To return the left endpoints of each Interval in the IntervalArray as an Index, use the array.left property.At first, import the required libraries −import pandas as pdCreate two Interval objects −nterval1 = pd.Interval(10, 25) interval2 = pd.Interval(15, 70)Display the intervals −print("Interval1...", interval1) print("Interval2...", interval2)Construct a new IntervalArray from Interval objects −array = pd.arrays.IntervalArray([interval1, interval2])Get the left endpoints −print("The left endpoints of each Interval in the IntervalArray as an Index...", array.left)ExampleFollowing is the code −import pandas as pd # Create two Interval objects interval1 = pd.Interval(10, 25) interval2 = pd.Interval(15, 70) # display the intervals print("Interval1...", interval1) print("Interval2...", interval2) ... Read More

To create a Pandas array for interval data, use the pandas.arrays.IntervalArray() method. At first, import the required libraries −import pandas as pdCreate two Interval objects −interval1 = pd.Interval(10, 30) interval2 = pd.Interval(30, 70)Display the intervals −print("Interval1...", interval1) print("Interval2...", interval2)Construct a new IntervalArray from Interval objects −array = pd.arrays.IntervalArray([interval1, interval2]) ExampleFollowing is the code −import pandas as pd # Create two Interval objects interval1 = pd.Interval(10, 30) interval2 = pd.Interval(30, 70) # display the intervals print("Interval1...", interval1) print("Interval2...", interval2) # Construct a new IntervalArray from Interval objects array = pd.arrays.IntervalArray([interval1, interval2]) # Display the IntervalArray print("Our IntervalArray...", ... Read More

To get the right bound for the interval, use the interval.right property. At first, import the required libraries −import pandas as pdUse Timestamps as the bounds to create a time interval. Closed interval set using the "closed" parameter with value "right" −interval = pd.Interval(pd.Timestamp('2020-01-01 00:00:00'), pd.Timestamp('2021-01-01 00:00:00'), closed='left')Get the right bound −print("The right bound for the Interval...", interval.right)ExampleFollowing is the code −import pandas as pd # Use Timestamps as the bounds to create a time interval # Closed interval set using the "closed" parameter with value "right" interval = pd.Interval(pd.Timestamp('2020-01-01 00:00:00'), pd.Timestamp('2021-01-01 00:00:00'), closed='left') ... Read More

To check whether two Interval objects that share an open endpoint overlap, use the overlaps() method.At first, import the required libraries −import pandas as pdTwo intervals overlap if they share a common point, including closed endpoints. Intervals that only have an open endpoint in common do not overlap.Create two Interval objects. Interval1 is closed from both sides. Interval2 is open from both sides −interval1 = pd.Interval(10, 30, closed='both') interval2 = pd.Interval(30, 50, closed='neither')Display the intervals −print("Interval1...", interval1) print("Interval2...", interval2)Check whether both the interval objects overlap −print("Do both the interval objects overlap?", interval1.overlaps(interval2))ExampleFollowing is the code −import pandas as pd ... Read More

Suppose we have a non-negative integer called num, we have to check whether it is a palindrome or not. We have to solve it without using stringsSo, if the input is like num = 25352, then the output will be TrueTo solve this, we will follow these steps −a := 0c := numwhile num > 0, dor := num mod 10num := floor of num / 10a :=(10 * a) + rif a is same as c, thenreturn Trueotherwise return FalseExampleLet us see the following implementation to get better understandingdef solve(num): a = 0 ... Read More

Suppose we have a list of numbers called nums. We have to find the number of index pairs i, j, where i < j such that nums[i] + nums[j] is equal to 2^k for some 0 >= k.So, if the input is like nums = [1, 2, 6, 3, 5], then the output will be 3, as there are three pairs sum like (6, 2): sum is 8, (5, 3): sum is 8 and (1, 3) sum is 4To solve this, we will follow these steps −res := 0c := a map containing frequencies of each elements present infor each ... Read More

Suppose we have a numeric string s. We have to check whether there is some arrangement where we can have one pair of the same character and the rest of the string form any number of triplets of the same characters.So, if the input is like s = "21133123", then the output will be True, because there are two 2s to form "22" as the pair and "111", "333" as two triplets.To solve this, we will follow these steps −d := a list containing frequencies of each elements present in sfor each k in d, dod[k] := d[k] - 2if ... Read More

Suppose we have alphanumeric string s. It can hold both uppercase or lowercase letters. We have to check whether s is a palindrome or not considering only the lowercase alphabet characters.So, if the input is like s = "rLacHEec0a2r8", then the output will be True because the string contains "racecar" in lowercase, which is a palindrome.To solve this, we will follow these steps −x := blank stringfor each character i in s, doif i is in lowercase, thenx := x concatenate ireturn true when x is palindrome, otherwise falseExampleLet us see the following implementation to get better understandingdef solve(s):   ... Read More

Suppose we have a list called relations. Where each element in relations list relations[i] contains two numbers [ai, bi] it indicates person ai is following bi on a social media platform. We have to find the list of people who follow someone and they follow them back, we have to return it in sorted sequence.So, if the input is like relations = [[0, 2], [2, 3], [2, 0], [1, 0]], then the output will be [0, 2].To solve this, we will follow these steps −ans := a new setseen := a new setfor each pair a and b in relations, ... Read More

Suppose we have a lowercase string s. We have to find the minimum numbers of contiguous substrings in which s is divided into parts such that each substring is either non-increasing or non-decreasing. So for example, if the string is like "pqqqr" is a non-decreasing string, and "qqqp" is a non-increasing string.So, if the input is like s = "pqrsrqp", then the output will be 2, because we can break s like "pqrs" and "rqp".To solve this, we will follow these steps −if s is empty, thenreturn 0last := s[0]direction := 1count := 1for each char in s, doif char ... Read More