Home > Java > javaTutorial > body text

LeetCode DayDynamic Programming Part 5

WBOY
Release: 2024-07-16 12:01:49
Original
677 people have browsed it

LeetCode DayDynamic Programming Part 5

518. Coin Change II

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
Original Page

    public int change(int amount, int[] coins) {
        int[][] dp = new int[coins.length+1][amount+1];
        for(int i=0; i<=coins.length; i++){
            dp[i][0] = 1;
        }
        for(int i=1; i<= coins.length; i++){
            for(int j=1; j<=amount; j++){
                if(j<coins[i-1]){
                    dp[i][j] = dp[i-1][j];
                }else{
                    dp[i][j] = dp[i-1][j] + dp[i][j-coins[i-1]];
                }

            }
        }
                    // Arrays.stream(dp).map(Arrays::toString).forEach(System.out::println);
        return dp[coins.length][amount];
    }
Copy after login
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount+1];
        dp[0] = 1;
        for(int i=0; i< coins.length; i++){
            for(int j=coins[i]; j<=amount; j++){
                dp[j] = dp[j] + dp[j-coins[i]];
            }
            System.out.println(Arrays.toString(dp));
        }
        return dp[amount];
    }
Copy after login

377. Combination Sum IV

Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

Example 1:

Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
Example 2:

Input: nums = [9], target = 3
Output: 0

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 1000
All the elements of nums are unique.
1 <= target <= 1000

Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for(int i=1; i<=target; i++){
            for(int j=0; j<nums.length; j++){
                if(nums[j] <= i){
                    dp[i] = dp[i] + dp[i-nums[j]];
                }
            }
            // System.out.println(Arrays.toString(dp));
        }
        return dp[target];
    }
Copy after login

The above is the detailed content of LeetCode DayDynamic Programming Part 5. For more information, please follow other related articles on the PHP Chinese website!

source:dev.to
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!