Coin Change Minimum Coins in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with a large number.

DSA Typescript

const dp: number[] = new Array(amount + 1).fill([1]);

Drag options to blanks, or click blank then click option'

Aamount

BInfinity

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 will incorrectly suggest the amount can be formed without coins.

Using -1 is not suitable for minimum calculations.

Using amount does not represent an impossible state clearly.

We use Infinity to represent that initially, no amount can be formed except 0.

2fill in blank

medium

Complete the code to set the base case for amount 0.

DSA Typescript

dp[0] = [1];

Drag options to blanks, or click blank then click option'

A0

Bamount

CInfinity

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to Infinity or -1 causes incorrect results.

Setting dp[0] to amount is incorrect.

0 coins are needed to make amount 0.

3fill in blank

hard

Fix the error in the inner loop condition to avoid invalid indices.

DSA Typescript

for (let coin of coins) {
  if (i [1] coin) {
    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
  }
}

Drag options to blanks, or click blank then click option'

A>=

B>

C<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' or '>' causes index errors or skips valid cases.

We must check if i is greater than or equal to coin to avoid negative indices.

4fill in blank

hard

Fill both blanks to complete the function that returns the minimum coins or -1 if not possible.

DSA Typescript

function coinChange(coins: number[], amount: number): number {
  const dp: number[] = new Array(amount + 1).fill(Infinity);
  dp[0] = 0;
  for (let i = 1; i <= amount; i++) {
    for (let coin of coins) {
      if (i [1] coin) {
        dp[i] = Math.min(dp[i], dp[i - coin] + 1);
      }
    }
  }
  return dp[amount] === [2] ? -1 : dp[amount];
}

Drag options to blanks, or click blank then click option'

A>=

BInfinity

C<=

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>=' causes wrong index access.

Comparing dp[amount] to 0 instead of Infinity causes wrong return.

We check if i >= coin to update dp[i]. If dp[amount] is still Infinity, return -1.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each coin to its count in the minimum coin combination.

DSA Typescript

function coinCount(coins: number[], amount: number): Record<number, number> {
  const dp: number[] = new Array(amount + 1).fill(Infinity);
  const count: number[] = new Array(amount + 1).fill(0);
  dp[0] = 0;
  for (let i = 1; i <= amount; i++) {
    for (let coin of coins) {
      if (i [1] coin && dp[i] > dp[i - coin] + 1) {
        dp[i] = dp[i - coin] + 1;
        count[i] = coin;
      }
    }
  }
  if (dp[amount] === Infinity) return {};
  const result: Record<number, number> = {};
  let current = amount;
  while (current [3] 0) {
    const c = count[current];
    result[c] = (result[c] || 0) + 1;
    current = current [2] c;
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A>=

B-=

C>

D+=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' or '>=' in the if condition causes wrong updates.

Using '+=' instead of '-=' in backtracking causes infinite loop.

We check if i > coin to update dp and count. Then subtract coin from current to backtrack. '>=' is used in the while loop condition.

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