Table of Contents
Coin Change II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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 <= 3001 <= coins[i] <= 5000- All the values of
coinsare unique. 0 <= amount <= 5000
C++ Coin Change II LeetCode Solution
class Solution {
public:
int coinChange(vector<int>& coins, int n, int amount)
{
if(n==0)
return 0;
if(amount == 0)
{
return 1;
}
if(coins[n-1] > amount)
{
return coinChange(coins, n-1, amount);
}
return coinChange(coins, n, amount-coins[n-1]) + coinChange(coins, n-1, amount);
}
int change(int amount, vector<int>& coins) {
int n = coins.size();
if(amount == 0) {
return 1;
}
if(n==0)
return 0;
return coinChange(coins, n, amount);
}
};
Java Coin Change II LeetCode Solution
class Solution {
public int change(int amount, int[] coins) {
Integer[][] dp = new Integer[coins.length][amount + 1];
return dfs(coins, 0, amount, dp);
}
int dfs(int[] coins, int i, int amount, Integer[][] dp) {
if (amount == 0) return 1;
if (i == coins.length) return 0;
if (dp[i][amount] != null) return dp[i][amount];
int ans = dfs(coins, i + 1, amount, dp); // skip ith coin
if (amount >= coins[i])
ans += dfs(coins, i, amount - coins[i], dp); // use ith coin
return dp[i][amount] = ans;
}
}
Python 3 Coin Change II LeetCode Solution
class Solution(object):
def change(self, amount, coins):
"""
:type amount: int
:type coins: List[int]
:rtype: int
"""
#If we found a way to create the desired amount
if amount == 0:
return 1
#If we went over our amount or we have no more coins left
if amount < 0 or len(coins) == 0:
return 0
#Our solutions can be divided into two sets,
# 1) Solutions that cointain the coin at the end of the coins array
# 2) Solutions that don't contain that coin
return self.change(amount - coins[-1], coins) + self.change(amount, coins[:-1])
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