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 ...
Input:nums = [1,1,1]Output:trueExplanation:[1,1,1] is the original sorted array. You can rotate any number of positions to make nums. Example 5: Input:nums = [2,1]Output:trueExplanation:[1,2] is the original sorted array. You can rotate the array by x = 5 positions to begin ...
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...
Thus, 5 is a majority element because 5 > 9/2 is true. Example 2: Input: nums =[10,100,101,101], target =101 Output:false Explanation: The value 101 appears 2 times and the length of the array is 4. Thus, 101 is not a majority element because 2 > 4/2 is false. Constraints:...
new raw data array with allocated memory can allocate int array under the hood, just not use its features start with 16, or if starting number is greater, use power of 2 - 16, 32, 64, 128 size() - number of items capacity() - number of items it can hold is_empty() at(index...
Thus, 5 is a majority element because 5 > 9/2 is true. Example 2: Input: nums =[10,100,101,101], target =101 Output:false Explanation: The value 101 appears 2 times and the length of the array is 4. Thus, 101 is not a majority element because 2 > 4/2 is false. ...
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...