Coin Change Total Number of Ways in DSA C

Mental Model

We want to find how many different ways we can make a total amount using given coin values, counting all combinations.

Analogy: Imagine you have different types of candies and want to fill a jar with a certain number of candies. You count all the ways to combine candies to reach that number.

coins: [1, 2, 3]
amount: 4

ways array (index = amount):
[1, 0, 0, 0, 0]
↑
Initially, 1 way to make 0 amount (use no coins).

Dry Run Walkthrough

Input: coins: [1, 2, 3], amount: 4

Goal: Find total number of ways to make amount 4 using coins 1, 2, and 3.

Step 1: Start with ways array: ways[0]=1 (one way to make 0), others 0

ways: [1, 0, 0, 0, 0]

Why: We know there is exactly one way to make amount 0: use no coins.

Step 2: Use coin 1: update ways for amounts 1 to 4

ways: [1, 1, 1, 1, 1]

Why: For each amount, add ways to make (amount - 1) because coin 1 can be used.

Step 3: Use coin 2: update ways for amounts 2 to 4

ways: [1, 1, 2, 2, 3]

Why: Add ways to make (amount - 2) to current ways, counting combinations using coin 2.

Step 4: Use coin 3: update ways for amounts 3 to 4

ways: [1, 1, 2, 3, 4]

Why: Add ways to make (amount - 3) to current ways, counting combinations using coin 3.

Result:

ways: [1, 1, 2, 3, 4]
Total ways to make 4 = 4

Annotated Code

DSA C

#include <stdio.h>

int coinChangeWays(int coins[], int n, int amount) {
    int ways[amount + 1];
    for (int i = 0; i <= amount; i++) {
        ways[i] = 0;
    }
    ways[0] = 1; // base case: one way to make amount 0

    for (int i = 0; i < n; i++) {
        int coin = coins[i];
        for (int j = coin; j <= amount; j++) {
            ways[j] += ways[j - coin];
        }
    }
    return ways[amount];
}

int main() {
    int coins[] = {1, 2, 3};
    int amount = 4;
    int n = sizeof(coins) / sizeof(coins[0]);
    int totalWays = coinChangeWays(coins, n, amount);
    printf("Total ways to make %d = %d\n", amount, totalWays);
    return 0;
}

ways[0] = 1; // base case: one way to make amount 0

Initialize ways array with one way to make zero amount

for (int i = 0; i < n; i++) {

Iterate over each coin to update ways

for (int j = coin; j <= amount; j++) {

For each amount from coin value to target, update ways

ways[j] += ways[j - coin];

Add ways to make (j - coin) to ways[j], counting new combinations

Output

Total ways to make 4 = 4

Complexity Analysis

Time: O(n * amount) because we update the ways array for each coin and each amount up to target

Space: O(amount) because we use a single array to store ways for all amounts up to target

vs Alternative: Compared to recursive solutions without memoization which can be exponential, this dynamic programming approach is efficient and avoids repeated calculations.

Edge Cases

amount = 0

Returns 1 because there is exactly one way to make zero amount (use no coins)

DSA C

ways[0] = 1; // base case: one way to make amount 0

coins array empty

Returns 0 because no coins means no way to make positive amount

DSA C

for (int i = 0; i < n; i++) {

coins with duplicates

Duplicates do not affect correctness; ways are counted correctly

DSA C

for (int i = 0; i < n; i++) {

When to Use This Pattern

When asked to count the number of ways to make a sum using given coins, use the Coin Change Total Number of Ways pattern with dynamic programming to efficiently count combinations.

Common Mistakes

Mistake: Updating ways array in the wrong order causing counting permutations instead of combinations
Fix: Iterate coins outer loop and amounts inner loop forward to count combinations correctly

Summary

Counts how many different combinations of coins sum up to a target amount.

Use when you need the total number of ways to make a sum from given coin denominations.

The key is to build up the count for each amount by adding ways from smaller amounts using each coin.

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