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])
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
