Why Coin Change Minimum Coins in DSA Typescript?

The Big Idea

What if you could instantly know the fewest coins needed for any payment without guessing?

The Scenario

Imagine you have a handful of coins of different values and you want to pay a certain amount using the fewest coins possible. You try to pick coins one by one, guessing which to use, but it takes forever and you often end up with too many coins.

The Problem

Manually trying every combination to find the minimum number of coins is slow and confusing. You might miss better combinations or waste time checking the same sets again and again. It's easy to make mistakes and get stuck.

The Solution

The Coin Change Minimum Coins algorithm uses a smart way to remember the best solutions for smaller amounts and builds up to the target amount. This saves time and ensures you find the fewest coins needed without guessing.

Before vs After

✗ Before

function minCoins(coins: number[], amount: number): number {
  // Try all combinations manually
  // Very slow and complex
  return -1; // placeholder
}

✓ After

function minCoins(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

This method lets you quickly find the smallest number of coins needed for any amount, making payments efficient and error-free.

Real Life Example

When a cashier needs to give change using the fewest coins possible, this algorithm helps decide which coins to hand out fast and correctly.

Key Takeaways

Manual guessing is slow and error-prone.

Dynamic programming remembers best solutions for smaller amounts.

Efficiently finds minimum coins needed for any amount.

Practice

1. What does the "Coin Change Minimum Coins" problem ask you to find?

easy

A. The maximum amount of money you can make with given coins

B. The total number of different coins available

C. The least number of coins needed to make a given amount

D. The sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of coins needed to make a specific amount.

2. Step 2: Differentiate from other coin problems

   It is not about counting coins or maximum amount, but minimizing coin count for exact amount.

3. Final Answer:

   The least number of coins needed to make a given amount -> Option C

4. Quick Check:

   Minimum coins = least number needed [OK]

Hint: Focus on minimum coins to form amount, not total coins [OK]

Common Mistakes:

• Confusing minimum coins with total coin types
• Thinking problem asks for maximum amount
• Ignoring the exact amount requirement

2. Which of the following is the correct TypeScript function signature for a coin change minimum coins function?

easy

A. function minCoins(coins: number[], amount: number): number

B. function minCoins(amount: number, coins: number[]): void

C. function minCoins(coins: number, amount: number[]): number[]

D. function minCoins(coins: string[], amount: string): number

Solution

1. Step 1: Check parameter types and order

   The function should take an array of coin values (number[]) and an amount (number).

2. Step 2: Check return type

   The function returns a number representing minimum coins needed.

3. Final Answer:

   function minCoins(coins: number[], amount: number): number -> Option A

4. Quick Check:

   Correct types and return number [OK]

Hint: Coins array first, amount second, returns number [OK]

Common Mistakes:

• Swapping parameter order
• Using wrong types like string instead of number
• Returning void or array instead of number

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

function minCoins(coins: number[], amount: number): number {
  const dp = Array(amount + 1).fill(amount + 1);
  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] > amount ? -1 : dp[amount];
}

console.log(minCoins([1, 2, 5], 11));

medium

A. 2

B. 3

C. -1

D. 5

Solution

1. Step 1: Understand dp array initialization

   dp array size is 12 (0 to 11), initialized with 12 (amount+1), dp[0] = 0.

2. Step 2: Calculate dp values for amount 11

   Using coins 1,2,5, minimum coins for 11 is 3 (5+5+1).

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Minimum coins for 11 with [1,2,5] = 3 [OK]

Hint: Use dp array to track min coins for each amount [OK]

Common Mistakes:

• Returning dp[amount] without checking if > amount
• Misunderstanding dp initialization
• Confusing coin values and indices

4. Identify the bug in this TypeScript code for minimum coins:

function minCoins(coins: number[], amount: number): number {
  const dp = new Array(amount + 1).fill(0);
  dp[0] = 0;
  for (let i = 1; i <= amount; i++) {
    dp[i] = Infinity;
    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];
}

medium

A. The condition should be coin <= i, not coin < i

B. dp array should be initialized with Infinity, not 0

C. dp[0] should be Infinity, not 0

D. Return statement should check dp[amount] > Infinity

Solution

1. Step 1: Check coin usage condition

   The condition 'coin < i' skips cases where coin equals i, missing exact matches.

2. Step 2: Correct condition to include equal coins

   Change condition to 'coin <= i' to allow using coin equal to current amount.

3. Final Answer:

   The condition should be coin <= i, not coin < i -> Option A

4. Quick Check:

   Include coin equal to amount [OK]

Hint: Use <= to include exact coin matches [OK]

Common Mistakes:

• Using < instead of <= in coin check
• Initializing dp array incorrectly
• Wrong return condition check

5. Given coins = [2, 3, 7] and amount = 13, what is the minimum number of coins needed? Use the coin change minimum coins approach.

hard

A. -1

B. 4

C. 2

D. 3

Solution

1. Step 1: Try combinations to form 13

   Possible combos: 7+3+3 = 3 coins, 2+2+2+7 = 4 coins, etc.

2. Step 2: Identify minimum coins

   Minimum coins is 3 using 7, 3, and 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   7+3+3 = 13 with 3 coins [OK]

Hint: Try largest coins first to minimize count [OK]

Common Mistakes:

• Assuming no solution (-1) without checking combos
• Counting coins incorrectly
• Ignoring combinations with repeated coins