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

Time Complexity: Coin Change Total Number of Ways

O(n * amount)

Understanding Time Complexity

We want to know how the time needed to find all ways to make change grows as we add more coins or increase the amount.

How does the work increase when the input size changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int countWays(int coins[], int n, int amount) {
    int dp[amount + 1];
    for (int i = 0; i <= amount; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}
    

This code counts how many ways we can make the target amount using the given coins.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop runs over each coin and the inner loop runs over amounts from coin value to target amount.
• How many times: Outer loop runs n times (number of coins), inner loop runs up to amount times.

How Execution Grows With Input

As we add more coins or increase the amount, the work grows by multiplying these two factors.

Input Size (coins n, amount)	Approx. Operations
n=5, amount=10	~50 operations
n=10, amount=100	~1,000 operations
n=20, amount=1000	~20,000 operations

Pattern observation: The total work grows roughly by multiplying the number of coins and the amount.

Final Time Complexity

Time Complexity: O(n * amount)

This means the time needed grows proportionally with both the number of coins and the target amount.

Common Mistake

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

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

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts.

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 target amount using given coin denominations.

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

C. The maximum value of coins that can be used to reach the target.

D. The total sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible ways to combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, not minimum coins.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking it finds maximum coin value
• Assuming it sums coin denominations

2. Which of the following is the correct way to declare the DP array for the Coin Change Total Number of Ways problem in C?

easy

A. int dp[target];

B. int dp[target+1];

C. int dp[target+1][coins];

D. int dp[coins+1];

Solution

1. Step 1: Understand DP array size

   We need to store ways for amounts from 0 to target, so size is target+1.

2. Step 2: Choose correct declaration

   One-dimensional array of size target+1 is used to store ways for each amount.

3. Final Answer:

   int dp[target+1]; -> Option B

4. Quick Check:

   Array size = target+1 [OK]

Hint: DP array size must cover 0 to target inclusive [OK]

Common Mistakes:

• Using dp[target] misses amount 0
• Using 2D array unnecessarily
• Using dp[coins+1] which is unrelated

3. What is the output of the following C code snippet for coins = {1, 2} and target = 3?

int coins[] = {1, 2};
int target = 3;
int dp[4] = {1, 0, 0, 0};
for (int i = 0; i < 2; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. 1

B. 3

C. 2

D. 4

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] means 1 way to make 0 amount.

2. Step 2: Process coin 1

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

4. Final Answer:

   2 -> Option C

5. Quick Check:

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

Hint: Add ways from smaller amounts using each coin [OK]

Common Mistakes:

• Not initializing dp[0] to 1
• Updating dp array incorrectly
• Mixing order of loops

4. Identify the error in this code snippet for calculating total ways to make change:

int coins[] = {1, 2, 3};
int target = 4;
int dp[5] = {0};
dp[0] = 1;
for (int i = 0; i <= 3; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. Loop i should run from 0 to 2, not 0 to 3.

B. dp array size is too small.

C. dp[0] should be initialized to 0.

D. Inner loop should start from 0, not coins[i].

Solution

1. Step 1: Check loop bounds for coins array

   coins has 3 elements indexed 0 to 2, but loop runs i=0 to 3 (4 iterations) causing out-of-bounds access.

2. Step 2: Correct loop range

   Change loop to i=0 to 2 to avoid accessing coins[3] which is invalid.

3. Final Answer:

   Loop i should run from 0 to 2, not 0 to 3. -> Option A

4. Quick Check:

   Array index out-of-bounds error [OK]

Hint: Loop over coins must match array length exactly [OK]

Common Mistakes:

• Using wrong loop bounds causing crashes
• Incorrect dp initialization
• Starting inner loop incorrectly

5. Given coins = {1, 3, 5} and target = 7, how many total ways are there to make change? Use the bottom-up DP approach.

hard

A. 6

B. 5

C. 7

D. 4

Solution

1. Step 1: Initialize dp array

   dp[0]=1, dp[1..7]=0 initially.

2. Step 2: Process coin 1

   Update dp for amounts 1 to 7 by adding dp[j-1]: dp=[1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   Update dp for j=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

   Update dp for j=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 D

6. Quick Check:

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

Hint: Add ways for each coin incrementally [OK]

Common Mistakes:

• Not updating dp correctly for each coin
• Forgetting dp[0] = 1 initialization
• Mixing order of coin processing