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Programming Articles - Page 1330 of 3366
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When it is required to create a class to get all the possible subsets of integers from a list, object oriented method is used. Here, a class is defined, and attributes are defined. Functions are defined within the class that perform certain operations. An instance of the class is created, and the functions are used to perform calculator operations.Below is a demonstration for the same −Example Live Democlass get_subset: def sort_list(self, my_list): return self. subset_find([], sorted(my_list)) def subset_find(self, curr, my_list): if my_list: return self. subset_find(curr, my_list[1:]) + self. subset_find(curr ... Read More
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When it is required to create a class that has a method that accepts a string from the user, and another method that prints the string, object oriented method is used. Here, a class is defined, and attributes are defined. Functions are defined within the class that perform certain operations. An instance of the class is created, and the functions are used to perform calculator operations.Below is a demonstration for the same −Example Live Democlass print_it(): def __init__(self): self.string = "" def get_data(self): self.string=input("Enter the string : ") def put_data(self): ... Read More
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When it is required to create a class that performs calculator operations, object oriented method is used. Here, a class is defined, and attributes are defined. Functions are defined within the class that perform certain operations. An instance of the class is created, and the functions are used to perform calculator operations.Below is a demonstration for the same −Example Live Democlass calculator_implementation(): def __init__(self, in_1, in_2): self.a=in_1 self.b=in_2 def add_vals(self): return self.a+self.b def multiply_vals(self): return self.a*self.b def divide_vals(self): return self.a/self.b def ... Read More
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In this problem, we are given a binary tree and an integer N. The task is to find the n-th node in Postorder traversal of a Binary Tree.A binary tree has a special condition that each node can have a maximum of two children.Traversal is a process to visit all the nodes of a tree and may print their values too.Let’s take an example to understand the problem, InputN = 6Output3ExplanationPost order traversal of tree − 4, 5, 2, 6, 7, 3, 1Solution ApproachThe idea is to use the post order traversal of the binary tree which is done by ... Read More
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In this problem, we are given an integer N, denoting a series of numbers consisting of 4 and 7 only.The series is 4, 7, 44, 47, 74, 77, …The task is to find the n-th element in a series with only 2 digits (and 7) allowed.Let’s take an example to understand the problem, InputN = 4, Output47ExplanationThe series is: 4, 7, 44, 47, ….Solution ApproachA simple solution to the problem is creating the series till Nth number. It’s simple, if the current number’s last digit is 7. Then the last digit of previous and next numbers is 4.So, we will ... Read More
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In this problem, we are given three values A, B and N. Our task is to Find n positive integers that satisfy the given equations.Problem Description − We need to find the N positive values that satisfy both the equations, x12 + x22 + … xn2 ≥ A x1 + x2 + … xn ≤ BPrint n values if present, otherwise print -1.Let’s take an example to understand the problem, InputN = 4, A = 65, B = 16Output1 1 1 8ExplanationThe equations are −12 + 12 + 12 + 82 = 1 + 1 + 1 + 64 = ... Read More
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In this problem, we are given a string num representing a large integer. Our task is to Find N % (Remainder with 4) for a large value of N.Problem Description − we will be finding the remainder of the number with 4.Let’s take an example to understand the problem, Inputnum = 453425245Output1Solution ApproachA simple solution to the problem is by using the fact that the remainder of the number with 4 can be found using the last two digits of the number. So, for any large number, we can find the remainder by dividing the number’s last two digits by ... Read More
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In this problem, we are given a number N. Our task is to Find Multiples of 2 or 3 or 5 less than or equal to N.Problem Description − We will be counting all elements from 1 to N that are divisible by 2 or 3 or 5.Let’s take an example to understand the problem, InputN = 7Output5ExplanationAll the elements from 1 to 7 are : 1, 2, 3, 4, 5, 6, 7. Elements divisible by 2/3/5 are 2, 3, 4, 5, 6Solution ApproachA simple approach to solve the problem is by traversing all numbers from 1 to N and ... Read More
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In this problem, we are given a singly linked list LL and a number k. Our task is to Find modular node in a linked list.Problem Description − we need to find the last node of the linked list for which the index is divisible by k i.e. i % k == 0.Let’s take an example to understand the problem, Inputll = 3 -> 1 -> 9 -> 6 -> 8 -> 2, k = 4Output6ExplanationThe element 6 has index 4, which is divisible by 4.Solution ApproachA simple solution to the problem is by creating a counter to count the ... Read More
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In this problem, we are given an array arr[] of size n and the start and end element denoting the range. Our task is to Find missing elements of a range.Problem Description − we will be finding the elements of the range that are not present in the range.Let’s take an example to understand the problem, Inputarr[] = {4, 6, 3, 7}, start = 3, end = 8Output5, 8ExplanationThe range is [3, 4, 5, 6, 7, 8]The array is {4, 6, 3, 7}Elements of range that are not present in array is 5, 8Solution ApproachYou can solve this problem in ... Read More