Coin Change Minimum Coins in DSA Java - 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 number of coins and the amount affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int coinChange(int[] coins, int amount) {
    int max = amount + 1;
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, max);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
        for (int coin : coins) {
            if (coin <= i) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
    }
    return dp[amount] > amount ? -1 : dp[amount];
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where for each amount from 1 to the target, we check all coin values.
• How many times: Outer loop runs amount times, inner loop runs coins.length times for each outer iteration.

How Execution Grows With Input

As the amount grows, the work grows by checking each coin for every 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 increase linearly with the amount and linearly with the number of coins.

Final Time Complexity

Time Complexity: O(amount x number_of_coins)

This means the time grows proportionally to the product of the amount and the number of coin types.

Common Mistake

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

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

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why dynamic programming is a good choice here.

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?
Given coins of different denominations and a total amount, find the minimum number of coins needed to make that amount.

easy

A. Minimum number of coins needed to make the amount

B. Maximum number of coins needed to make the amount

C. Number of ways to make the amount

D. Sum of all coin denominations

Solution

1. Step 1: Understand the problem statement

   The problem asks for the least number of coins needed to reach a given amount using available coin denominations.

2. Step 2: Differentiate from similar problems

   It is not about counting ways or maximum coins, but the minimum coins required.

3. Final Answer:

   Minimum number of coins needed to make the amount -> Option A

4. Quick Check:

   Minimum coins = minimum coins needed [OK]

Hint: Focus on 'minimum coins' needed, not ways or sums [OK]

Common Mistakes:

• Confusing minimum coins with number of ways
• Thinking about maximum coins instead of minimum
• Ignoring the total amount requirement

2. Which of the following Java method signatures correctly represents a function to find minimum coins for a given amount?

public static int minCoins(int[] coins, int amount)

easy

A. public int minCoins(int coins, int[] amount)

B. public static int minCoins(int[] coins, int amount)

C. static void minCoins(int[] coins, int amount)

D. public static int minCoins(int amount, int[] coins)

Solution

1. Step 1: Check parameter types and order

   The method should accept an array of coin denominations and an integer amount, in that order.

2. Step 2: Check return type and modifiers

   It should return an int representing minimum coins and be static for utility use.

3. Final Answer:

   public static int minCoins(int[] coins, int amount) -> Option B

4. Quick Check:

   Correct signature = public static int minCoins(int[] coins, int amount) [OK]

Hint: Array first, amount second; return int minimum coins [OK]

Common Mistakes:

• Swapping parameter order
• Wrong return type (void instead of int)
• Missing static keyword if utility method

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

public class CoinChange {
    public static int minCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        java.util.Arrays.fill(dp, max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 3, 4};
        int amount = 6;
        System.out.println(minCoins(coins, amount));
    }
}

medium

A. -1

B. 3

C. 1

D. 2

Solution

1. Step 1: Understand the DP array filling

   dp[i] stores minimum coins for amount i. Initialized with max (7 here) except dp[0]=0.

2. Step 2: Calculate dp values for amount=6

   Using coins {1,3,4}: - dp[1] = 1 (1 coin of 1) - dp[2] = 2 (1+1) - dp[3] = 1 (coin 3) - dp[4] = 1 (coin 4) - dp[5] = 2 (4+1) - dp[6] = min(dp[6-1]+1=dp[5]+1=3, dp[6-3]+1=dp[3]+1=2, dp[6-4]+1=dp[2]+1=3) = 2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array updates for each amount [OK]

Common Mistakes:

• Not initializing dp array correctly
• Returning dp[amount] without checking if solution exists
• Confusing dp indices or coin subtraction

4. Find the error in the following Java code snippet for minimum coin change:

public static int minCoins(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, Integer.MAX_VALUE);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
        for (int coin : coins) {
            if (coin <= i) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
    }
    return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}

medium

A. Using Integer.MAX_VALUE causes overflow when adding 1

B. dp array size should be amount, not amount + 1

C. dp[0] should be Integer.MAX_VALUE, not 0

D. The for loop should start from 0, not 1

Solution

1. Step 1: Identify the problem with Integer.MAX_VALUE

   Adding 1 to Integer.MAX_VALUE causes integer overflow, leading to incorrect dp values.

2. Step 2: Explain correct initialization

   Use a large number less than Integer.MAX_VALUE (like amount+1) to avoid overflow when adding 1.

3. Final Answer:

   Using Integer.MAX_VALUE causes overflow when adding 1 -> Option A

4. Quick Check:

   Integer.MAX_VALUE + 1 overflows [OK]

Hint: Avoid Integer.MAX_VALUE for dp initialization to prevent overflow [OK]

Common Mistakes:

• Ignoring overflow when adding 1 to max value
• Wrong dp array size
• Incorrect dp[0] initialization

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

medium

A. 1

B. 3

C. 2

D. -1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins: 2, 5, 10 - 2+5=7 (2 coins) - 5+2=7 (2 coins) - 10 is too big But 2+5=7 is valid, so minimum coins = 2.

2. Step 2: Verify if amount 7 can be formed

   Actually, 2+5=7 is valid, so minimum coins = 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   7 = 5 + 2 = 2 coins [OK]

Hint: Try largest coin first, then smaller coins [OK]

Common Mistakes:

• Assuming no solution without checking combinations
• Ignoring order of coins
• Returning -1 prematurely