# Posts tagged as “hard”

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).

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.

Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th job is jobDifficulty[i].

Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.

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


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.


Example 3:

Input: jobDifficulty = [1,1,1], d = 3
Output: 3
Explanation: The schedule is one job per day. total difficulty will be 3.


Example 4:

Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15


Example 5:

Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843


Constraints:

• 1 <= jobDifficulty.length <= 300
• 0 <= jobDifficulty[i] <= 1000
• 1 <= d <= 10

## Solution: DP

dp[i][k] := min difficulties to schedule jobs 1~i in k days.

Schedule 1 ~ j in k – 1 days and schedule j + 1 ~ i in 1 day.

Init: dp[0][0] = 0
Transition: dp[i][k] := min(dp[j][k -1] + max(jobs[j + 1 ~ i]), k – 1 <= j < i

Time complexity: O(n^2*d)
Space complexity: O(n)

## C++

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

## Solution 1: Brute Force

r[i] = min(max(h[0:i+1]), max(h[i:n]))
ans = sum(r[i])

Time complexity: O(n^2)
Space complexity: O(1)

## Solution 2: DP

l[i] := max(h[0:i+1])
r[i] := max(h[i:n])
ans = sum(min(l[i], r[i]) – h[i])

Time complexity: O(n)
Space complexity: O(n)

## Solution 3: Two Pointers

Time complexity: O(n)
Space complexity: O(1)

## C++

You have a keyboard layout as shown above in the XY plane, where each English uppercase letter is located at some coordinate, for example, the letter A is located at coordinate (0,0), the letter B is located at coordinate (0,1), the letter P is located at coordinate (2,3) and the letter Z is located at coordinate (4,1).

Given the string word, return the minimum total distance to type such string using only two fingers. The distance between coordinates (x1,y1) and (x2,y2) is |x1 – x2| + |y1 – y2|

Note that the initial positions of your two fingers are considered free so don’t count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.

Example 1:

Input: word = "CAKE"
Output: 3
Explanation:
Using two fingers, one optimal way to type "CAKE" is:
Finger 1 on letter 'C' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2
Finger 2 on letter 'K' -> cost = 0
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1
Total distance = 3


Example 2:

Input: word = "HAPPY"
Output: 6
Explanation:
Using two fingers, one optimal way to type "HAPPY" is:
Finger 1 on letter 'H' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'H' to letter 'A' = 2
Finger 2 on letter 'P' -> cost = 0
Finger 2 on letter 'P' -> cost = Distance from letter 'P' to letter 'P' = 0
Finger 1 on letter 'Y' -> cost = Distance from letter 'A' to letter 'Y' = 4
Total distance = 6


Example 3:

Input: word = "NEW"
Output: 3


Example 4:

Input: word = "YEAR"
Output: 7


Constraints:

• 2 <= word.length <= 300
• Each word[i] is an English uppercase letter.

## Solution: DP

Top down: O(n*27^2)

## C++

Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself.

Example 1:

Input: text = "abcabcabc"
Output: 3
Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".


Example 2:

Input: text = "leetcodeleetcode"
Output: 2
Explanation: The 2 substrings are "ee" and "leetcodeleetcode".


Constraints:

• 1 <= text.length <= 2000
• text has only lowercase English letters.

## Solution 1: Brute Force + HashSet

Try all possible substrings

Time complexity: O(n^3)
Space complexity: O(n^2)

## C++

Given a string s. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make s palindrome.

Palindrome String is one that reads the same backward as well as forward.

Example 1:

Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.


Example 2:

Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".


Example 3:

Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".


Example 4:

Input: s = "g"
Output: 0


Example 5:

Input: s = "no"
Output: 1


Constraints:

• 1 <= s.length <= 500
• All characters of s are lower case English letters.

## Solution: DP

dp[i][j] := min chars to insert
dp[j][j] = dp[i-1][j+1] if s[i] == s[j] else min(dp[i+1][j] , dp[i][j-1]) + 1
base case: dp[i][i] = 0
ans: dp[0][n-1]

Time complexity: O(n^2)
Space complexity: O(n^2)

## C++

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