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Coin Change 2 in C++
Suppose we have coins of different denominations and a total amount of money. we have to Write a module to compute the number of combinations that make up that amount. we can assume that we have infinite number of each kind of coin. So if the amount is 5 and coins are [1, 2, 5], then there are four combinations. (1+1+1+1+1), (1+1+1+2), (1+2+2), (5)
To solve this, we will follow these steps −
- create one array dp of size amount + 1
- dp[0] := 1
- n := size of coins array
- for i in range 0 to n – 1
- for j in range coins[i] to amount
- dp[j] := dp[j – coins[i]]
- for j in range coins[i] to amount
- return dp[amount]
Example(C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int change(int amount, vector<int>& coins) {
vector <int> dp(amount + 1);
dp[0] = 1;
int n = coins.size();
for(int i = 0; i < n; i++){
for(int j = coins[i]; j <= amount; j++){
dp[j] += dp[j - coins[i]];
}
}
return dp[amount];
}
};
main(){
Solution ob;
vector<int> v = {1,2,5};
cout << (ob.change(5, v));
}Input
5 [1,2,5]
Output
4
Updated on: 29-Apr-2020
339 Views
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