leetcode.cn 2025-12-05
🟢3432.count-partitions-with-even-sum-difference
🏷️ Tags
#array #math #prefix_sum
🟢3432.count-partitions-with-even-sum-difference
🏷️ Tags
#array #math #prefix_sum
Telegraph
count-partitions-with-even-sum-difference
给你一个长度为 n 的整数数组 nums 。 分区 是指将数组按照下标 i (0 <= i < n - 1)划分成两个 非空 子数组,其中:
leetcode.com 2025-12-05
🟢3432.count-partitions-with-even-sum-difference
🏷️ Tags
#array #math #prefix_sum
🟢3432.count-partitions-with-even-sum-difference
🏷️ Tags
#array #math #prefix_sum
Telegraph
count-partitions-with-even-sum-difference
You are given an integer array nums of length n. A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that:
leetcode.cn 2025-12-06
🟡3578.count-partitions-with-max-min-difference-at-most-k
🏷️ Tags
#queue #array #dynamic_programming #prefix_sum #sliding_window #monotonic_queue
🟡3578.count-partitions-with-max-min-difference-at-most-k
🏷️ Tags
#queue #array #dynamic_programming #prefix_sum #sliding_window #monotonic_queue
Telegraph
count-partitions-with-max-min-difference-at-most-k
给你一个整数数组 nums 和一个整数 k。你的任务是将 nums 分割成一个或多个 非空 的连续子段,使得每个子段的 最大值 与 最小值 之间的差值 不超过 k。
leetcode.com 2025-12-06
🟡3578.count-partitions-with-max-min-difference-at-most-k
🏷️ Tags
#queue #array #dynamic_programming #prefix_sum #sliding_window #monotonic_queue
🟡3578.count-partitions-with-max-min-difference-at-most-k
🏷️ Tags
#queue #array #dynamic_programming #prefix_sum #sliding_window #monotonic_queue
Telegraph
count-partitions-with-max-min-difference-at-most-k
You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k. Return the total number of…
leetcode.cn 2025-12-10
🟡3577.count-the-number-of-computer-unlocking-permutations
🏷️ Tags
#brainteaser #array #math #combinatorics
🟡3577.count-the-number-of-computer-unlocking-permutations
🏷️ Tags
#brainteaser #array #math #combinatorics
Telegraph
count-the-number-of-computer-unlocking-permutations
给你一个长度为 n 的数组 complexity。 在房间里有 n 台 上锁的 计算机,这些计算机的编号为 0 到 n - 1,每台计算机都有一个 唯一 的密码。编号为 i 的计算机的密码复杂度为 complexity[i]。 编号为 0 的计算机密码已经 解锁 ,并作为根节点。其他所有计算机必须通过它或其他已经解锁的计算机来解锁,具体规则如下:
leetcode.com 2025-12-10
🟡3577.count-the-number-of-computer-unlocking-permutations
🏷️ Tags
#brainteaser #array #math #combinatorics
🟡3577.count-the-number-of-computer-unlocking-permutations
🏷️ Tags
#brainteaser #array #math #combinatorics
Telegraph
count-the-number-of-computer-unlocking-permutations
You are given an array complexity of length n. There are n locked computers in a room with labels from 0 to n - 1, each with its own unique password. The password of the computer i has a complexity complexity[i]. The password for the computer labeled 0 is…