We are given two strings. We need to find the count it of the characters in the first string which are also present in a second string.With setThe set function gives us unique values all the elements in a string. We also use the the & operator which finds the common elements between the two given strings.Example Live DemostrA = 'Tutorials Point' uniq_strA = set(strA) # Given String print("Given String", strA) strB = 'aeio' uniq_strB = set(strB) # Given String print("Search character strings", strB) common_chars = uniq_strA & uniq_strB print("Count of matching characters are : ", len(common_chars))OutputRunning the above code gives ... Read More

Suppose we have one number n, we have to count how many strings of length n can be formed using these rules − Each character is a lower case vowel Each vowel 'a' may only be followed by an 'e'. Each vowel 'e' may only be followed by an 'a' or 'i'. Each vowel 'i' may not be followed by another 'i'. Each vowel 'o' may only be followed by an 'i' or 'u'. Each vowel 'u' may only be followed by an 'a'. The answer may be too large, so we will find the answer modulo 10^9 + 7.So, ... Read More

In this article we will see how to get some selected elements out of a Python list. So we need to design some condition and only the elements satisfying that condition should be picked and their count to be printed.Wit in and sumIn this approach we use in condition to pick the elements and use some to get their count. 1 is used if the element is present else 0 is used for the result of in condition.Example Live DemoAlist = ['Mon', 'Wed', 'Mon', 'Tue', 'Thu'] # Given list print("Given list:", Alist) cnt = sum(1 for i in Alist if i ... Read More

In this article we are given a list and a string. We are required to find how many times the given string is present as an element in the list.With CounterThe counter function from collections module will give us the count of each element present in the list. From the counting result we can extract only that account fair the index matches with the value of the element we are searching for.Example Live Demofrom collections import Counter Alist = ['Mon', 'Wed', 'Mon', 'Tue', 'Thu'] elem = 'Mon' # Given list and element print("Given list:", Alist) print("Given element:", elem) cnt = Counter(Alist) ... Read More

Suppose there are n servers. And these are numbered from 0 to n-1 connected by an undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. All servers are connected directly or through some other servers. Now, a critical connection is a connection that, if that is removed, it will make some server unable to reach some other server. We have to find all critical connections.So, if the input is like n = 4 and connection = [[0, 1], [1, 2], [2, 0], [1, 3]], then the output will be [[1, ... Read More

We are given a string and a character. We want to find out how many times the given character is repeated in a given string.With range and lenWe design a for loop to match the character with every character present in the string which are accessed using index. The range and len function helps us determine how many times the matching has to be done when moving from left to right of the string.Example Live DemoAstr = "How do you do" char = 'o' # Given String and Character print("Given String:", Astr) print("Given Character:", char) res = 0 for i in ... Read More

Suppose we have two arrays arr1 and arr2, these can store integer numbers. We have to find the minimum number of operations needed to make arr1 strictly increasing. Here, we can choose two indices 0 size of arr2, then -1, otherwise ret - 1)Let us see the following implementation to get better understanding −Example Live Demo#include using namespace std; class Solution {    public:    int solve(vector& arr1, vector& arr2, int i, int j, int prev, vector& dp){       if (i >= arr1.size())          return 1;       j = upper_bound(arr2.begin() + j, ... Read More

Suppose we have s as string, we have to find the last substring of s in lexicographic order.So, if the input is like "abbbcabbc", then the output will be "cabbc"To solve this, we will follow these steps −i := 0, j := 1, k := 0while j + k < size of s, do &minsu;if s[i + k] is same as s[j + k], then −(increase k by 1)Ignore following part, skip to the next iterationif s[i + k] < s[j + k], then −i := j(increase j by 1)Otherwisej := j + k + 1k := 0return substring of ... Read More

We have a list and tuple. We match the elements of the list with the elements of the tuple and account the number of elements in the table matching with the elements of the list.With CounterWe use the counter function from collections to get the count of every element in the tuple. Again design a for and in condition find those elements which are present in the the list and part of the counting result from the tuple.Example Live Demofrom collections import Counter Atup = ('Mon', 'Wed', 'Mon', 'Tue', 'Thu') Alist = ['Mon', 'Thu'] # Given Tuple and list print("Given tuple ... Read More

Suppose we have a text. We have to find the largest possible k such that there exists a[1], a[2], ..., a[k] such that: Each a[i] is a non-empty string; Their concatenation a[1] + a[2] + ... + a[k] is equal to the given text; For all i in range 1 to k, a[i] = a[{k+1 - i}].So, if the input is like "antaprezatepzapreanta", then the output will be 11, because we can split it like "(a)(nt)(a)(pre)(za)(tpe)(za)(pre)(a)(nt)(a)".To solve this, we will follow these steps −start := 0, end := length of text - 1initialize temp1 and temp2 with empty stringsans = ... Read More