Coin Change Total Number of Ways in DSA Java - Time & Space Complexity

Time Complexity: Coin Change Total Number of Ways

O(c * n)

Understanding Time Complexity

We want to understand how the time needed to find the total number of ways to make change grows as the amount or number of coins increases.

How does the work grow when we have more coins or a bigger amount?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int countWays(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    dp[0] = 1;
    for (int coin : coins) {
        for (int x = coin; x <= amount; x++) {
            dp[x] += dp[x - coin];
        }
    }
    return dp[amount];
}

This code calculates how many ways we can make the given amount using the coins available.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where for each coin, we loop through amounts from coin value up to the target amount.
• How many times: Outer loop runs once per coin (let's say c coins), inner loop runs once per amount from coin to amount (up to n times).

How Execution Grows With Input

As the number of coins or the amount grows, the total operations grow roughly by multiplying these two.

Input Size (coins c, amount n)	Approx. Operations
c=5, n=10	About 50 operations
c=5, n=100	About 500 operations
c=10, n=1000	About 10,000 operations

Pattern observation: Doubling coins or amount roughly doubles the work, so work grows like coins times amount.

Final Time Complexity

Time Complexity: O(c * n)

This means the time needed grows proportionally to the number of coins times the amount we want to make.

Common Mistake

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

[OK] Correct: Each coin adds a loop over the amounts, so more coins mean more work overall.

Interview Connect

Understanding this helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts, a skill valued in many coding challenges.

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 Total Number of Ways" problem ask you to find?

easy

A. The total number of ways to make a given amount using unlimited coins of given denominations

B. The minimum number of coins needed to make a given amount

C. The maximum value of coins that can be used to make a given amount

D. The total number of coins available in the input array

Solution

1. Step 1: Understand the problem statement

   The problem asks for counting all possible ways to form a target amount using unlimited coins of given denominations.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, here we count ways, not minimize coins.

3. Final Answer:

   The total number of ways to make a given amount using unlimited coins of given denominations -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins needed [OK]

Common Mistakes:

• Confusing with minimum coins problem
• Thinking about maximum coin value
• Counting coins instead of ways

2. Which of the following Java code snippets correctly initializes the DP array for the Coin Change Total Number of Ways problem for amount 5?

easy

A. int[] dp = new int[6]; dp[0] = 1;

B. int[] dp = new int[5]; dp[0] = 1;

C. int[] dp = new int[6]; dp[1] = 1;

D. int[] dp = new int[5]; dp[1] = 1;

Solution

1. Step 1: Determine DP array size

   For amount 5, dp array size should be 6 to include index 0 to 5.

2. Step 2: Initialize base case

   dp[0] = 1 means one way to make amount 0 (using no coins).

3. Final Answer:

   int[] dp = new int[6]; dp[0] = 1; -> Option A

4. Quick Check:

   dp size = amount+1, dp[0] = 1 [OK]

Hint: DP array size = amount + 1; dp[0] = 1 base case [OK]

Common Mistakes:

• Using dp array size equal to amount instead of amount+1
• Initializing dp[1] instead of dp[0]
• Forgetting to set base case dp[0] = 1

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

int[] coins = {1, 2};
int amount = 3;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
  for (int i = coin; i <= amount; i++) {
    dp[i] += dp[i - coin];
  }
}
System.out.println(dp[amount]);

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] initially, dp[0] = 1.

2. Step 2: Process coin 1

   For i=1 to 3: dp[1]+=dp[0]=1, dp[2]+=dp[1]=1, dp[3]+=dp[2]=1; dp = [1,1,1,1]

3. Step 3: Process coin 2

   For i=2 to 3: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; dp = [1,1,2,2]

4. Step 4: Final dp[3]

   dp[3] = 2 means 2 ways: {1+1+1, 1+2}

5. Final Answer:

   2 -> Option C

6. Quick Check:

   Ways to make 3 with coins 1,2 = 2 [OK]

Hint: Add ways for each coin from coin to amount [OK]

Common Mistakes:

• Not updating dp correctly inside nested loops
• Confusing dp[i] with dp[i-1]
• Miscounting combinations

4. Identify the error in the following Java code for Coin Change Total Number of Ways:

int[] coins = {1, 2, 3};
int amount = 4;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int i = 1; i <= amount; i++) {
  for (int coin : coins) {
    if (i >= coin) {
      dp[i] += dp[i - coin];
    }
  }
}
System.out.println(dp[amount]);

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition i >= coin is incorrect

D. The outer loop should iterate over coins, inner over amounts

Solution

1. Step 1: Analyze loop order

   Outer loop iterates over amounts, inner over coins, which counts permutations, not combinations.

2. Step 2: Correct loop order

   Outer loop should iterate over coins, inner over amounts to count combinations correctly.

3. Final Answer:

   The outer loop should iterate over coins, inner over amounts -> Option D

4. Quick Check:

   Loop order affects counting ways [OK]

Hint: Outer loop coins, inner loop amounts for combinations [OK]

Common Mistakes:

• Swapping loop order causing wrong counts
• Incorrect dp initialization
• Misusing conditions inside loops

5. Given coins = {1, 3, 5} and amount = 7, how many total ways are there to make amount 7 using unlimited coins? Use the Coin Change Total Number of Ways approach.

hard

A. 5

B. 4

C. 6

D. 7

Solution

1. Step 1: Initialize dp array

   dp size = 8, dp[0] = 1, others 0.

2. Step 2: Process coin 1

   Update dp for i=1 to 7: dp[i] += dp[i-1], dp = [1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   For i=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3; dp = [1,1,1,2,2,2,3,3]

4. Step 4: Process coin 5

   For i=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4; dp = [1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option B

6. Quick Check:

   Count ways with coins 1,3,5 for amount 7 = 4 [OK]

Hint: Add ways coin by coin updating dp array [OK]

Common Mistakes:

• Miscounting dp updates for each coin
• Forgetting base case dp[0] = 1
• Mixing permutations with combinations