The progressbar is a common GUI element which is used to show the progress of certain task. In tis article we will see how to create a progressbar using the Python tkinter GUI library.In the below program we have imported the progressbar sub-module of tkinter.ttk module. Then used the style object to create the style options and supply the value for the length of the button as well as value of the progress.Exampleimport tkinter as tk from tkinter.ttk import Progressbar from tkinter import ttk canv = tk.Tk() canv.title("Tkinter Progressbar") canv.geometry('250x100') style = ttk.Style() style.theme_use('default') style.configure("grey.Horizontal.TProgressbar", background='blue') bar = Progressbar(canv, length=180, ... Read More

Every programming language has a feature to create scripts and run them from the terminal or being called by other programs. When running such scripts we often need to pass on arguments needed by the script for various functions to be executed inside the script. In this article we will see what are the various ways to pass arguments into a python script.Using sys.argvThis is a inbuilt module sys.argv can process arguments that are passed on with the script. By default the first argument which is considered at sys.argv[0] is the file name. the rest of the arguments are indexed ... Read More

Tkinter has great support for creating the GUI programs based on python. It offers different ways of styling a button on the Tkinter canvas based on its font, size, colour etc. In this article we will see how to apply style to specific buttons or all buttons in general on the canvas.Applying to Specific ButtonsLets consider the case when we have two buttons in the canvas and we want to apply some styling only to the first button. We use the W.TButton as part of the configuration along with the font and the foreground colour.Examplefrom tkinter import * from tkinter.ttk ... Read More

Suppose we have a number N, we have to find all factors of N and return the product of four factors of N such that: The sum of the four factors is same as N. The product of the four factors is maximum. All four factors can be equal to each other to maximize the product.So, if the input is like N = 60, then the output will be All the factors are -> 1 2 3 4 5 6 10 12 15 20 30 60 and product is 50625, as we 15 has been selected four times to make ... Read More

Suppose we have a sorted doubly linked list of unique positive numbers; we have to find pairs in the doubly linked list whose product is same as a given value x. We have to keep in mind that, this will be solved without consuming any extra space.So, if the input is like L = 1 ⇔ 2 ⇔ 4 ⇔ 5 ⇔ 6 ⇔ 8 ⇔ 9 and x = 8, then the output will be (1, 8), (2, 4)To solve this, we will follow these steps −curr := head, nxt := headwhile nxt.next is not None is non-zero, donxt ... Read More

Suppose we have an array A where GCD of every possible pair of elements of another array is given, we have to find the original numbers which are used to compute the given GCD array.So, if the input is like A = [6, 1, 1, 13], then the output will be [13, 6] as gcd(13, 13) is 13, gcd(13, 6) is 1, gcd(6, 13) is 1, gcd(6, 6) is 6To solve this, we will follow these steps −n := size of Asort array A in descending orderoccurrence := an array of size A[0] and fill with 0for i in range ... Read More

Suppose we have a sequence of numbers called bn, this is represented using a recurrence relation like b1=1 and bn+1/bn=2n . We have to find the value of log2(bn) for a given n.So, if the input is like 6, then the output will be 5 as log2(bn) = (n * (n - 1)) / 2 = (6*(6-1))/2 = 15We can solve this problem by solving this relation as follows −bn+1/bn = 2nbn/bn-1 = 2n-1…b2/b1 = 21 , If we multiply all of above, we can get(bn+1/bn).(bn/bn-1)……(b2/b1) = 2n + (n-1)+……….+1So, bn+1/b1 = 2n(n+1)/2As 1 + 2 + 3 + ………. ... Read More

Suppose we have a string whose length is m, and this string is containing only lowercase letters, we have to find the n-th permutation of string lexicographically.So, if the input is like string = "pqr", n = 3, then the output will be "qpr" as all permutations are [pqr, prq, qpr, qrp, rpq, rqp], they are in sorted order.To solve this, we will follow these steps −MAX_CHAR := 26MAX_FACT := 20factorials := an array of size MAX_FACTfactorials[0] := 1for i in range 1 to MAX_FACT, dofactorials[i] := factorials[i - 1] * isize := size of stringoccurrence := an array of ... Read More

Suppose we have two integers N and K; we have to find N unique values whose bit-wise OR is same as K. If there is no such result, then return -1So, if the input is like N = 4 and K = 6, then the output will be [6, 0, 1, 2].To solve this, we will follow these steps −MAX := 32visited := a list of size MAX and fill with Falseres := a new listDefine a function add() . This will take numpoint := 0value := 0for i in range 0 to MAX, doif visited[i] is non-zero, thengo for ... Read More

Suppose we have an array A of n unique numbers, these n elements are present in the array in ascending order, but there is one missing element. We have to find the missing element.So, if the input is like A = [1, 2, 3, 4, 5, 6, 7, 9], then the output will be 8.To solve this, we will follow these steps −n := size of Aleft := 0right := n - 1mid := 0while right > left , domid := left +(right - left) / 2if A[mid] - mid is same as A[0], thenif A[mid + 1] - A[mid] ... Read More