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Coin Change - Minimum number of coins

Helper answered on January 2, 2022 Popularity 8/10 Helpfulness 1/10

answer Coin Change - Minimum number of coins

related Coin Change - Minimum number of coins

Coin Change - Minimum number of coins

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Recursion:
def minimumNumberOfCoins(coins,numberOfCoins,value):
    
    def recur(coins, n, value):
        
        if n == 0:
            return float("inf")
            
        if value == 0:
            return 0
            
        if coins[n-1] > value:
            return recur(coins, n-1, value)
        else:
            return min(1 + recur(coins, n, value-coins[n-1]), recur(coins, n-1, value))
            
    res = recur(coins,numberOfCoins,value)
    
    if res == float("inf"):
        return 0
    return res

Bottom Up:
def coinChange(coins, amount):
        
    n = len(coins)
    
    def recur(amount,coins,n,dp):
        
        if dp[n][amount] != -1:
            return dp[n][amount]

        if coins[n-1] > amount:
            dp[n][amount] =  recur(amount,coins,n-1,dp)

        if coins[n-1] <= amount:
            dp[n][amount] =  min(1 + recur(amount - coins[n-1],coins,n,dp), recur(amount,coins,n-1,dp))
        
        return dp[n][amount]
        
    dp = [[-1 for i in range(amount+1)] for j in range(n+1)]
    

    for i in range(amount+1):
        dp[0][i] = float("inf")
    
    for i in range(1, n+1):
        dp[i][0] = 0

    recur(amount,coins,n,dp)
    
    if dp[n][amount] == float("inf"):
        return -1
    return dp[n][amount]

Top Down:
def coinChange(coins, amount):
    n = len(coins)
    def recur(amount,coins,n,dp):
        
        for i in range(1, n+1):
            for j in range(1, amount+1):
                
                if coins[i-1] > j:
                    dp[i][j] = dp[i-1][j]
                else:
                    dp[i][j] = min(1 + dp[i][j-coins[i-1]], dp[i-1][j])
                    
    dp = [[-1 for i in range(amount+1)] for j in range(n+1)]
    

    for i in range(1, amount+1):
        dp[0][i] = float("inf")
    
    for i in range(n+1):
        dp[i][0] = 0
        
    recur(amount,coins,n,dp)
    
    if dp[n][amount] == float("inf"):
        return -1
    return dp[n][amount]

Popularity 8/10 Helpfulness 1/10 Language whatever

Source: practice.geeksforgeeks.org

Tags: coin-change minimum whatever

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Contributed on Jan 02 2022

Helper

0 Answers Avg Quality 2/10

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Coin Change - Minimum number of coins

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class Solution
{
    //Function to find the minimum number of coins to make the change 
    //for value using the coins of given denominations.
    public long minimumNumberOfCoins(int coins[],int n,int value)
    {
        //long res=method1(coins,0,n,value,0);
        //return res==Integer.MAX_VALUE?-1:res;
        return method2(coins,n,value);
    }
    long method1(int coins[],int i,int n,int val,int no){
        if(val<0) return Integer.MAX_VALUE;
        if(val==0) return no;
        if(i==n){
           return Integer.MAX_VALUE;   
        }
        return Math.min(method1(coins,i+1,n,val,no),method1(coins,i,n,val-coins[i],no+1));
    }
    long method2(int coins[],int n,int value){
        long dp[]=new long[value+1];
        for(int i=1;i<=value;++i){
            dp[i]=Integer.MAX_VALUE;
            for(int j=0;j<n;++j) if(i-coins[j]>=0) dp[i]=Math.min(dp[i],dp[i-coins[j]]+1);
        }
        long res=dp[value];
        return res==Integer.MAX_VALUE?-1:res;
    }
    long method3(int coins[],int n,int value){
        int dp[]=new int[value+1];
        dp[0]=0;
        for(int i=1;i<=value;i++) dp[i]=Integer.MAX_VALUE;
        for(int i=1;i<=value;i++){
            for(int j=0;j<n;j++){
                if(coins[j]<=i){
                    int sub_res=dp[i-coins[j]];
                    if(sub_res!=Integer.MAX_VALUE){
                        dp[i]=Math.min(dp[i],sub_res+1);
                    }
                }
            }
        }
        if(dp[value]==Integer.MAX_VALUE) return -1;
        return (long)dp[value];
    }
}

Popularity 7/10 Helpfulness 2/10 Language whatever

Source: Grepper

Tags: minimum whatever coin-change m

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Contributed on Oct 19 2022

Sumit Rawal

0 Answers Avg Quality 2/10

Coin Change - Minimum number of coins

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