Coin Change Minimum Coins in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimum coins for amount 11

What is the output of the following TypeScript code that finds the minimum number of coins needed to make amount 11 using coins [1, 2, 5]?

DSA Typescript

function minCoins(coins: number[], amount: number): number {
  const dp = new 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];
}

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

A 3

B 4

C 5

D -1

💡 Hint

Think about how many coins of 5, 2, and 1 can sum to 11 with minimum count.

The minimum coins to make 11 using [1, 2, 5] are 3 coins: 5 + 5 + 1.

❓ Predict Output

intermediate

Output when no solution exists

What is the output of this TypeScript code when trying to make amount 3 with coins [2, 4]?

DSA Typescript

function minCoins(coins: number[], amount: number): number {
  const dp = new 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];
}

console.log(minCoins([2, 4], 3));

A 1

B -1

C 0

D 3

💡 Hint

Check if amount 3 can be formed by any combination of 2 and 4.

Amount 3 cannot be formed using coins 2 and 4, so the function returns -1.

🔧 Debug

advanced

Identify the bug in coin change code

The following TypeScript code is intended to find the minimum number of coins needed for a given amount. What is the bug causing incorrect results?

DSA Typescript

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

console.log(minCoins([1, 3, 4], 6));

A Return statement should check for dp[0] instead of dp[amount]

B Loop should start from i = 0 instead of 1

C dp array initialized with zeros causes wrong base case

D Coins array should be sorted before processing

💡 Hint

Check how dp array is initialized and how it affects calculations.

The dp array is initialized with zeros, so dp[0] = 0 but other dp[i] are also zero initially, which is incorrect. They should be initialized to Infinity except dp[0].

🧠 Conceptual

advanced

Time complexity of coin change minimum coins algorithm

What is the time complexity of the dynamic programming solution for the coin change minimum coins problem with n coins and amount m?

A O(n * m)

B O(n + m)

C O(n^m)

D O(m^n)

💡 Hint

Consider nested loops over coins and amounts.

The algorithm uses two nested loops: one over the amount (m) and one over the coins (n), resulting in O(n * m) time complexity.

🚀 Application

expert

Minimum coins with unlimited and limited coin usage

Given coins [1, 3, 4] and amount 6, what is the minimum number of coins needed if coin 3 can only be used once, but others unlimited?

A 2

B -1

C 4

D 3

💡 Hint

Think about using coin 3 once and then filling the rest with 1 or 4.

Using coin 3 once, the remaining amount is 3. Minimum coins for 3 using 1 and 4 is 3 (1+1+1). Total coins = 1 + 3 = 4 is wrong. But better is using 4 once and 1 twice: 4 + 1 + 1 = 6 with 3 coins total. Since coin 3 can be used once, best is 3 coins.

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