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Posts tagged as “hard”

花花酱 LeetCode 630. Course Schedule III

There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuouslyfor t days and must be finished before or on the dth day. You will start at the 1st day.

Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.

Example:

Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
Explanation: 
There're totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day. 
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day. 
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Note:

  1. The integer 1 <= d, t, n <= 10,000.
  2. You can’t take two courses simultaneously.

Solution: Priority queue

  1. Sort courses by end date
  2. Use a priority queue (Max-Heap) to store the course lengths or far
  3. Swap with a longer course if we could not take the current one

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

C++

花花酱 LeetCode 1000. Minimum Cost to Merge Stones

There are N piles of stones arranged in a row.  The i-th pile has stones[i] stones.

move consists of merging exactly K consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these K piles.

Find the minimum cost to merge all piles of stones into one pile.  If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], K = 2
Output: 20
Explanation: 
We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], K = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore.  So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], K = 3
Output: 25
Explanation: 
We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.

Note:

  • 1 <= stones.length <= 30
  • 2 <= K <= 30
  • 1 <= stones[i] <= 100

Solution: DP

dp[i][j][k] := min cost to merge subarray i ~ j into k piles
Init: dp[i][j][k] = 0 if i==j and k == 1 else inf
ans: dp[0][n-1][1]
transition:
1. dp[i][j][k] = min{dp[i][m][1] + dp[m+1][j][k-1]} for all i <= m < j
2. dp[i][j][1] = dp[i][j][K] + sum(A[i]~A[j])

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

C++

C++/top down

Solution 2: DP

dp[l][i] := min cost to merge [i, i + l) into as less piles as possible. Number of merges will be (l-1) / (K – 1) and
Transition: dp[l][i] = min(dp[m][i] + dp[l – m][i + m]) for 1 <= m < l
if ((l – 1) % (K – 1) == 0) [i, i + l) can be merged into 1 pile, dp[l][i] += sum(A[i:i+l])

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

C++

C++/Top-Down

花花酱 LeetCode 30. Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in sthat is a concatenation of each word in words exactly once and without any intervening characters.

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []

Solution1: HashTable + Brute Force

Time complexity: O((|S| – |W|*l) * |W|*l))
Space complexity: O(|W|*l)

C++

花花酱 LeetCode 1001. Grid Illumination

On a N x N grid of cells, each cell (x, y) with 0 <= x < N and 0 <= y < N has a lamp.

Initially, some number of lamps are on.  lamps[i] tells us the location of the i-th lamp that is on.  Each lamp that is on illuminates every square on its x-axis, y-axis, and both diagonals (similar to a Queen in chess).

For the i-th query queries[i] = (x, y), the answer to the query is 1 if the cell (x, y) is illuminated, else 0.

After each query (x, y) [in the order given by queries], we turn off any lamps that are at cell (x, y) or are adjacent 8-directionally (ie., share a corner or edge with cell (x, y).)

Return an array of answers.  Each value answer[i] should be equal to the answer of the i-th query queries[i].

Example 1:

Input: N = 5, lamps = [[0,0],[4,4]], queries = [[1,1],[1,0]]
Output: [1,0]
Explanation: 
Before performing the first query we have both lamps [0,0] and [4,4] on.
The grid representing which cells are lit looks like this, where [0,0] is the top left corner, and [4,4] is the bottom right corner:
1 1 1 1 1
1 1 0 0 1
1 0 1 0 1
1 0 0 1 1
1 1 1 1 1
Then the query at [1, 1] returns 1 because the cell is lit.  After this query, the lamp at [0, 0] turns off, and the grid now looks like this:
1 0 0 0 1
0 1 0 0 1
0 0 1 0 1
0 0 0 1 1
1 1 1 1 1
Before performing the second query we have only the lamp [4,4] on.  Now the query at [1,0] returns 0, because the cell is no longer lit.

Note:

  1. 1 <= N <= 10^9
  2. 0 <= lamps.length <= 20000
  3. 0 <= queries.length <= 20000
  4. lamps[i].length == queries[i].length == 2

Solution: HashTable

use lx, ly, lp, lq to track the # of lamps that covers each row, col, diagonal, antidiagonal

Time complexity: O(|L| + |Q|)
Space complexity: O(|L|)

C++

C++ v2

花花酱 LeetCode 995. Minimum Number of K Consecutive Bit Flips

In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.

Return the minimum number of K-bit flips required so that there is no 0 in the array.  If it is not possible, return -1.

Example 1:

Input: A = [0,1,0], K = 1
Output: 2
Explanation: Flip A[0], then flip A[2].

Example 2:

Input: A = [1,1,0], K = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].

Example 3:

Input: A = [0,0,0,1,0,1,1,0], K = 3
Output: 3
Explanation:
Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]

Note:

  1. 1 <= A.length <= 30000
  2. 1 <= K <= A.length

Solution: Greedy

From left most, if there is a 0, that bit must be flipped since the right ones won’t affect left ones.

Time complexity: O(nk) -> O(k)
Space complexity: O(1)

C++ / O(nk)

C++ / O(n)