🟨 Coin Change (#322)
🔗 Problem Link
📋 Problem Statement
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.
💡 Examples
Example 1
Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example 2
Input: coins = [2], amount = 3
Output: -1
🔑 Key Insights & Approach
Core Observation: dp[amount] = min coins needed. For each amount, try all coins and take minimum.
Why DP?
- O(amount * coins) time
- O(amount) space
- Bottom-up unbounded knapsack
Pattern: "Unbounded Knapsack DP" pattern.
🐍 Solution: Python
Approach: Bottom-Up DP
Time Complexity: O(amount * n) | Space Complexity: O(amount)
from typing import List
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [float('inf')] * (amount + 1)
dp[0] = 0
for amt in range(1, amount + 1):
for coin in coins:
if amt >= coin:
dp[amt] = min(dp[amt], dp[amt - coin] + 1)
return dp[amount] if dp[amount] != float('inf') else -1
🔵 Solution: Golang
Approach: DP
Time Complexity: O(amount * n) | Space Complexity: O(amount)
import "math"
func coinChange(coins []int, amount int) int {
dp := make([]int, amount+1)
for i := range dp {
dp[i] = math.MaxInt32
}
dp[0] = 0
for amt := 1; amt <= amount; amt++ {
for _, coin := range coins {
if amt >= coin && dp[amt-coin] != math.MaxInt32 {
dp[amt] = min(dp[amt], dp[amt-coin]+1)
}
}
}
if dp[amount] == math.MaxInt32 {
return -1
}
return dp[amount]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}