Python Program Divisibility with 5

Write a python program to print and count all numbers from 0 to n which are divisible by 5 and having none of the digits being repeated.

def checkRepeatingDigits(numb):

    checkString = str(numb)

    flag = True

    for i in range(len(checkString)-1):

        for j in range(i+1,len(checkString)):

            if checkString[i] == checkString[j]:

                flag = False

                break

        if flag == False:

            break

    return(flag)

n = int(input("Enter the limit:"))

counter = 0

for i in range(n):

    numb = i

    if numb % 5 == 0:

        if checkRepeatingDigits(numb) == True:

            counter +=1

            print(numb)

print("Total number of numbers divisible by 5 and having non-repeating characters =",counter)

INCOME TAX CALCULATOR : New Tax Regime FY 2023-24 Onwards

INCOME TAX CALCULATOR - NEW REGIME

CALCULATE YOUR INCOME TAX UNDER NEW REGIME HERE: 

Enter your gross salary in ₹

Enter your total income from other sources (Savings Interest, Dividend, Interest on FDs) in ₹

Enter your total Professional Tax (P.Tax) in ₹

Compare with the Old Tax Regime

Income Tax Slabs & Rates 2023-24

The Finance Minister introduced new tax regime in Union Budget, 2020 wherein there is

 an option for individuals and HUF (Hindu Undivided Family) to pay taxes at lower 

rates without claiming deductions under various sections. 

The following Income Tax slab rates are notified in new tax regime:

Income Tax Slab Tax Rates As Per New Regime 

₹0 - ₹3,00,000  Nil 

₹3,00,000 -  ₹6,00,000      5%  

₹6,00,001 -  ₹9,00,000      ₹15000 + 10% of total income exceeding ₹5,00,000    

₹9,00,001 -  ₹12,00,000     ₹45000 + 15% of total income exceeding ₹7,50,000     

₹12,00,001 - ₹15,00,000     ₹90000 + 20% of total income exceeding ₹12,50,000  

Above        ₹15,00,000     ₹140000 + 30% of total income exceeding ₹15,00,000  

Above rates does not include Surcharge and Cess.

4% Health & Education Cess is applicable on the income tax and applicable surcharge.

 

Banking ICSE Interest Calculator

Calculation of interest on a savings bank account

Mini Python project for middle and high school students

Example

Run the code on google colaboratory

Python Code:

class Solution(object):

    def Banking(self, numberOfMonths):

        transactions = dict()

        for i in range(numberOfMonths):

            month = str(input("Enter the month:"))

            transactions_for_month = int(

                input("Enter the number of transactions for this month:"))

            tempList = []

            for j in range(transactions_for_month):

                tempList.append(int(input("Enter the date of the month:")))

                # put a negative sign before the number if the amount is withdrawn

                tempList.append(

                    float(input("Enter the amount deposited or withdrawn:")))

            transactions[month] = tempList

        #print(transactions)

        self.MinimumBalance(transactions)

    def MinimumBalance(self, transactions):

        key = list(transactions.keys())

        balance = []

        balanceFinal = 0

        date = []

        monthWiseMBList = []

        for i in range(len(transactions)):

            for j in range(len(transactions[key[i]])):

                if j % 2 == 0:

                    date.append(transactions[key[i]][j])

                else:

                    balanceFinal += (transactions[key[i]][j])

                    balance.append(balanceFinal)

            balanceFinal = balance[-1]

            #print(balance)

            #print(date)

            if i != 0:

                move_date += int((len(transactions[key[i - 1]]) / 2))

            else:

                move_date = 0

            #print(move_date)

            minimumBalanceMonthly = float('inf')

            flagTransactionLessThanTenth = False

           

            for k in range(len(date) - move_date):

                if date[k+move_date] <= 10:

                    minimumBalanceMonthly = balance[k + move_date]

                    flagTransactionLessThanTenth = True

                else:

                    if minimumBalanceMonthly > balance[k + move_date]:

                        minimumBalanceMonthly = balance[k + move_date]

            if len(transactions[key[i]]) == 0:

                monthWiseMBList.append(monthWiseMBList[i - 1])

            else:

                monthWiseMBList.append(minimumBalanceMonthly)

            if flagTransactionLessThanTenth == False:

                if i == 0:

                    print(monthWiseMBList[i], i, minimumBalanceMonthly)

                    monthWiseMBList[i] = 0

                if i!=0:

                    if monthWiseMBList[i] > monthWiseMBList[i - 1] and len(transactions[key[i]]) != 0:

                        monthWiseMBList[i] = monthWiseMBList[i - 1]

                    else:

                        monthWiseMBList[i] = balance[-1]

            if i == len(key) - 1:

                print("Press Y/y is the account was closed in the month of ", key[-1])

                choice = str(input())

                if choice == "Y" or choice == "y":

                    monthWiseMBList[-1] = 0

            print("Minimum Monthly Balance for the month of ", key[i], "= ",

                  monthWiseMBList[i])

        #calculation of interest

        qualifyingAmount = 0

        for j in (monthWiseMBList):

            qualifyingAmount += j

        print("Total of balances ", qualifyingAmount)

        rate_of_interest = float(input("Enter the rate of interest:"))

        interest = (qualifyingAmount * rate_of_interest * (1 / 12)) / 100.0

        print("Total interest earned = ", round(interest, 2))

obj = Solution()

numberOfmonths = int(input("Enter the number of months:"))

obj.Banking(numberOfmonths)

Leet Code Rotate Image

Leet Code Rotate Image 

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

 

Constraints:

• n == matrix.length == matrix[i].length
• 1 <= n <= 20
• -1000 <= matrix[i][j] <= 1000

class Solution:

    def rotate(self, matrix):

        order = len(matrix[0])

        newmatrix = []

        for i in range(order):

            newmatrix.append([])

            for j in range(order):

                newmatrix[i].append(0)

        newColoumn = []

        for i in range(order):

            newColoumn.append(order-1-i)

        for i in range(order):

            for j in range(order):

                newmatrix[j][newColoumn[i]] = matrix[i][j]

        for i in range(order):

            for j in range(order):

                matrix[i][j] = int(newmatrix[i][j])

        print(matrix)

obj = Solution()

#change the matrix here

matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]

obj.rotate(matrix)

Leet Code Longest Substring Without Repeating Characters

Given a string s, find the length of the longest 

substring

 without repeating characters.

 

Example 1:

Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

 

Constraints:

• 0 <= s.length <= 5 * 104
• s consists of English letters, digits, symbols and spaces.

class Solution:

    def lengthOfLongestSubstring(self, s: str) -> int:

        maxLength = 0

        for i in range(len(s)):

            substring = []

            for j in range(i,len(s)):

                if s[j] in substring:

                    break

                else:

                    substring.append(s[j])

            if len(substring) > maxLength:

                maxLength = len(substring)

        return maxLength

obj = Solution()

s = str(input("Enter a string:"))

print(obj.lengthOfLongestSubstring(s))

Leet Code ZigZag Conversion

 

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

 

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

Input: s = "A", numRows = 1
Output: "A"

 

Constraints:

• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000

import numpy as np

class Solution(object):

    def convert(self, s, numRows):

        numCols = len(s)

        arr_characters = np.zeros(numRows * numCols).reshape(numRows, numCols)

        #print(arr_characters)

        if numRows == 1:

            return s

        else:

            counter = 0

            pos = 0

            vertical_shift = 0

            for i in range(len(s)):

                if counter <= numRows:

                    _zero = 0

                    for j in range(numRows):

                        if pos >= len(s):

                            break

                        else:

                            #print(s[pos],end=' ')

                            #print("(",_zero,vertical_shift,")",end= ' ') # = ord(s[pos])

                            arr_characters[_zero][vertical_shift] = ord(s[pos])

                            pos += 1

                            _zero += 1

                    counter = numRows + 1

                    vertical_shift += (numRows - 1)

                    #print(vertical_shift,"***************")

                else:

                    diagonal_shift_col = vertical_shift - (numRows - 2)

                    diagonal_shift_row = numRows - 2

                    for j in range(numRows - 2):

                        if pos >= len(s):

                            break

                        else:

                            #print(s[pos],end = ' ')

                            #print("(",diagonal_shift_row,diagonal_shift_col,")",end= ' ') # = ord(s[pos])

                            arr_characters[diagonal_shift_row][diagonal_shift_col] = ord(s[pos])

                            diagonal_shift_col += 1

                            diagonal_shift_row -= 1

                            pos += 1

                    #print("**************")

                    counter = 0

            #print(arr_characters)

            output = ""

            for i in range(numRows):

                for j in range(numCols):

                    if arr_characters[i][j] != 0:

                        output += chr(int(arr_characters[i][j]))

                    else:

                        output += ""

            #output.encode('utf-8')

            return output

obj = Solution()

s = ""

numRows = int(input("Enter the number of rows:"))

s = str(input("Enter the string:"))

print(obj.convert(s, numRows))