Coin Change
/**
* @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
}
About this Algorithm
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 thatdp[0]stored a value of 1. Thus, dp[2] will store the value of1 + 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}}
