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Suppose we have a binary string s. We have to find the number of substrings that contain only "1"s. If the answer is too large, mod the result by 10^9+7. So, if the input is like s = "100111", then the output will be 7, because the substrings containing only "1"s are ["1", "1", "1", "1", "11", "11" and "111"]. Algorithm To solve this, we will follow these steps − Initialize a counter a to 0 for tracking consecutive 1s Initialize count to 0 for total substrings ... Read More
A square submatrix is a subset of elements from a matrix arranged in a square pattern. In a binary matrix, we want to count square submatrices where all elements are 1. This problem uses dynamic programming to efficiently count all possible squares. Problem Understanding Given a binary matrix, we need to find the total number of square submatrices containing only 1s. For example: 0 1 1 0 1 1 This matrix has 5 square submatrices: one (2×2) square and four (1×1) squares. Algorithm Approach We use dynamic programming ... Read More
Suppose we have a list of intervals where each interval[i] has [start, end] values. We need to find the number of intervals that are completely contained inside other intervals. If an interval is contained by multiple other intervals, it should only be counted once. An interval [s0, e0] is inside another interval [s1, e1] when s1 ≤ s0 and e0 ≤ e1. So, if the input is like intervals = [[2, 6], [3, 4], [4, 7], [5, 5]], then the output will be 2, because [3, 4] is inside [2, 6] and [5, 5] is inside [4, 7]. ... Read More
To extract hour from the DateTimeIndex with specific time series frequency, use the DateTimeIndex.hour property. This is particularly useful when working with time series data that has hourly frequency patterns. Creating DateTimeIndex with Hourly Frequency First, let's create a DateTimeIndex with hourly frequency using pd.date_range() ? import pandas as pd # DatetimeIndex with period 6 and frequency as H i.e. hour # The timezone is Australia/Sydney datetimeindex = pd.date_range('2021-10-20 02:35:55', periods=6, tz='Australia/Sydney', freq='H') # Display DateTimeIndex print("DateTimeIndex...") print(datetimeindex) DateTimeIndex... DatetimeIndex(['2021-10-20 02:35:55+11:00', '2021-10-20 03:35:55+11:00', ... Read More
Suppose we have two numbers start and end, we have to find a sorted list of integers such that every number e in range [start, end] both inclusive and the digits of e are contiguously increasing. An example of continuously increasing number is 5678, but 169 is not. So, if the input is like start = 10 end = 150, then the output will be [12, 23, 34, 45, 56, 67, 78, 89, 123] Algorithm To solve this, we will follow these steps − s := all 9 digits as a ... Read More
We need to find the number of ways to play n games such that we never win more than k consecutive games. Each game can result in a win (W) or loss (L), and we must ensure no sequence of k+1 consecutive wins occurs. For example, if n = 3 and k = 2, the valid sequences are: "LLL", "WLL", "LWL", "LLW", "WWL", "LWW", "WLW" (7 ways total). Approach We'll use dynamic programming with memoization. The key insight is to track the current position in the game sequence and the number of consecutive wins so far ? ... Read More
To return an IntervalArray identical to the current one but closed on the left side, use the set_closed() method with value left. This method allows you to change the closure type of intervals without modifying the actual interval boundaries. Understanding Interval Closure An interval can be closed on different sides: right (default): (a, b] includes b but excludes a left: [a, b) includes a but excludes b both: [a, b] includes both a and b neither: (a, b) excludes both a and b Creating an IntervalArray First, let's create an IntervalArray using from_breaks() ? ... Read More
The get_loc() method in Pandas IntervalIndex returns the integer position of a specified label within the index. This is useful when you need to find where a particular value falls within your interval structure. Syntax IntervalIndex.get_loc(key, method=None, tolerance=None) Parameters key: The label to locate method: Method for inexact matches (None, 'pad', 'backfill', 'nearest') tolerance: Maximum distance for inexact matches Basic Example Let's create an IntervalIndex and find the location of a specific value ? import pandas as pd # Create two Interval objects interval1 = pd.Interval(50, 75) interval2 = ... Read More
To check if intervals that only have an open endpoint in common overlap or not, use the IntervalIndex.is_overlapping property. This property returns True if any intervals overlap, and False if they only share open endpoints. Understanding Interval Overlapping Intervals with open endpoints like [0, 1) and [1, 2) share the point 1, but since it's open in the first interval and closed in the second, they don't actually overlap ? Creating IntervalIndex with Left-Closed Intervals First, let's create an IntervalIndex with left-closed intervals ? import pandas as pd # Create IntervalIndex with left-closed ... Read More
To check if intervals that share closed endpoints overlap in Pandas, use the IntervalIndex.is_overlapping property. This property returns True when intervals share endpoints and both endpoints are closed. Understanding Interval Overlapping When intervals are created with closed='both', adjacent intervals share endpoints. For example, intervals [0, 1] and [1, 2] both include the point 1, making them overlapping ? import pandas as pd # Create IntervalIndex with both endpoints closed interval = pd.interval_range(0, 4, closed='both') print("IntervalIndex with closed='both':") print(interval) print("Does it overlap?", interval.is_overlapping) IntervalIndex with closed='both': IntervalIndex([[0, 1], [1, 2], [2, 3], [3, ... Read More