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| 1 | +/** |
| 2 | + * [Problem]: [322] Coin Change |
| 3 | + * |
| 4 | + * (https://leetcode.com/problems/coin-change/description/) |
| 5 | + */ |
| 6 | +function coinChange(coins: number[], amount: number): number { |
| 7 | + // 시간복잡도: O(c^a) |
| 8 | + // 공간복잡도: O(a) |
| 9 | + // Time Exceed |
| 10 | + function dfsFunc(coins: number[], amount: number): number { |
| 11 | + if (!amount) return 0; |
| 12 | + let result = dfs(amount); |
| 13 | + |
| 14 | + return result <= amount ? result : -1; |
| 15 | + |
| 16 | + function dfs(remain: number): number { |
| 17 | + if (remain === 0) return 0; |
| 18 | + if (remain < 0) return amount + 1; |
| 19 | + |
| 20 | + let min_count = amount + 1; |
| 21 | + |
| 22 | + for (let coin of coins) { |
| 23 | + const result = dfs(remain - coin); |
| 24 | + min_count = Math.min(min_count, result + 1); |
| 25 | + } |
| 26 | + |
| 27 | + return min_count; |
| 28 | + } |
| 29 | + } |
| 30 | + // 시간복잡도: O(ca) |
| 31 | + // 공간복잡도: O(a) |
| 32 | + function dpFunc(coins: number[], amount: number): number { |
| 33 | + const dp = new Array(amount + 1).fill(amount + 1); |
| 34 | + dp[0] = 0; |
| 35 | + |
| 36 | + for (let coin of coins) { |
| 37 | + for (let i = coin; i <= amount; i++) { |
| 38 | + dp[i] = Math.min(dp[i], dp[i - coin] + 1); |
| 39 | + } |
| 40 | + } |
| 41 | + |
| 42 | + return dp[amount] <= amount ? dp[amount] : -1; |
| 43 | + } |
| 44 | + |
| 45 | + // 시간복잡도: O(ca) |
| 46 | + // 공간복잡도: O(a) |
| 47 | + function memoizationFunc(coins: number[], amount: number): number { |
| 48 | + const memo: Record<number, number> = {}; |
| 49 | + |
| 50 | + const result = dfs(amount); |
| 51 | + return result <= amount ? result : -1; |
| 52 | + |
| 53 | + function dfs(remain: number): number { |
| 54 | + if (remain === 0) return 0; |
| 55 | + if (remain < 0) return amount + 1; |
| 56 | + if (remain in memo) return memo[remain]; |
| 57 | + |
| 58 | + let min_count = amount + 1; |
| 59 | + |
| 60 | + for (let coin of coins) { |
| 61 | + const res = dfs(remain - coin); |
| 62 | + min_count = Math.min(min_count, res + 1); |
| 63 | + } |
| 64 | + |
| 65 | + memo[remain] = min_count; |
| 66 | + |
| 67 | + return min_count; |
| 68 | + } |
| 69 | + } |
| 70 | + |
| 71 | + return memoizationFunc(coins, amount); |
| 72 | +} |
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