Coin Change Minimum Coins in DSA C++ - Time & Space Complexity

Time Complexity: Coin Change Minimum Coins

O(amount x number_of_coins)

Understanding Time Complexity

We want to understand how the time needed to find the minimum coins changes as the amount grows.

How does the work increase when we try to make bigger amounts with given coins?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int minCoins(vector<int>& coins, int amount) {
    vector<int> dp(amount + 1, amount + 1);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
        for (int c : coins) {
            if (c <= i) {
                dp[i] = min(dp[i], dp[i - c] + 1);
            }
        }
    }
    return dp[amount] > amount ? -1 : dp[amount];
}
    

This code finds the minimum number of coins needed to make a given amount using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops: outer loop runs for each amount from 1 to amount, inner loop runs for each coin.
• How many times: Outer loop runs amount times, inner loop runs coins.size() times for each outer iteration.

How Execution Grows With Input

As the amount grows, the code checks each coin for every smaller amount up to the target.

Input Size (amount)	Approx. Operations
10	10 x number_of_coins
100	100 x number_of_coins
1000	1000 x number_of_coins

Pattern observation: The operations grow linearly with the amount and the number of coins.

Final Time Complexity

Time Complexity: O(amount x number_of_coins)

This means the time needed grows proportionally with the amount and how many coin types we have.

Common Mistake

[X] Wrong: "The time depends only on the amount, not the number of coins."

[OK] Correct: Each coin is checked for every amount, so more coins mean more work.

Interview Connect

Understanding this helps you explain how dynamic programming improves efficiency over simple recursion.

It shows you can analyze nested loops and relate them to input sizes clearly.

Self-Check

"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"

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