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  • + 0 comments

    JS Javascript solution passes all tests:

    function getWays(n, c) {
        // Write your code here
        const dp = new Array(c.length).fill(new Array(n).fill(0));
        const sorted = c.sort((a,b)=>a-b);
        for (let i = 0; i<sorted.length; i++){
            for(let j=0; j<=n; j++){
            if (j === 0) {
                dp[i][j] = 1;
                continue;
            }
            if (i === 0){
                dp[i][j] = j%sorted[i] === 0 ? 1 : 0;
                continue;
            }
            if (sorted[i] > j) {
                dp[i][j] = dp[i-1][j];
            } else {
                dp[i][j] = dp[i][j-sorted[i]] + dp[i-1][j];
            }
            }
            if (sorted[i]>n) {
            return dp[i][n];
            }
        }
        return dp[sorted.length-1][n];
    }
    
  • + 0 comments
    def getWays(n, c):
        ways = [0] * (n + 1)
        ways[0] = 1
        for coin in c:
            for i in range(coin, n + 1):
                ways[i] += ways[i - coin]
        return ways[n]
        
    
  • + 0 comments
    def getWays(n, c):
        count = [0]*(n+1)
        count[0] = 1
        c.sort(reverse = True) 
        for coin in c:
            for i in range(coin, n +1):
                count[i] += count[i-coin]
        return count[n]
    		
    

    sorting is done in order to understand how denomination is formed

  • + 0 comments

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  • + 0 comments

    The Coin Change Problem involves finding the number of ways to make a given amount using a set of coins with specified denominations. It's commonly solved using dynamic programming to optimize calculations z7. Let me know if you need further clarification or code examples!