To extract the seconds from the DateTimeIndex with specific time series frequency, use the DateTimeIndex.second property.At first, import the required libraries −import pandas as pdCreate a DatetimeIndex with period 6 and frequency as S i.e. seconds. The timezone is Australia/Sydney −datetimeindex = pd.date_range('2021-10-20 02:30:50', periods=6, tz='Australia/Sydney', freq='S') Display DateTimeIndex −print("DateTimeIndex...", datetimeindex)Get the seconds −print("Getting the seconds..", datetimeindex.second) ExampleFollowing is the code −import pandas as pd # DatetimeIndex with period 6 and frequency as S i.e. seconds # The timezone is Australia/Sydney datetimeindex = pd.date_range('2021-10-20 02:30:50', periods=6, tz='Australia/Sydney', freq='S') # display DateTimeIndex print("DateTimeIndex...", datetimeindex) # display DateTimeIndex frequency ... Read More

To extract the minute from the DateTimeIndex with specific time series frequency, use the DateTimeIndex.minute property.At first, import the required libraries −import pandas as pdCreate a DatetimeIndex with period 6 and frequency as T i.e. minute. The timezone is Australia/Sydney −datetimeindex = pd.date_range('2021-10-20 02:30:55', periods=6, tz='Australia/Sydney', freq='T') Display DateTimeIndex −print("DateTimeIndex...", datetimeindex)Get the minute −print("Getting the minute..", datetimeindex.minute) ExampleFollowing is the code −import pandas as pd # DatetimeIndex with period 6 and frequency as T i.e. minute # The timezone is Australia/Sydney datetimeindex = pd.date_range('2021-10-20 02:30:55', periods=6, tz='Australia/Sydney', freq='T') # display DateTimeIndex print("DateTimeIndex...", datetimeindex) # display DateTimeIndex frequency ... Read More

Suppose we have a lowercase string s. Now consider an operation, where we can delete, insert or update any character in s. We have to count minimum number of operations required to make s = (t concatenate t) for any string t.So, if the input is like s = "pqrxqsr", then the output will be 2, because, we can update the "x" with "p" and delete "s", then s is "pqrpqr", this is s = t concatenate t, for t = "pqr".To solve this, we will follow these steps −Define a function edit_distance(). This will take s1, s2m := size ... Read More

Suppose we have a list with positive values, called nums and also have a positive number k. We have to find the length of the shortest sublist (may be empty) that we can delete from nums, such that sum of the remaining elements is divisible by k. But we cannot remove the entire list. If there is no such sublist to delete, return -1.So, if the input is like nums = [5, 8, 6, 3] k = 8, then the output will be 1, because current sum of the elements of [5, 8, 6, 3] is 22. If we remove ... Read More

Suppose we have two lowercase strings s and t, now consider an operation where we can delete any character in any of these two strings. We have to find the minimum number of operations needed to make s and t equal.So, if the input is like s = "pipe" t = "ripe", then the output will be 2, because we can delete "p" from s and "r" from t to make these strings same "ipe"To solve this, we will follow these steps −m := size of sn := size of tDefine a function dp() . This will take i, jif ... Read More

Suppose we have a list of rod lengths called rodLen. We also have another two integers called profit and cost, represents profit per length and cost per cut. We can make gain profit per unit length of a rod but we can only sell rods that are all of the same length. We can also cut a rod into two pieces such that their lengths are integers, but we have to pay cost amount for each cut. We can cut a rod as many times as we want. We have to find the maximum profit that we can make.So, if ... Read More

Suppose we have a list of positive numbers, representing ribbons length and also have one value k. We can cut the ribbons as many times as we want, we have to find the largest length r such that we can have k ribbons of length r. If we cannot find such solution, return -1.So, if the input is like ribbons = [1, 2, 5, 7, 15] k = 5, then the output will be 5, as we can cut the ribbon of size 15 into 3 pieces of length 5 each. Then cut the ribbon of size 7 into size ... Read More

Suppose we have a list of positive numbers called nums. Now consider an operation where we remove any two values a and b where a ≤ b and if a < b is valid, then insert back b-a into the list nums. If we can perform any number of operations, we have to find the smallest remaining number we can get. If the list becomes empty, then simply return 0.So, if the input is like nums = [2, 4, 5], then the output will be 1, because, we can select 4 and 5 then insert back 1 to get [2, ... Read More

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"]To solve this, we will follow these steps −a := 0count := 0for i in range 0 to size of s - 1, doif s[i] is same as "0", thena := 0otherwise, a := a + 1count := count + areturn countExampleLet ... Read More

Suppose we have a 2D binary matrix. We have to find the total number of square submatrices present in matrix, where all elements are 1.So, if the input is like011011then the output will be 5, because there is one (2 × 2) square, and four (1 × 1) squaresTo solve this, we will follow these steps −if mat is empty, thenreturn 0c := 0for i in range 0 to row count of mat, dofor j in range 0 to column count of mat, doif mat[i, j] is 1, thenif i is 0 or j is 0, thenc := c + ... Read More