Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false. There may be duplicates in the original array. Note: An array A rotated by x positions results in an array B of ...
Given an arraynums, returntrueif the array was originally sorted in non-decreasing order, then rotatedsomenumber of positions (including zero). Otherwise, returnfalse. There may beduplicatesin the original array. Note:An arrayArotated byxpositions results in an arrayBof the same length such tha...
Hello, LeetCode team 😄 If we go through the examples of the problem and stop at the very first explanation: Explanation:[1,2,3,4,5] is the original sorted array. You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2]. The definit...
The array must be sorted, ifArrays.binarySearch()method is used. In this case, the array is not sorted, therefore, it should not be used. Actually, if you need to check if a value is contained in some array/collection efficiently, a sorted list or tree can do it inO(log(n))or has...
Given an arraynumssorted in non-decreasing order, and a numbertarget, returnTrueif and only iftargetis a majority element. Amajority elementis an element that appears more thanN/2times in an array of lengthN. Example 1: Input: nums =[2,4,5,5,5,5,5,6,6], target =5 ...
Given an arraynumssorted in non-decreasing order, and a numbertarget, returnTrueif and only iftargetis a majority element. Amajority elementis an element that appears more thanN/2times in an array of lengthN. Example 1: Input: nums =[2,4,5,5,5,5,5,6,6], target =5 ...