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  1. Prepare
  2. Algorithms
  3. Dynamic Programming
  4. The Coin Change Problem
  5. Discussions

The Coin Change Problem

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746 Discussions

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  • mrweisu
     + 0 comments

    My super non-slick code. Passed all tests though.

    def getWays(n, c): # Write your code here

    rows = len(c)
cols = n + 1

dp = [[0 for _ in range(cols)] for _ in range(rows)]

for row in range(rows):
    for col in range(cols):
        if col == 0:
            dp[row][col] = 1
        elif row == 0:
            if col < c[row]:
                dp[row][col] = 0
            else:
                dp[row][col] = dp[row][col-c[row]]
        else:
            if col < c[row]:
                dp[row][col] = dp[row-1][col]
            else:
                dp[row][col] = dp[row-1][col] + dp[row][col-c[row]]

return dp[rows-1][cols-1]

  • isralop26
     + 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];
}

  • C51_22002029155
     + 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]

  • harsh_k_arora
     + 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

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     + 0 comments

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