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# Posts tagged as “priority queue”

You are given an array nums of n positive integers.

You can perform two types of operations on any element of the array any number of times:

• If the element is evendivide it by 2.
• For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
• If the element is oddmultiply it by 2.
• For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4].

The deviation of the array is the maximum difference between any two elements in the array.

Return the minimum deviation the array can have after performing some number of operations.

Example 1:

Input: nums = [1,2,3,4]
Output: 1
Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.


Example 2:

Input: nums = [4,1,5,20,3]
Output: 3
Explanation: You can transform the array after two operations to [4,2,5,5,3], then the deviation will be 5 - 2 = 3.


Example 3:

Input: nums = [2,10,8]
Output: 3


Constraints:

• n == nums.length
• 2 <= n <= 105
• 1 <= nums[i] <= 109

## Solution: Priority Queue

If we double an odd number it becomes an even number, then we can only divide it by two which gives us back the original number. So we can pre-double all the odd numbers and only do division in the following process.

We push all numbers including pre-doubled odd ones onto a priority queue, and track the difference between the largest and smallest number.

Each time, we pop the largest number out and divide it by two then put it back to the priority queue, until the largest number becomes odd. We can not discard it and divide any other smaller numbers by two will only increase the max difference, so we can stop here.

ex1: [3, 5, 8] => [6, 8, 10] (pre-double) => [5, 6, 8] => [4, 5, 6] => [3, 4, 5] max diff is 5 – 3 = 2
ex2: [4,1,5,20,3] => [2, 4, 6, 10, 20] (pre-double) => [2, 4, 6, 10] => [2, 4, 5, 6] => [2,3,4,5] max diff = 5-2 = 3

Time complexity: O(n*logm*logn)
Space complexity: O(n)

## C++/PQ

You are given an m * n matrix, mat, and an integer k, which has its rows sorted in non-decreasing order.

You are allowed to choose exactly 1 element from each row to form an array. Return the Kth smallest array sum among all possible arrays.

Example 1:

Input: mat = [[1,3,11],[2,4,6]], k = 5
Output: 7
Explanation: Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.  

Example 2:

Input: mat = [[1,3,11],[2,4,6]], k = 9
Output: 17


Example 3:

Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
Output: 9
Explanation: Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.


Example 4:

Input: mat = [[1,1,10],[2,2,9]], k = 7
Output: 12


Constraints:

• m == mat.length
• n == mat.length[i]
• 1 <= m, n <= 40
• 1 <= k <= min(200, n ^ m)
• 1 <= mat[i][j] <= 5000
• mat[i] is a non decreasing array.

## Solution 1: Priority Queue

Generate the arrays in order.

Each node is {sum, idx_0, idx_1, …, idx_m},

Start with {sum_0, 0, 0, …, 0}.

For expansion, pick one row and increase its index

Time complexity: O(k * m ^ 2* log k)
Space complexity: O(k)

## C++

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++

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.


Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.


Note:

1. 2 <= N <= 50.
2. grid[i][j] is a permutation of [0, …, N*N – 1].

## Solution 1: Dijkstra’s Algorithm

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

## Solution 2: Binary Search + BFS

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

## C++

Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k.

Example:

Input: n = 12, primes = [2,7,13,19] Output: 32  Explanation: [1,2,4,7,8,13,14,16,19,26,28,32] is the sequence of the first 12               super ugly numbers given primes = [2,7,13,19] of size 4.

Note:

• 1 is a super ugly number for any given primes.
• The given numbers in primes are in ascending order.
• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
• The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

## Solution 1: Set

Maintain an ordered set of super ugly numbers, each time extract the smallest one, and multiply it with all primes and insert the new number into set.

Time complexity: O(n*k*logn)
Space complexity: O(n)

## Solution 2: Priority Queue

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