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 origi
Suppose we have an array called nums, we have to check whether the array was originally sorted in non-decreasing order, and then rotated some number of positions (may be zero) or not. Duplicates may also present in the array.So, if the input is like nums = [12,15,2,5,6,9], ...
Input:nums = [2,1,3,4]Output:falseExplanation:There is no sorted array once rotated that can make nums. Example 3: Input:nums = [1,2,3]Output:trueExplanation:[1,2,3] is the original sorted array. You can rotate the array by x = 0 positions (i.e. no rotation) to make nums....
in certain situations such as binary searching, the task requires the collection to be sorted. In such cases, it’s good to check if the collection is sorted before proceeding with further operations.
LeetCode Username anacondrai Problem Number, Title, and Link Check if Array Is Sorted and Rotated https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/ Bug Category Problem examples Bug Description Hello, LeetCode team 😄 If...
Python program to check if a Pandas dataframe's index is sorted# Importing pandas package import pandas as pd # Creating two dictionaries d1 = {'One':[i for i in range(10,100,10)]} # Creating DataFrame df = pd.DataFrame(d1) # Display the DataFrame print("Original DataFrame:\n",df...
// Golang program to check a specified slice of strings// is sorted or notpackagemainimport"fmt"import"sort"funcmain() {varstatusbool=falseslice:=[]string{"honesty ","is ","the ","best ","policy"} status = sort.StringsAreSorted(slice)ifstatus==true{ ...
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 Output:true Explanation: The value 5 appears 5 times and the length of the array is 9. ...
In this case, we verify, if array corresponding to level l is sorted or not. It should be noted that this solution has large memory requirements that can be reduced.According to efficient solution,we perform vertical level order traversal of the binary tree and maintain track of node values ...
If all the values in vector are less than or equal to M then "i" will point to v.end(). auto i = upper_bound(v.begin(), v.end(), M)- v.begin(); Now if I want to do certain action if "i" points to the end of vector, I tried ...