Greedy vs DP How to Know Which to Apply in DSA Javascript - Complexity Comparison

Time Complexity: Greedy vs DP How to Know Which to Apply

O(amount x coins)

Understanding Time Complexity

We want to understand how to decide between greedy and dynamic programming (DP) approaches.

Which method runs faster and when should we use each?

Scenario Under Consideration

Analyze the time complexity difference between greedy and DP for a simple problem.

// Greedy approach example
function greedyCoins(coins, amount) {
  let count = 0;
  for (let coin of coins) {
    while (amount >= coin) {
      amount -= coin;
      count++;
    }
  }
  return count;
}

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

The greedy function tries the largest coins first, while the DP function tries all possibilities to find the minimum coins.

Identify Repeating Operations

Look at the loops and repeated checks in each approach.

• Greedy primary operation: Outer loop over coins, inner while loop subtracting coin value repeatedly.
• Greedy how many times: The inner loop runs roughly amount / coin times for each coin.
• DP primary operation: Nested loops: outer loop from 1 to amount, inner loop over coins.
• DP how many times: Approximately amount x number of coins iterations.

How Execution Grows With Input

Compare how the number of operations grows as amount increases.

Input Size (amount)	Greedy Approx. Operations	DP Approx. Operations
10	Depends on coin sizes, roughly 10 steps	10 x coins steps
100	Up to 100 steps	100 x coins steps
1000	Up to 1000 steps	1000 x coins steps

Greedy steps grow roughly linearly with amount but can be faster if coins are large; DP always grows linearly with amount times coins.

Final Time Complexity

Time Complexity: O(amount x coins) for DP, O(amount x coins) for greedy in worst cases

DP checks all possibilities, so it takes more time but finds the best answer; greedy is faster but may fail if problem needs full exploration.

Common Mistake

[X] Wrong: "Greedy always gives the best answer because it picks the biggest coin first."

[OK] Correct: Some problems need checking all options because greedy choices can miss better combinations.

Interview Connect

Knowing when to use greedy or DP shows you understand problem structure and efficiency trade-offs, a key skill in coding challenges.

Self-Check

"What if the coin denominations are not sorted or have duplicates? How would that affect the time complexity and choice between greedy and DP?"

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 results of subproblems

B. When the problem requires checking all possible solutions

C. When overlapping subproblems are present

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, known as the greedy choice property.

2. Step 2: Compare with dynamic programming conditions

   Dynamic programming is used when problems have overlapping subproblems and optimal substructure, but greedy algorithms require only the greedy choice property and optimal substructure.

3. Final Answer:

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

4. Quick Check:

   Greedy choice property = Use greedy [OK]

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

Common Mistakes:

• Confusing overlapping subproblems as greedy condition
• Using dynamic programming when greedy suffices
• Ignoring greedy choice property

2. Which syntax correctly initializes a memoization array for dynamic programming in JavaScript?

easy

A. const memo = new Array(n).fill(-1);

B. const memo = Array(n).map(() => 0);

C. const memo = new Array(n).map(() => 0);

D. const memo = Array.fill(n, -1);

Solution

1. Step 1: Understand Array initialization

   Using new Array(n).fill(-1) creates an array of length n filled with -1, suitable for memoization since -1 indicates uncomputed.

2. Step 2: Check other options

   A and C fill with 0, which conflicts if 0 is a valid result; map is also unnecessary. D is invalid syntax.

3. Final Answer:

   const memo = new Array(n).fill(-1); -> Option A

4. Quick Check:

   Use new Array(n).fill(value) for memo [OK]

Hint: Use new Array(n).fill(value) to initialize arrays [OK]

Common Mistakes:

• Using map on uninitialized array
• Wrong order of fill arguments
• Using Array.fill which is invalid

3. What is the output of the following JavaScript code?

function maxSum(arr) {
  let incl = arr[0];
  let excl = 0;
  for (let i = 1; i < arr.length; i++) {
    let new_excl = Math.max(incl, excl);
    incl = excl + arr[i];
    excl = new_excl;
  }
  return Math.max(incl, excl);
}
console.log(maxSum([3, 2, 7, 10]));

medium

A. 17

B. 13

C. 20

D. 22

Solution

1. Step 1: Trace variables for each element

   Start: incl=3, excl=0 i=1: new_excl=max(3,0)=3; incl=0+2=2; excl=3 i=2: new_excl=max(2,3)=3; incl=3+7=10; excl=3 i=3: new_excl=max(10,3)=10; incl=3+10=13; excl=10

2. Step 2: Calculate final max

   Return max(incl, excl) = max(13, 10) = 13

3. Final Answer:

   17 -> Option A

4. Quick Check:

   Max sum non-adjacent = 17 [OK]

Hint: Track incl and excl sums carefully [OK]

Common Mistakes:

• Adding adjacent elements mistakenly
• Mixing incl and excl updates
• Misreading problem as max sum of all elements

4. Identify the error in this greedy algorithm code for coin change problem and fix it:

function coinChange(coins, amount) {
  coins.sort();
  let count = 0;
  for (let i = 0; i < coins.length; i++) {
    while (amount >= coins[i]) {
      amount -= coins[i];
      count++;
    }
  }
  return amount === 0 ? count : -1;
}
console.log(coinChange([1, 3, 4], 6));

medium

A. Return count without checking amount

B. Change while loop to if condition

C. Sort coins in descending order using coins.sort((a,b) => b - a)

D. Use recursion instead of loop

Solution

1. Step 1: Check sorting order

   coins.sort() sorts ascending (smallest first), but loop from i=0 processes small coins first, not greedy.

2. Step 2: Fix sorting for greedy approach

   Use coins.sort((a,b) => b - a) to sort descending numerically, so largest coins used first.

3. Final Answer:

   Sort coins in descending order using coins.sort((a,b) => b - a) -> Option C

4. Quick Check:

   Correct numeric descending sort fixes greedy coin change [OK]

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

Common Mistakes:

• Using default sort causing wrong order
• Changing loop logic unnecessarily
• Ignoring amount check in return

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

const tasks = [
  {start: 1, end: 3},
  {start: 2, end: 5},
  {start: 4, end: 7},
  {start: 6, end: 9}
];

Choose the best method to solve this scheduling problem.

hard

A. Use recursion without memoization

B. Use a greedy algorithm by sorting tasks by earliest end time

C. Use dynamic programming to check all subsets of tasks

D. Sort tasks by start time and pick the longest tasks first

Solution

1. Step 1: Understand problem type

   This is the classic activity selection problem, where we want the maximum number of non-overlapping intervals.

2. Step 2: Identify best approach

   Greedy algorithm works best by sorting tasks by earliest end time and selecting tasks that start after the last selected task ends.

3. Final Answer:

   Use a greedy algorithm by sorting tasks by earliest end time -> Option B

4. Quick Check:

   Greedy by earliest end time solves scheduling [OK]

Hint: Sort by end time, pick earliest finishing tasks [OK]

Common Mistakes:

• Trying all subsets with DP unnecessarily
• Sorting by start time instead of end time
• Picking longest tasks first causes overlaps