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1335.Minimum-Difficulty-of-a-Job-Schedule

1335. Minimum Difficulty of a Job Schedule

题目地址

题目描述

You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the i-th job, you have to finish all the jobs j where 0 <= j < i).
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You have to finish at least one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the d days. The difficulty of a day is the maximum difficulty of a job done in that day.
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Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th job is jobDifficulty[i].
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Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.
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Example 1:
Input: jobDifficulty = [6,5,4,3,2,1], d = 2
Output: 7
Explanation: First day you can finish the first 5 jobs, total difficulty = 6.
Second day you can finish the last job, total difficulty = 1.
The difficulty of the schedule = 6 + 1 = 7
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Example 2:
Input: jobDifficulty = [9,9,9], d = 4
Output: -1
Explanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.
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Example 3:
Input: jobDifficulty = [1,1,1], d = 3
Output: 3
Explanation: The schedule is one job per day. total difficulty will be 3.
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Example 4:
Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15
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Example 5:
Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843
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Constraints:
1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10

代码

Approach #1 DFS

Time: O(nnd) && Space: O(nd)
class Solution {
public int minDifficulty(int[] jobDifficulty, int d) {
int n = jobDifficulty.length;
if (d > n) {
return -1;
}
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int[][] memo = new int[d + 1][n];
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for (int i = 0; i <= d; i++) {
Arrays.fill(memo[i], -1);
}
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return dfs(d, 0, jobDifficulty, memo);
}
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private int dfs(int day, int index, int[] jobs, int[][] memo) {
final int n = jobs.length;
if (day == 0 && index == n) {
return 0;
}
if (day == 0 || index == n) {
return Integer.MAX_VALUE;
}
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if (memo[day][index] != -1) {
return memo[day][index];
}
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int curMax = jobs[index];
int min = Integer.MAX_VALUE;
for (int i = index; i < n; i++) {
curMax = Math.max(jobs[i], curMax);
int temp = dfs(day - 1, i + 1, jobs, memo);
if (temp != Integer.MAX_VALUE) {
min = Math.min(min, temp + curMax);
}
}
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memo[day][index] = min;
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return min;
}
}

Approach #2 DP

Time: O(nnd) && Space: O(nd)
class Solution {
public int minDifficulty(int[] jobDifficulty, int d) {
int n = jobDifficulty.length;
if (d > n) return -1;
int[][] dp = new int[d + 1][n + 1];
​
// 初始化条件很重要
for (int i = 0; i <= d; i++) {
Arrays.fill(dp[i], Integer.MAX_VALUE);
}
dp[0][0] = 0;
for (int day = 1; day <= d; day++) {
for (int task = day; task <= n; task++) {
int localMax = jobDifficulty[task-1];
for (int s = task; s >= day; s--) {
localMax = Math.max(localMax, jobDifficulty[s-1]);
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if (dp[day-1][s-1] != Integer.MAX_VALUE) { // 注意越界
dp[day][task] = Math.min(dp[day][task], dp[day-1][s-1] + localMax);
}
​
}
}
}
​
return dp[d][n];
}
}