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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
Suppose we have a list of intervals. In this list interval[i] has [start, end] values. We have to find the number of intervals are contained by another interval. If there is an interval that is contained by multiple other intervals that should only be counted once. An interval [s0, e0] is inside another interval [s1, e1] when s0 ≤ s1 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] and [5, 5] are inside [2, 6] and [4, 7] respectively.To solve ... 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]To solve this, we will follow these steps −s := all 9 digits as a string "123456789"a := a new listfor i in range 0 to 8, dofor j in range i + 1 to 9, dox := (substring of s from index i to j-1) as numberif start
To extract hour from the DateTimeIndex with specific time series frequency, use the DateTimeIndex.hour property.At first, import the required libraries −import pandas as pdDatetimeIndex 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...", datetimeindex)Get the hour −print("Getting the hour..", datetimeindex.hour) ExampleFollowing is the code −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...", datetimeindex) # display DateTimeIndex frequency print("DateTimeIndex frequency...", datetimeindex.freq) ... Read More
Suppose we have two numbers n and k. Here n represents the number of games we are going to play. We have to find in how many ways we can win k or fewer games consecutively. If the answer is too large then mod the result by 10^9 + 7.So, if the input is like n = 3 k = 2, then the output will be 7, as the possible ways in which we can win 2 or fewer times consecutively, are ["LLL", "WLL", "LWL", "LLW", "WWL", "LWW", "WLW"]To solve this, we will follow these steps −m := 1^9 + ... Read More
To return an IntervalArray identical to the current one but closed on the specified side, use the set_closed() method with parameter set as both.At first, import the required libraries −import pandas as pdCreate IntervalArray −index = pd.arrays.IntervalArray.from_breaks(range(6)) Display the interval −print("IntervalIndex...", index)Return an IntervalArray identical to the current one but closed on specified side i.e. "both" here −print("Result...", index.set_closed('both')) ExampleFollowing is the code −import pandas as pd # Create IntervalArray index = pd.arrays.IntervalArray.from_breaks(range(6)) # Display the interval print("IntervalIndex...", index) # Display the interval length print("IntervalIndex length...", index.length) # the left bound print("The left bound for the ... 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.At first, import the required libraries −import pandas as pdCreate IntervalArray −index = pd.arrays.IntervalArray.from_breaks(range(5))Display the interval −print("IntervalIndex...", index)Return an IntervalArray identical to the current one but closed on specified side i.e. "left" here −print("Result...", index.set_closed('left')) ExampleFollowing is the code −import pandas as pd # Create IntervalArray index = pd.arrays.IntervalArray.from_breaks(range(5)) # Display the interval print("IntervalIndex...", index) # Display the interval length print("IntervalIndex length...", index.length) # the left bound print("The left bound for the IntervalIndex...", index.left) ... Read More