Minimum Number of K Consecutive Bit FlipsIn 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....
In an arrayAcontaining only 0s and 1s, a *K-bit flip *consists of choosing a (contiguous) subarray of lengthKand simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0. Return the minimum number ofK-bit flips required so that there is no 0 in the ...
技术标签: leetcodeIn 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 ...
Can you solve this real interview question? Minimum Increment to Make Array Unique - You are given an integer array nums. In one move, you can pick an index i where 0 <= i < nums.length and increment nums[i] by 1. Return the minimum number of moves to m
You can do the following operation on the array any number of times: Select an index i such that 0 <= i < n - 1 and replace either of nums[i] or nums[i+1] with their gcd value. Return the minimum number of operations to make all elements of nums equal to 1. If it is ...
来自专栏 · LeetCode·力扣·300首 读题 解法一:入度为0的节点 简单来说,我们要找的是图中所有入度为0的节点,因为只有这些节点不是任何其他节点的目的地,所以从这些节点出发,可以到达图中的任何其他节点。 在解决这个问题时,我们可以通过统计每个节点的入度来解决。具体的,我们可以初始化一个长度等于节点数量的列...
You are given an integer arraynums(0-indexed). In one operation, you can choose an element of the array and increment it by1. For example, ifnums = [1,2,3], you can choose to incrementnums[1]to makenums = [1,3,3]. Return the minimum number of operations needed to makenumsstric...
题目要求 Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. Example: Input: [1,2,3] Output: 3 Explanation: ...
2. Solution **解析:**Version 1,这道题跟Leetcode 560的解法很像,首先计算数组的总和total,如果total < x,则无论如何也不会将x减到0,如果total = x,则需要移除所有元素才能将x变为0,由于x一直是从最左或最右移除,因此问题可以变为:找到一个最大连续子数组,使得其和为total - x,这样可以保证剩下的元...
首发于Leetcode HOT 100 切换模式写文章 登录/注册 Leetcode 871 Minimum Number of Refueling Stops Cloudy 一个喜欢篮球的半吊子程序员。 来自专栏 · Leetcode HOT 100 Leetcode 871 Minimum Number of Refueling Stops 算法设计:发布于 2019-10-17 09:27 力扣(LeetCode) Online Judge...