g0501_0600.s0518_coin_change_2.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0518_coin_change_2;

// #Medium #Array #Dynamic_Programming #Dynamic_Programming_I_Day_20
// #2022_07_25_Time_4_ms_(84.67%)_Space_42.2_MB_(60.30%)

/**
 * 518 - Coin Change 2\.
 *
 * Medium
 *
 * 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.
 *
 * **Example 1:**
 *
 * **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
 *
 * **Example 2:**
 *
 * **Input:** amount = 3, coins = [2]
 *
 * **Output:** 0
 *
 * **Explanation:** the amount of 3 cannot be made up just with coins of 2.
 *
 * **Example 3:**
 *
 * **Input:** amount = 10, coins = [10]
 *
 * **Output:** 1
 *
 * **Constraints:**
 *
 * *   `1 <= coins.length <= 300`
 * *   `1 <= coins[i] <= 5000`
 * *   All the values of `coins` are **unique**.
 * *   `0 <= amount <= 5000`
**/
public class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for (int coin : coins) {
            for (int i = coin; i <= amount; i++) {
                dp[i] += dp[i - coin];
            }
        }
        return dp[amount];
    }
}