Solution for Rosetta Code: Count the coins

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1

What is your hint or solution suggestion?

This isn’t the fastest solution, but it’s one of the simplest solutions. The key equation for coin counting is equal to 25*q + 10*d + 5*n + 5*p = 100. In this case, we can see that pennies can only be counted in batches of five for this problem, hence why there is a coefficient of 5 for p.

The limits for (q,d,n,p) are (0-3,0-9,0-19,and 0-19). All possible subarrays of [q,d,n,p] are created, then Array.map is used to plug them into the equation 25*q + 10*d + 5*n + 5*p. Then, the values are filtered so that only instances where the equation is equal to 100 remain. There are 238 instances, as I left off the 4 cases where all coins are the same, and 238+4 = 242.

Note: If you want to try something faster, learn about what is “Dynamic Programming”.

Solution 1 
function countCoins() {
  var quarter = Array.from(Array(4).keys());
  var dime = Array.from(Array(10).keys());
  var nickel = Array.from(Array(20).keys());
  var penny = Array.from(Array(20).keys());

  var parts = [quarter,dime,nickel,penny],
      result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));

  var result1 = result.map((sub) => sub[0]*25 + sub[1]*10 + sub[2]*5 + sub[3]*5);

  const result2 = new Map([...new Set(result1)].map(
      x => [x, result1.filter(y => y === x).length]
  ));

  return result2.get(100)+4;
}

Challenge: Count the coins

Link to the challenge: