Suppose we have a list of numbers called nums. If the frequency of a most frequent number in nums is k. We have to find the length of a shortest sublist such that the frequency of its most frequent item is also k.So, if the input is like nums = [10, 20, 30, 40, 30, 10], then the output will be 3, because here the most frequent numbers are 10 and 30 , here k = 2. If we select the sublist [30, 40, 30] this is the shortest sublist where 30 is present and its frequency is also 2.To ... Read More

Suppose we have a list of numbers called nums, where n elements are present. We have to chesk whether we can make a list with first n natural numbers either in increasing or decreasing fashion, like [1, 2, ..., n] or [n, n - 1, ..., 1] by shifting nums to the right any number of times or not.So, if the input is like nums = [5, 6, 1, 2, 3, 4], then the output will be True, because we can shift them four times to make the array [1, 2, 3, 4, 5, 6]To solve this, we will follow ... Read More

Suppose we want to implement a set data structure with the following methods −Constructor to construct a new instance of a setadd(val) to insert integer val into the setexists(val) to check whether val is in the set or notremove(val) to remove the val from the setSo, if we construct a set s, then call s.add(10), s.add(20), s.add(10), s.exists(10), s.remove(10), s.exists(10), s.exists(20), then the output will befor s.add(10) it will insert 10for s.add(20) it will insert 2010 is already in s, so nothing will happens.exists(10) will return true as 10 is thereDelete 10 by s.remove(10)s.exists(10) will return false because 10 is ... Read More

Suppose we have a number n, there are n number of people searching for a seat, we also have a list of bits where 1 represents an already occupied seat and 0 represents empty seat. No two people can sit next to each other, so we have to check whether all n people can find a seat or not.So, if the input is like n = 2 seats = [1, 0, 0, 0, 1, 0, 0], then the output will be True, because they can seat at index 2 and 6.To solve this, we will follow these steps −insert 0 ... Read More

Suppose we want to define an iterator class that constructs with a run-length encoded lowercase string say s, there are two functions for this iterator they are −next() this finds the next element in the iteratorhasnext() this checks whether the next element is present or notSo, if the input is like s = "2b1a", then construct an object with s, then call next(), hasnext(), next(), next(), hasnext(), then the output will be "b", True, "b", "a", False.To solve this, we will follow these steps −Define a constructor. This will take soutput := a new listnum := blank stringfor each i ... Read More

Suppose we have a binary string s, now let us consider an operation where we select a bit and flip its value from 0 to 1 or vice-versa. We have to find the minimum number of operations needed to get a string with no three identical consecutive bits.So, if the input is like s = "10011100", then the output will be 1, because we can flip 1 to 0 the bit at index 4 to make the string "10010100" there are no three consecutive identical bits.To solve this, we will follow these steps −l := 0, count := 0while l ... Read More

Suppose we want to develop a data structure that can build up with an expire time, and supports a function that takes user id and a timestamp. This will check whether a user with given user_id at time given timestamp the request fails or not. It will fail only when the user had a successful request less than the given expire time ago.So, if the input is like expire = 6 then construct an object obj, and call functions obj.limit(0, 10), obj.limit(0, 16), obj.limit(0, 17) and obj.limit(1, 20), then the output will be False, False, True and False respectively because ... Read More

Suppose we want to develop a data structure that can build up with a list of integers, and there is a function to find sum of elements from index i to index j-1 whenever we require in an efficient way. There are two functions.Constructor that constructs a new instance with the integer array.get_sum(i, j) returns the sum of integers of array elements from starting index i and ending index j-1.So, if the input is like array = [5, 2, 3, 6, 4, 7, 8, 9, 3, 2] then construct an object obj, and call functions obj.get_sum(1, 5) and obj.get_sum(4, 8), ... Read More

To calculate the left slice bound that corresponds to given label, use the index.get_slice_bound(). Set the side parameter to left.At first, import the required libraries −import pandas as pdCreating Pandas index −index = pd.Index([10, 20, 30, 40, 50, 60, 70])Display the Pandas index −print("Pandas Index...", index)Get the left slice bound. Returns the left slice bound of given label if "side" parameter is set to "left"print("Get the left slice bound...", index.get_slice_bound(30, side='left', kind ='getitem')) ExampleFollowing is the code −import pandas as pd # Creating Pandas index index = pd.Index([10, 20, 30, 40, 50, 60, 70]) # Display the Pandas ... Read More

To calculate the right slice bound that corresponds to given label, use the index.get_slice_bound(). Set the side parameter to right.At first, import the required libraries −import pandas as pdCreating Pandas index −index = pd.Index([10, 20, 30, 40, 50, 60, 70])Display the Pandas index −print("Pandas Index...", index)Get the right slice bound. Returns the right slice bound of given label if "side" parameter is set to "right" −print("Get the right slice bound...", index.get_slice_bound(30, side='right', kind ='getitem')) ExampleFollowing is the code −import pandas as pd # Creating Pandas index index = pd.Index([10, 20, 30, 40, 50, 60, 70]) # Display the ... Read More