🤖 leetcode 518. Coin Change II | medium | dynamic-programming | javascript

🗒️ Problems

518. Coin Change II - Leetcode

You are given an integer array coins representing coins of different denominations and
an integer amount representing a total amount of money.

Return the number of combinations that make up that amount.
If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.
The answer is guaranteed to fit into a signed 32-bit integer.

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

🤔 Understand problem

• 322. Coin Change asks us to answer the minimum number of coins usage.
• 518. Coin Change II asks us to answer the number of combinations that make up that amount.

⬇️ top-down: dp(index, money) function

🥳⬇️ Think differently

• Is it possible to use coins upto the given amount ?

basecase

• If the money that I summed up money is the same as the given amount, I found one case.
• If the money that I summed up money exceeds the given amount, it's the wrong case.

  const N = coins.length

  const dfs = (index, money) => {
    if (money === amount) return 1
    if (money > amount) return 0
    if (index >= N) return 0

    ...
  }

main-code

• Only choose the same or larger coin among coins[] to prevent duplication.
• For not taking this coin, dfs(index + 1, money).
• For taking this coin, dfs(index, money + coins[index])
  • because I can use the same again later if money is available.

const dfs = (index, money) => {
  // basecase

  let res = 0;
  // use
  res += dfs(index, money + coins[index]);
  // not use
  res += dfs(index + 1, money);

  return res;
};

⬇️ top-down without memoization

• No memoization causes TLE.

var change = function (amount, coins) {
  const N = coins.length;

  const dfs = (index, money) => {
    if (money === amount) return 1;
    if (money > amount) return 0;
    if (index >= N) return 0;

    let res = 0;
    // use
    res += dfs(index, money + coins[index]);
    // not use
    res += dfs(index + 1, money);

    return res;
  };

  return dfs(0, 0);
};

🔥⬇️ top-down with memoization

/**
 * @param {number} amount
 * @param {number[]} coins
 * @return {number}
 */
var change = function (amount, coins) {
  const N = coins.length;
  const cache = {};
  const genKey = (index, money) => `${index}:${money}`;

  const dfs = (index, money) => {
    if (money === amount) return 1;
    if (money > amount) return 0;
    if (index >= N) return 0;
    if (cache[genKey(index, money)] !== undefined)
      return cache[genKey(index, money)];

    let res = 0;
    // use
    res += dfs(index, money + coins[index]);
    // not use
    res += dfs(index + 1, money);

    return (cache[genKey(index, money)] = res);
  };

  return dfs(0, 0);
};

🙆‍♂️ Time complexity: O(N*m)