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Define a function string_bits(str1) that returns a new string made of every other (every two) charachter starting with the first.

Question

Define a function string_bits(str1) that returns a new string made of every other (every two) charachter starting with the first.

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Solution

Sure, here is a Python function that does exactly that:

def string_bits(str1):
    return str1[::2]

This function uses Python's slice notation. Here's how it works:

  • str1[::2] means "start at the beginning of str1, go until the end, and step by 2". In other words, it takes every other character from str1.

For example, if you call string_bits('Hello'), the function will return 'Hlo'.

This problem has been solved

Similar Questions

How many 4-bit string contain at least one โ€œ1โ€

You are given a positive integer n.A binary string x is valid if all substrings of x of length 2 contain at least one "1".Return all valid strings with length n, in any order.ย Example 1:Input: n = 3Output: ["010","011","101","110","111"]Explanation:The valid strings of length 3 are: "010", "011", "101", "110", and "111".Example 2:Input: n = 1Output: ["0","1"]Explanation:The valid strings of length 1 are: "0" and "1".

You are given a binary string s๐‘  of length n๐‘›, consisting of zeros and ones. You can perform the following operation exactly once:Choose an integer p๐‘ (1โ‰คpโ‰คn1โ‰ค๐‘โ‰ค๐‘›).Reverse the substring s1s2โ€ฆsp๐‘ 1๐‘ 2โ€ฆ๐‘ ๐‘. After this step, the string s1s2โ€ฆsn๐‘ 1๐‘ 2โ€ฆ๐‘ ๐‘› will become spspโˆ’1โ€ฆs1sp+1sp+2โ€ฆsn๐‘ ๐‘๐‘ ๐‘โˆ’1โ€ฆ๐‘ 1๐‘ ๐‘+1๐‘ ๐‘+2โ€ฆ๐‘ ๐‘›.Then, perform a cyclic shift of the string s๐‘  to the left p๐‘ times. After this step, the initial string s1s2โ€ฆsn๐‘ 1๐‘ 2โ€ฆ๐‘ ๐‘› will become sp+1sp+2โ€ฆsnspspโˆ’1โ€ฆs1๐‘ ๐‘+1๐‘ ๐‘+2โ€ฆ๐‘ ๐‘›๐‘ ๐‘๐‘ ๐‘โˆ’1โ€ฆ๐‘ 1.For example, if you apply the operation to the string 110001100110 with p=3๐‘=3, after the second step, the string will become 011001100110, and after the third step, it will become 001100110011.A string s๐‘  is called k๐‘˜-proper if two conditions are met:s1=s2=โ€ฆ=sk๐‘ 1=๐‘ 2=โ€ฆ=๐‘ ๐‘˜;si+kโ‰ si๐‘ ๐‘–+๐‘˜โ‰ ๐‘ ๐‘– for any i๐‘– (1โ‰คiโ‰คnโˆ’k1โ‰ค๐‘–โ‰ค๐‘›โˆ’๐‘˜).For example, with k=3๐‘˜=3, the strings 000, 111000111, and 111000 are k๐‘˜-proper, while the strings 000000, 001100, and 1110000 are not.You are given an integer k๐‘˜, which is a divisor of n๐‘›. Find an integer p๐‘ (1โ‰คpโ‰คn1โ‰ค๐‘โ‰ค๐‘›) such that after performing the operation, the string s๐‘  becomes k๐‘˜-proper, or determine that it is impossible. Note that if the string is initially k๐‘˜-proper, you still need to apply exactly one operation to it.InputEach test consists of multiple test cases. The first line contains one integer t๐‘ก (1โ‰คtโ‰ค1041โ‰ค๐‘กโ‰ค104)ย โ€” the number of test cases. The description of the test cases follows.The first line of each test case contains two integers n๐‘› and k๐‘˜ (1โ‰คkโ‰คn1โ‰ค๐‘˜โ‰ค๐‘›, 2โ‰คnโ‰ค1052โ‰ค๐‘›โ‰ค105)ย โ€” the length of the string s๐‘  and the value of k๐‘˜. It is guaranteed that k๐‘˜ is a divisor of n๐‘›.The second line of each test case contains a binary string s๐‘  of length n๐‘›, consisting of the characters 0 and 1.It is guaranteed that the sum of n๐‘› over all test cases does not exceed 2โ‹…1052โ‹…105.OutputFor each test case, output a single integerย โ€” the value of p๐‘ to make the string k๐‘˜-proper, or โˆ’1โˆ’1 if it is impossible.If there are multiple solutions, output any of them.ExampleinputCopy78 4111000014 2111012 31110001000115 5000006 11010018 40111000112 2110001100110outputCopy3-1754-13NoteIn the first test case, if you apply the operation with p=3๐‘=3, after the second step of the operation, the string becomes 11100001, and after the third step, it becomes 00001111. This string is 44-proper.In the second test case, it can be shown that there is no operation after which the string becomes 22-proper.In the third test case, if you apply the operation with p=7๐‘=7, after the second step of the operation, the string becomes 100011100011, and after the third step, it becomes 000111000111. This string is 33-proper.In the fourth test case, after the operation with any p๐‘, the string becomes 55-proper.

You are given a binary string s that contains at least one '1'.You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination.Return a string representing the maximum odd binary number that can be created from the given combination.Note that the resulting string can have leading zeros.

ow many binary strings of length 8 are there that contain two or fewer 1s?

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