Minimum Coins for Change

mediumcodingAlgorithmsDynamic Programming~30 min

Asked at Google, Stripe, Meta

Overview

Given an array of coin denominations and a target amount, return the fewest number of coins needed to make up that amount. If the amount cannot be made up by any combination of the given coins, return -1.

You have an infinite supply of each coin denomination.

Constraints

1 <= coins.length <= 12
1 <= coins[i] <= 2^31 - 1
0 <= amount <= 10,000
Each coin denomination is a positive integer.
Coins array may not be sorted.

Examples

minCoins([1, 5, 10, 25], 36);
// => 3  (25 + 10 + 1)

minCoins([2], 3);
// => -1  (impossible with only 2-cent coins)

minCoins([1], 0);
// => 0  (no coins needed for amount 0)

Solution

Reveal solution

function minCoins(coins, amount) {
  const dp = new Array(amount + 1).fill(Infinity);
  dp[0] = 0;

  for (let i = 1; i <= amount; i++) {
    for (const coin of coins) {
      if (coin <= i && dp[i - coin] + 1 < dp[i]) {
        dp[i] = dp[i - coin] + 1;
      }
    }
  }

  return dp[amount] === Infinity ? -1 : dp[amount];
}

Classic bottom-up DP. For each amount i, try subtracting each coin and take the minimum.

Resources

minimum-coins-for-change.js

Minimum Coins for Change

mediumcodingAlgorithmsDynamic Programming

Overview

Given an array of coin denominations and a target amount, return the fewest number of coins needed to make up that amount. If the amount cannot be made up by any combination of the given coins, return -1.

You have an infinite supply of each coin denomination.

Constraints

1 <= coins.length <= 12
1 <= coins[i] <= 2^31 - 1
0 <= amount <= 10,000
Each coin denomination is a positive integer.
Coins array may not be sorted.

Examples

minCoins([1, 5, 10, 25], 36);
// => 3  (25 + 10 + 1)

minCoins([2], 3);
// => -1  (impossible with only 2-cent coins)

minCoins([1], 0);
// => 0  (no coins needed for amount 0)

Solution

Reveal solution

function minCoins(coins, amount) {
  const dp = new Array(amount + 1).fill(Infinity);
  dp[0] = 0;

  for (let i = 1; i <= amount; i++) {
    for (const coin of coins) {
      if (coin <= i && dp[i - coin] + 1 < dp[i]) {
        dp[i] = dp[i - coin] + 1;
      }
    }
  }

  return dp[amount] === Infinity ? -1 : dp[amount];
}

Classic bottom-up DP. For each amount i, try subtracting each coin and take the minimum.

Resources

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