Why Coin Change Minimum Coins in DSA C++?

The Big Idea

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

The Scenario

Imagine you have a pile of coins of different values and you want to pay a certain amount using the fewest coins possible. Without a method, you try every combination by hand, which takes forever and is confusing.

The Problem

Trying all coin combinations manually is slow and easy to mess up. You might pick too many small coins or miss a better combination, wasting time and money.

The Solution

The Coin Change Minimum Coins method uses a smart way to find the smallest number of coins needed. It checks smaller amounts first and builds up to the target, saving time and avoiding mistakes.

Before vs After

✗ Before

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

✓ After

#include <climits>
#include <algorithm>

int minCoins(int coins[], int n, int amount) {
  int dp[amount+1];
  dp[0] = 0;
  for (int i = 1; i <= amount; i++) {
    dp[i] = INT_MAX;
    for (int j = 0; j < n; j++) {
      if (coins[j] <= i && dp[i - coins[j]] != INT_MAX) {
        dp[i] = std::min(dp[i], dp[i - coins[j]] + 1);
      }
    }
  }
  return dp[amount] == INT_MAX ? -1 : dp[amount];
}

What It Enables

This method lets you quickly find the least number of coins needed to make any amount, saving time and effort.

Real Life Example

When a cashier needs to give change using the fewest coins possible, this method helps find the best way fast.

Key Takeaways

Manual trial of coin combinations is slow and error-prone.

Dynamic programming builds solutions from smaller amounts efficiently.

Minimum coins needed can be found quickly for any amount.

Practice

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

easy

A. The fewest number of coins needed to make a given amount

B. The total number of different coins available

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

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 using given coin denominations.

2. Step 2: Identify the correct description

   The fewest number of coins needed to make a given amount correctly describes this goal, while others describe unrelated concepts.

3. Final Answer:

   The fewest number of coins needed to make a given amount -> Option A

4. Quick Check:

   Minimum coins = fewest coins needed [OK]

Hint: Focus on 'minimum coins' needed for the amount [OK]

Common Mistakes:

• Confusing minimum coins with total coin types
• Thinking about maximum amount instead of minimum coins
• Mixing sum of coins with count of coins

2. Which of the following is the correct C++ syntax to declare a vector named dp of size amount+1 initialized with INT_MAX?

easy

A. vector<int> dp(amount, 0);

B. std::vector dp(amount+1, INT_MAX);

C. std::vector<int> dp(amount+1, INT_MAX);

D. vector<int> dp = new vector(amount+1, INT_MAX);

Solution

1. Step 1: Check vector declaration syntax

   Correct syntax uses std::vector<int> dp(size, initial_value); std::vector<int> dp(amount+1, INT_MAX); matches this exactly.

2. Step 2: Identify errors in other options

   vector<int> dp(amount, 0); initializes with 0 and wrong size; std::vector dp(amount+1, INT_MAX); misses template brackets; vector<int> dp = new vector(amount+1, INT_MAX); uses incorrect new syntax.

3. Final Answer:

   std::vector<int> dp(amount+1, INT_MAX); -> Option C

4. Quick Check:

   Correct vector declaration = std::vector<int> dp(amount+1, INT_MAX); [OK]

Hint: Use std::vector<Type> name(size, value) for initialization [OK]

Common Mistakes:

• Omitting template brackets <>
• Using wrong initialization syntax
• Confusing size and initial value parameters

3. What is the output of the following C++ code snippet?

int coins[] = {1, 2, 5};
int amount = 11;
std::vector<int> dp(amount+1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
  for (int c : coins) {
    if (i - c >= 0 && dp[i - c] != INT_MAX) {
      dp[i] = std::min(dp[i], dp[i - c] + 1);
    }
  }
}
std::cout << dp[amount] << std::endl;

medium

A. 11

B. 3

C. 5

D. INT_MAX

Solution

1. Step 1: Understand the DP approach

   The code finds minimum coins to make amount 11 using coins {1,2,5} by building dp array.

2. Step 2: Calculate minimum coins for 11

   11 = 5 + 5 + 1 uses 3 coins, which is minimum. dp[11] = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Minimum coins for 11 = 3 [OK]

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

Common Mistakes:

• Assuming dp[amount] remains INT_MAX
• Counting total coin value instead of count
• Ignoring dp initialization and updates

4. Identify the error in this C++ code snippet for Coin Change Minimum Coins:

int coins[] = {1, 3, 4};
int amount = 6;
std::vector<int> dp(amount+1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
  for (int c : coins) {
    if (i - c >= 0) {
      dp[i] = std::min(dp[i], dp[i - c] + 1);
    }
  }
}
std::cout << dp[amount] << std::endl;

medium

A. dp[0] should be initialized to INT_MAX

B. Wrong initialization of dp vector size

C. Incorrect coin denominations array syntax

D. Missing check if dp[i - c] is INT_MAX before adding 1

Solution

1. Step 1: Analyze dp update condition

   The code updates dp[i] without checking if dp[i - c] is INT_MAX, which can cause overflow.

2. Step 2: Identify missing condition

   Must check dp[i - c] != INT_MAX before adding 1 to avoid invalid calculations.

3. Final Answer:

   Missing check if dp[i - c] is INT_MAX before adding 1 -> Option D

4. Quick Check:

   Check dp[i - c] != INT_MAX before update [OK]

Hint: Always check dp[i-c] != INT_MAX before adding 1 [OK]

Common Mistakes:

• Skipping dp[i-c] validity check
• Incorrect dp vector size
• Wrong initialization of dp[0]

5. Given coin denominations {2, 5, 10} and amount 7, what is the minimum number of coins needed? If it is not possible, return -1.

hard

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins are 2, 5, 10. 10 > 7 so ignore. Try 5 + 2 = 7 (2 coins). But 5 + 2 is valid, so minimum coins = 2.

2. Step 2: Verify if 5 + 2 is allowed

   Since 5 and 2 are in denominations, 7 can be made with 2 coins.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   7 = 5 + 2 uses 2 coins [OK]

Hint: Try combinations starting from largest coin [OK]

Common Mistakes:

• Assuming no solution because 7 is odd
• Ignoring valid combinations with smaller coins
• Returning -1 incorrectly