Created
April 19, 2021 00:42
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Saved from https://leetcode.com/problems/coin-change/
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class Solution { | |
// The unbounded knapsack problem | |
// dp[i][j]: the fewest # coins among 0 to i-1 to make up to amount j | |
// dp[i][j] = min(dp[i-1][j], dp[i-1][j-coins[i]] + 1, dp[i][j-coins[i]] + 1) | |
// since dp[i][j-coins[i]] includes dp[i-1][j-coins[i]], thus: | |
// dp[i][j] = min(dp[i-1][j], dp[i][j-coins[i]] + 1) | |
// dp[i][j] < 0 means there is no solution. | |
// | |
// Time: O(n * amount) | |
// Space: O(n * amount) | |
public int coinChange(int[] coins, int amount) { | |
int[][] dp = new int[coins.length + 1][amount + 1]; | |
int MAX_VALUE = Integer.MAX_VALUE-1; | |
Arrays.fill(dp[0], MAX_VALUE); | |
dp[0][0] = 0; | |
for (int i = 1; i < coins.length + 1; i++) { | |
int currCoin = coins[i-1]; | |
for (int j = 0; j <= amount; j++) { | |
int min = dp[i-1][j]; | |
if (j >= currCoin) { | |
min = Math.min(min, dp[i][j-currCoin] + 1); | |
} | |
dp[i][j] = min; | |
} | |
} | |
return dp[coins.length][amount] == MAX_VALUE ? -1 : dp[coins.length][amount]; | |
} | |
} |
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