/**
 * @params {Array} coins
 * @params {Number} amount
 */
export const change = (coins, amount) => {
  // Create and initialize the storage
  const combinations = new Array(amount + 1).fill(0)
  combinations[0] = 1
  // Determine the direction of smallest sub-problem
  for (let i = 0; i < coins.length; i++) {
    // Travel and fill the combinations array
    for (let j = coins[i]; j < combinations.length; j++) {
      combinations[j] += combinations[j - coins[i]]
    }
  }
  return combinations[amount]
}
/**
 * @params {Array} coins
 * @params {Number} amount
 */
export const coinChangeMin = (coins, amount) => {
  const map = { 0: 1 }
  for (let i = 1; i <= amount; i++) {
    let min = Infinity
    for (const coin of coins) {
      if (i < coin) continue
      min = Math.min(min, 1 + map[i - coin])
    }
    map[i] = min
  }
  return map[amount] === Infinity ? -1 : map[amount] - 1
}

Problem Statement

Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = {S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.

Approach

Let the dp[i] be the length of the coin change of prefix N[1..i]. Our answer is dp[N]. We fill dp[0] as 1 because there is only one way to get 0 coins (We pick no coins).

Now let's try calculate dp[i] for any i. dp[0..i] will store each sub problems from 0 to N. That's why the answer will be dp[N]. First, we need to iterate each coin types to pick them one-by-one. Then we iterate the sub problems from current coin that we pick before to N cents. That's why we must make dp table with N columns.

This is the codes for the Coin Change algorithm:

    for coin_val in S:
        for i in range(coin_val, n + 1):
            dp[i] += dp[i - coin_val]

In the second iteration, for every cent that can be exchanged, we take it by subtracting the i-th column by the value of the coin we take and adding it into the current column. So dp[i] will store the current sub problem.

Time Complexity

O(N * S) in any case

Space Complexity

O(N) - simple implementation. We only need 1D array to store the answer.

Example

Let's say we have 3 coin types [1,2,3] and we want to change for 7 cents. So we will define our table like this.

[1, 0, 0, 0, 0, 0, 0, 0]

0th column will store 1 since there is only one way to get 0 cents.

• For the first iteration we take a coin that has a value of 1. Then for all sub problems, there is only one way to make change. For 7 cents, the only way is {1,1,1,1,1,1,1}. On the final iteration, our table be like:

[1, 1, 1, 1, 1, 1, 1, 1]

• For the second iteration, we take a coin that has a value of 2. From here, all sub problems that can be divided by 2 will store another new way to make change. So, when the iteration stopped at 2nd column it will be like dp[2] += dp[0]. We know that dp[0] stored a value of 1. Thus, dp[2] will store the value of 1 + 1 = 2. From here we know that for 2 cents, there are 2 ways {1,1} and {2}. And this operation will continue. Now our table be like:

[1, 1, 2, 2, 3, 3, 4, 4]

4 ways to make 7 cents using value of 1 and 2. {{1,1,1,1,1,1,1}, {1,1,1,1,1,2}, {1,1,1,2,2}, {1,2,2,2}}

• For the final iteration (3rd iteration), we take a coin that has a value of 3. Like before, now all the columns that can be divided by 3 will store another new way. And the final result will be like:

[1, 1, 2, 3, 4, 5, 7, 8]

So the final answer is 8. 8 ways to make change of 7 cents using all coin types. {{1,1,1,1,1,1,1}, {1,1,1,1,1,2}, {1,1,1,2,2}, {1,2,2,2}, {1,1,1,1,3}, {1,3,3}, {2,2,3}, {1,1,2,3}}

Code Implementation Link

Python

Video Explanation

Total Unique Ways To Make Change by Back To Back SWE