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Posts tagged as “medium”

LeetCode 2033. Minimum Operations to Make a Uni-Value Grid

You are given a 2D integer grid of size m x n and an integer x. In one operation, you can add x to or subtract x from any element in the grid.

uni-value grid is a grid where all the elements of it are equal.

Return the minimum number of operations to make the grid uni-value. If it is not possible, return -1.

Example 1:

Input: grid = [[2,4],[6,8]], x = 2
Output: 4
Explanation: We can make every element equal to 4 by doing the following: 
- Add x to 2 once.
- Subtract x from 6 once.
- Subtract x from 8 twice.
A total of 4 operations were used.

Example 2:

Input: grid = [[1,5],[2,3]], x = 1
Output: 5
Explanation: We can make every element equal to 3.

Example 3:

Input: grid = [[1,2],[3,4]], x = 2
Output: -1
Explanation: It is impossible to make every element equal.

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 105
  • 1 <= m * n <= 105
  • 1 <= x, grid[i][j] <= 104

Solution: Median

To achieve minimum operations, the uni-value must be the median of the array.

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

C++

花花酱 LeetCode 2039. The Time When the Network Becomes Idle

There is a network of n servers, labeled from 0 to n - 1. You are given a 2D integer array edges, where edges[i] = [ui, vi] indicates there is a message channel between servers ui and vi, and they can pass any number of messages to each other directly in one second. You are also given a 0-indexed integer array patience of length n.

All servers are connected, i.e., a message can be passed from one server to any other server(s) directly or indirectly through the message channels.

The server labeled 0 is the master server. The rest are data servers. Each data server needs to send its message to the master server for processing and wait for a reply. Messages move between servers optimally, so every message takes the least amount of time to arrive at the master server. The master server will process all newly arrived messages instantly and send a reply to the originating server via the reversed path the message had gone through.

At the beginning of second 0, each data server sends its message to be processed. Starting from second 1, at the beginning of every second, each data server will check if it has received a reply to the message it sent (including any newly arrived replies) from the master server:

  • If it has not, it will resend the message periodically. The data server i will resend the message every patience[i] second(s), i.e., the data server i will resend the message if patience[i] second(s) have elapsed since the last time the message was sent from this server.
  • Otherwise, no more resending will occur from this server.

The network becomes idle when there are no messages passing between servers or arriving at servers.

Return the earliest second starting from which the network becomes idle.

Example 1:

example 1
Input: edges = [[0,1],[1,2]], patience = [0,2,1]
Output: 8
Explanation:
At (the beginning of) second 0,
- Data server 1 sends its message (denoted 1A) to the master server.
- Data server 2 sends its message (denoted 2A) to the master server.

At second 1,
- Message 1A arrives at the master server. Master server processes message 1A instantly and sends a reply 1A back.
- Server 1 has not received any reply. 1 second (1 < patience[1] = 2) elapsed since this server has sent the message, therefore it does not resend the message.
- Server 2 has not received any reply. 1 second (1 == patience[2] = 1) elapsed since this server has sent the message, therefore it resends the message (denoted 2B).

At second 2,
- The reply 1A arrives at server 1. No more resending will occur from server 1.
- Message 2A arrives at the master server. Master server processes message 2A instantly and sends a reply 2A back.
- Server 2 resends the message (denoted 2C).
...
At second 4,
- The reply 2A arrives at server 2. No more resending will occur from server 2.
...
At second 7, reply 2D arrives at server 2.

Starting from the beginning of the second 8, there are no messages passing between servers or arriving at servers.
This is the time when the network becomes idle.

Example 2:

example 2
Input: edges = [[0,1],[0,2],[1,2]], patience = [0,10,10]
Output: 3
Explanation: Data servers 1 and 2 receive a reply back at the beginning of second 2.
From the beginning of the second 3, the network becomes idle.

Constraints:

  • n == patience.length
  • 2 <= n <= 105
  • patience[0] == 0
  • 1 <= patience[i] <= 105 for 1 <= i < n
  • 1 <= edges.length <= min(105, n * (n - 1) / 2)
  • edges[i].length == 2
  • 0 <= ui, vi < n
  • ui != vi
  • There are no duplicate edges.
  • Each server can directly or indirectly reach another server.

Solution: Shortest Path

Compute the shortest path from node 0 to rest of the nodes using BFS.

Idle time for node i = (dist[i] * 2 – 1) / patince[i] * patience[i] + dist[i] * 2 + 1

Time complexity: O(E + V)
Space complexity: O(E + V)

C++

花花酱 LeetCode 2044. Count Number of Maximum Bitwise-OR Subsets

Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b. Two subsets are considered different if the indices of the elements chosen are different.

The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] (0-indexed).

Example 1:

Input: nums = [3,1]
Output: 2
Explanation: The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3:
- [3]
- [3,1]

Example 2:

Input: nums = [2,2,2]
Output: 7
Explanation: All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 23 - 1 = 7 total subsets.

Example 3:

Input: nums = [3,2,1,5]
Output: 6
Explanation: The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7:
- [3,5]
- [3,1,5]
- [3,2,5]
- [3,2,1,5]
- [2,5]
- [2,1,5]

Constraints:

  • 1 <= nums.length <= 16
  • 1 <= nums[i] <= 105

Solution: Brute Force

Try all possible subsets

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

C++

花花酱 LeetCode 2043. Simple Bank System

You have been tasked with writing a program for a popular bank that will automate all its incoming transactions (transfer, deposit, and withdraw). The bank has n accounts numbered from 1 to n. The initial balance of each account is stored in a 0-indexed integer array balance, with the (i + 1)th account having an initial balance of balance[i].

Execute all the valid transactions. A transaction is valid if:

  • The given account number(s) are between 1 and n, and
  • The amount of money withdrawn or transferred from is less than or equal to the balance of the account.

Implement the Bank class:

  • Bank(long[] balance) Initializes the object with the 0-indexed integer array balance.
  • boolean transfer(int account1, int account2, long money) Transfers money dollars from the account numbered account1 to the account numbered account2. Return true if the transaction was successful, false otherwise.
  • boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise.
  • boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise.

Example 1:

Input
["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
[[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
Output

[null, true, true, true, false, false]

Explanation Bank bank = new Bank([10, 100, 20, 50, 30]); bank.withdraw(3, 10); // return true, account 3 has a balance of $20, so it is valid to withdraw $10. // Account 3 has $20 – $10 = $10. bank.transfer(5, 1, 20); // return true, account 5 has a balance of $30, so it is valid to transfer $20. // Account 5 has $30 – $20 = $10, and account 1 has $10 + $20 = $30. bank.deposit(5, 20); // return true, it is valid to deposit $20 to account 5. // Account 5 has $10 + $20 = $30. bank.transfer(3, 4, 15); // return false, the current balance of account 3 is $10, // so it is invalid to transfer $15 from it. bank.withdraw(10, 50); // return false, it is invalid because account 10 does not exist.

Constraints:

  • n == balance.length
  • 1 <= n, account, account1, account2 <= 105
  • 0 <= balance[i], money <= 1012
  • At most 104 calls will be made to each function transferdepositwithdraw.

Solution: Simulation

Time complexity: O(1) per operation
Space complexity: O(n) for n accounts

C++

花花酱 LeetCode 1906. Minimum Absolute Difference Queries

The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1.

  • For example, the minimum absolute difference of the array [5,2,3,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different.

You are given an integer array nums and the array queries where queries[i] = [li, ri]. For each query i, compute the minimum absolute difference of the subarray nums[li...ri] containing the elements of nums between the 0-based indices li and ri (inclusive).

Return an array ans where ans[i] is the answer to the ith query.

subarray is a contiguous sequence of elements in an array.

The value of |x| is defined as:

  • x if x >= 0.
  • -x if x < 0.

Example 1:

Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]]
Output: [2,1,4,1]
Explanation: The queries are processed as follows:
- queries[0] = [0,1]: The subarray is [1,3] and the minimum absolute difference is |1-3| = 2.
- queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is |3-4| = 1.
- queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is |4-8| = 4.
- queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is |3-4| = 1.

Example 2:

Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]]
Output: [-1,1,1,3]
Explanation: The queries are processed as follows:
- queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the
  elements are the same.
- queries[1] = [0,2]: The subarray is [4,5,2] and the minimum absolute difference is |4-5| = 1.
- queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is |4-5| = 1.
- queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is |7-10| = 3.

Constraints:

  • 2 <= nums.length <= 105
  • 1 <= nums[i] <= 100
  • 1 <= queries.length <= 2 * 104
  • 0 <= li < ri < nums.length

Solution: Binary Search

Since the value range of num is quiet small [1~100], we can store the indices for each value.
[2, 1, 2, 2, 3] => {1: [1], 2: [0, 2, 3]: 3: [4]}.

For each query, we try all possible value b. Check whether b is the query range using binary search, we also keep tracking the previous available value a, ans will be min{b – a}.

Time complexity: O(n + q * 100 * log(n))
Space complexity: O(n)

C++