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lets understand coins problem ,Given N and an array(say coins[]) that contains some numbers(coins in rupees). N is a coin, and the array contains various coins. The task is to make the change of N using the coins of the array. Make a change in such a way that a minimum number of coins are used.

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coins+problem-in-c++

lets understand coins problem ,Given N and an array(say coins[]) that contains some numbers(coins in rupees). N is a coin, and the array contains various coins. The task is to make the change of N using the coins of the array. Make a change in such a way that a minimum number of coins are used.

#include <bits/stdc++.h> using namespace std; int coins[] = {10, 25, 5}; // coins array int dp[1000] = {0}; // array for memoisation int minCoins(int N, int M) // N is the sum , M is total_coins {

for (int i = 0; i <= N; i++) dp[i] = INT_MAX; // Initialise all dp[] value to INT_MAX dp[0] = 0; // Base case if sum becomes zero min rupees = 0 // Iterating in the outer loop for possible values of sum between 1 to N // Since our final solution for sum = N might depend upon any of these values for (int i = 1; i <= N; i++) { // Inner loop denotes the index of coin array. // For each value of sum, to obtain the optimal solution. for (int j = 0; j < M; j++) { // i ?> sum // j ?> next coin index // If we can include this coin in our solution if (coins[j] <= i) { // Solution might include the newly included coin dp[i] = min(dp[i], 1 + dp[i - coins[j]]); } } } return dp[N]; 

}

int main() {

int sum = 30; // the money to convert int total_coins = 3; // total availability of coins cout << "Minimum coins needed are " << minCoins(sum, total_coins); 

}

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lets understand coins problem ,Given N and an array(say coins[]) that contains some numbers(coins in rupees). N is a coin, and the array contains various coins. The task is to make the change of N using the coins of the array. Make a change in such a way that a minimum number of coins are used.

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