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To check elementwise if an Interval overlaps the values in the IntervalArray, use the overlaps() method in Pandas. Two intervals overlap if they share a common point, including closed endpoints. Intervals that only have an open endpoint in common do not overlap. Syntax IntervalArray.overlaps(other) Parameters other − An Interval object to check overlap against each interval in the array. Creating an IntervalArray First, let's create an IntervalArray from tuples − import pandas as pd # Create an IntervalArray intervals = pd.arrays.IntervalArray.from_tuples([(10, 20), (15, 35)]) # Display the IntervalArray ... Read More
Suppose we have two lists called walks and target. At the beginning we are at position 0 in a one-dimensional line. Now |walks[i]| represents the number of steps we have walked. When walks[i] is positive it indicates we walked right, and negative for left. When we walk, we move one block at a time to the next or previous integer position. We have to find the number of blocks that have been walked on at least target number of times. So, if the input is like walks = [3, -7, 2] and target = 2, then the output will ... Read More
Suppose we have two binary matrices mat1 and mat2. Here 1 represents land and 0 represents water. If there is a group of 1s (land) surrounded by water, it's called an island. We need to find the number of islands that exist in both mat1 and mat2 at the exact same coordinates. Problem Understanding Given two matrices, we need to count overlapping islands where both matrices have land (1) at the same positions. For example, if mat1 = 101 100 100 And mat2 = 101 100 101 ... Read More
Finding the correct insertion index to maintain sorted order is a common problem that can be efficiently solved using binary search. We need to find the rightmost position where we can insert the target element while keeping the list sorted in ascending order. Given a sorted list nums and a target value, we want to find the index where the target should be inserted. If the target already exists, we return the largest possible index (after all existing occurrences). Problem Example For nums = [1, 5, 6, 6, 8, 9] and target = 6, the output should ... Read More
Suppose we have a lowercase string s whose length is even. We have to find the minimum number of characters that need to be updated such that one of the following three conditions is satisfied for all i, where 0 ≤ i < n/2 and j, n/2 ≤ j < n − s[i] > s[j] (left half characters greater than right half) s[i] < s[j] (left half characters less than right half) s[i] == s[j] (left half characters equal to right half) So, if the input is ... Read More
To round the DateTimeIndex with minute frequency, use the DateTimeIndex.round() method. For minute frequency, use the freq parameter with value 'T'. Creating a DateTimeIndex First, let's create a DateTimeIndex with seconds frequency to demonstrate the rounding operation ? import pandas as pd # DatetimeIndex with period 5 and frequency as 45 seconds # timezone is Australia/Adelaide datetimeindex = pd.date_range('2021-09-29 07:00', periods=5, tz='Australia/Adelaide', freq='45s') # display DateTimeIndex print("DateTimeIndex...", datetimeindex) print("DateTimeIndex frequency...", datetimeindex.freq) The output of the above code is ? DateTimeIndex... DatetimeIndex(['2021-09-29 07:00:00+09:30', '2021-09-29 07:00:45+09:30', ... Read More
To round the DateTimeIndex with hourly frequency, use the DateTimeIndex.round() method. For hourly frequency, use the freq parameter with value 'H'. Creating a DateTimeIndex At first, import the required libraries and create a DateTimeIndex with period 5 and frequency as 35 minutes − import pandas as pd # Create DatetimeIndex with period 5 and frequency as 35 minutes datetimeindex = pd.date_range('2021-09-29 07:00', periods=5, tz='Australia/Adelaide', freq='35T') # Display DateTimeIndex print("DateTimeIndex...", datetimeindex) DateTimeIndex... DatetimeIndex(['2021-09-29 07:00:00+09:30', '2021-09-29 07:35:00+09:30', '2021-09-29 ... Read More
Suppose we have a list called fruits and two values k and cap. Each fruits[i] contains three values: [c, s, t], where fruit i costs c each, has size s, and there are t total fruits available. The k represents number of fruit baskets with capacity cap. We want to fill the fruit baskets with the following constraints in this order ? Each basket can only hold same type fruits Each basket should be as full as possible Each basket should be as cheap as possible ... Read More
To snap time stamps in DateTimeIndex to nearest occurring frequency, use the DateTimeIndex.snap() method. This method rounds timestamps to the nearest frequency boundary, such as month-end, week-start, or any other valid frequency. Syntax DateTimeIndex.snap(freq) Parameters freq: A frequency string (e.g., 'M' for month-end, 'W' for week, 'D' for day) Basic Example Let's create a DateTimeIndex and snap timestamps to the nearest month-end ? import pandas as pd # Create DatetimeIndex with period 6 and frequency as D i.e. day # The timezone is Australia/Adelaide datetimeindex = pd.date_range('2021-10-20 ... Read More
Suppose we have a list called costs, where costs[i] has [c1, c2] indicating that for person i it costs c1 amount to reach city 0 and c2 amount to reach city 1. We want the same number of people to go to city 0 as city 1, and we need to find the minimum cost required. So, if the input is like costs = [[2, 6], [10, 3], [4, 9], [5, 8]], then the output will be 17, because person 0 and 2 will go to city 0 and person 1 and 3 to city 1. For city 0, ... Read More