In one operation, you can select two indicesxandywhere0 <= x, y < nand subtract1fromarr[x]and add1toarr[y](i.e. performarr[x] -=1andarr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made e...
In one operation, you can select two indicesxandywhere0 <= x, y < nand subtract1fromarr[x]and add1toarr[y](i.e. performarr[x] -=1andarr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made e...
贪心解。尽可能的选择最前面的两个不相同的元素，得到的就是最长的美丽数组。然后用总长度减去该长度即为答案。还有
not a// value. Unlike new, make's return type is the same as the type of its// argument, not a pointer to it. The specification of the result depends on// the type:// Slice: The size specifies the length. The capacity of the slice is// equal to its length. A second integer a...
Iterate through the array, each time all elements to the left are smaller (or equal) to all elements to the right, there is a new chunck. Use two arrys to store the left max and right min to achieve O(n) time complexity. Space complexity is O(n) too. ...
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 ...
LeetCode #1460. Make Two Arrays Equal by Reversing Sub-arrays 题目1460. Make Two Arrays Equal by Reversing Sub-arrays解题方法看看两个数组是否包含了完全相同的元素,且元素个数也相等。这是因为根据题意允许无限交换相邻元素位置,类似于冒泡排序,所以只要包含数量和种类完全相同的元素,两个数组就可以变成完全...
原题链接在这里:https://leetcode.com/problems/steps-to-make-array-non-decreasing/ 题目: You are given a 0-indexed integer arraynums. In one step, remove all elementsnums[i]wherenums[i - 1] > nums[i]for all0 < i < nums.length. ...
Algorithm: Iterate through the array, each time all elements to the left are smaller (or equal) to all elements to the right, there is anewchunck. Use two arrays to store the left max and right min to achieve O(n) time complexity. Space complexity is O(n) too. ...
Algorithm: Iterate through the array, each time all elements to the left are smaller (or equal) to all elements to the right, there is a new chunck. Use two arrays to store the left max and right min to achieve O(n) time complexity. Space complexity is O(n) too. ...