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Hello, I need your support, i have home work and seems my below is not correct ,

ID: 3707493 • Letter: H

Question

Hello,

I need your support, i have home work and seems my below is not correct , please can you check it ;

Option 1: Numbers. Your algorithm should:

Search a string of at least five numbers (for example, 37540)

Identify all of the substrings that form numbers that are divisible by 3.

For example, applying the algorithm on the string 37540 should produce the following substrings (not necessarily in this order): 0; 3; 75; 54; 375; 540.

Solution:

To accomplish the task, modulus is typically calculated using the following formula

Example number: 37540

Example divider: 3

Read String (input value)

For example, String = "37540" input string

Initial a to 1

If LEN(string) >= 5 then

The length of the string must be equal or more than 5

While a <= LEN(string) do

Loop from 1 to the length of the string

Set j to LEN(string) - a + 1

While b >=1 do

Loop from the length from the string point to the end (-1)

If MID (string,a,b) MOD 3 =0 then

The substring value dividable by 3 (using MOD function)

Print MID (string,a,b)

If MOD =0 then print the substring value

Set b to b -1

end while

Increase a

end while

End If

Stop

Explanation / Answer

Define function permute() which will return all possible permutations of a number. Use this function in main program.

permutedNumbers = []

void permute(string, start,end):

            if start equals end

                        add string[start] to permutedNumbers

            else

                        for i in start to end:

                                    swap string[start] and string[i]

                                    permute(string, l+1, r)

                                    swap string[start] and string[i]

end permute

Read String (input value)

For example, String = "37540" input string

If LEN(string) >= 5 then

The length of the string must be equal or more than 5

call permute(string, 0 , LEN(string))

for i in 0 to LEN(permutedNumbers):

            if permutedNumbers[i]%3 == 0:

                        print permutedNumbers[i]

Stop

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