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 ...
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...
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. Returnthe number of steps performed untilnumsbecomes a ...
51. 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. This 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 arrys 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. ...
原题链接在这里:https://leetcode.com/problems/max-chunks-to-make-sorted-ii/ 题目: This question is the same as "Max Chunks to Make Sorted" except the integers of the given array are not necessarily distinct, the input array could be up to length2000, and the elements could be up to10...
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. This algorithm can be used...
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 ...
Given an integern, the length of the array. Returnthe minimum number of operationsneeded to make all the elements of arr equal. Example 1: Input: n =3Output:2Explanation: arr = [1,3,5] First operationchoosex=2andy=0,thisleads arr to be [2,3,4] ...