  let count = 0;
  for (let i = coins.length - 1; i >= 0; i--) {
    while (amount >= coins[i]) {
      amount -= coins[i];
      count++;
    }
  }
  return amount === 0 ? count : -1;
}

medium

A. Return statement should always return count

B. Missing base case for amount = 0

C. Using while loop inside for loop is incorrect

D. coins.sort() sorts ascending, should sort descending

Solution

1. Step 1: Check sorting order for greedy coin change

   Greedy coin change requires coins sorted descending to pick largest first.

2. Step 2: Identify sorting mistake

   coins.sort() sorts ascending by default, so it should be coins.sort((a,b) => b - a).

3. Final Answer:

   coins.sort() sorts ascending, should sort descending -> Option D

4. Quick Check:

   Sort descending for greedy coin change [OK]

Hint: Sort coins descending for greedy coin change [OK]

Common Mistakes:

• Using default ascending sort
• Ignoring sorting affects greedy choice
• Assuming while loop is wrong here

5. You want to find the minimum number of jumps to reach the end of an array where each element represents max jump length at that position. Which approach is best and why?

const arr = [2,3,1,1,4];

hard

A. Greedy, because it chooses the farthest reachable index at each step

B. Recursion without memoization, to explore all paths

C. Dynamic Programming, to store minimum jumps for each index

D. Brute force, to try every possible jump

Solution

1. Step 1: Understand the problem nature

   Minimum jumps problem can be solved efficiently by greedy approach choosing the farthest reachable index.

2. Step 2: Compare approaches

   Recursion without memoization and brute force are inefficient; DP is possible but greedy is optimal and simpler here.

3. Final Answer:

   Greedy, because it chooses the farthest reachable index at each step -> Option A

4. Quick Check:

   Minimum jumps = Greedy approach [OK]

Hint: Pick farthest reachable index greedily for minimum jumps [OK]

Common Mistakes:

• Using recursion without memoization causing timeout
• Overcomplicating with DP when greedy suffices
• Trying brute force leading to inefficiency