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C++ Articles
Page 76 of 597
Queries to check if it is possible to join boxes in a circle in C++
In this tutorial, we will be discussing a program to find queries to check if it is possible to join boxes in a circle.For this we will be provided with a circle of boxes running from 1 to n. Our task is to find whether box i can be connected to box j with a rod without intersecting the previous rodes.Example#include using namespace std; //checking if making a circle from boxes is possible void isPossible(int n, int q, int queryi[], int queryj[]) { int arr[50]; for (int i = 0; i queryj[k]) { check = 1; break; } } } } if (check == 0) { cout
Read MoreQueries to count the number of unordered co-prime pairs from 1 to N in C++
In this problem, we are given Q queries each contains a number N. Our task is to create a program to solve Queries to count the number of unordered coprime pairs from 1 to N in C++.Co-prime also known as relatively prime or mutually prime are the pair of numbers that have only one factor i.e. 1.Let’s take an example to understand the problem, Input: Q = 2, queries = [5, 6] Output: 10ExplanationThe pairs are : (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5), (4, 5)Solution ApproachThe most promising solution ...
Read MoreSum of the series 0.7, 0.77, 0.777 … upto n terms in C++
In this problem, we are given n terms of a number. The series is 0.7, 0.77, 0.777…. Our task is to create a program to find the sim of the series 0.7, 0.77, 0.777 … upto n terms.Let’s take an example to understand the problem, Input 4Output Explanation −0.7 + 0.77 + 0.777 + 0.7777 = 3.0247To solve this problem, we will derive the formula for sum of series. Let’s find the general formula for it, sum = 0.7 + 0.77 + 0.777 + ... upto n terms sum = 7 (0.1 + 0.11 + 0.111 + … upto n ...
Read MoreSum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)) in C++
In this problem, we are given an integer n. Our task is to create a program to find the sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)).From this series, we can observe that ith term of the series is the sum of first i odd numbers.Let’s take an example to understand the problem, Inputn = 3Output 14Explanation −(1) + (1+3) + (1+3+5) = 14A simple solution to this problem is using a nested loop and then add all odd numbers to a sum variable. Then return the sum.ExampleProgram to illustrate the working of our solution, ...
Read MoreSum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + ... in C++\\n
In this problem, we are given a number n which is the nth term of the series 1/(1*2) + 1/(2*3) +…+ 1/(n*(n+1)). Our task is to create a program to find the sum of the series.Let’s take an example to understand the problem, Input n = 3Output 0.75Explanation − sum = 1/(1*2) + 1/(2*3) + 1/(3*4) = ½ + ⅙+ 1/12 = (6+2+1)/12 = 9/12 = ¾ = 0.75A simple solution to the problem is using the loop. And commuting value for each element of the series. Then add them to the sum value.AlgorithmInitialize sum = 0 Step 1: Iterate from i = ...
Read MoreMaximum bishops that can be placed on N*N chessboard in C++
We are given an input N which denotes the size of the chessboard. The task here is to find for any value of N, how many bishops can be placed on the NXN chessboard such that no two bishops can attack each other. Let’s understand with examples.Input − N=2Output− Maximum bishops that can be placed on N*N chessboard − 2 ( as shown above )Explanation − As depicted above the only non-contradictory positions are where the bishops are placed. Bishops at-most for 2X2 chessboard.Input − N=5Output− Maximum bishops that can be placed on N*N chessboard: 8 ( as shown above ...
Read MoreMaximum consecutive repeating character in string in C++
We are given a string of alphabets. The task is to find the character which has the longest consecutive repetitions occurring in string. Let’s understand with examples.Input− String[] = “abbbabbbbcdd”Output − bExplanation − In the above string, the longest consecutive sequence is of character ‘b’. Count of consecutive b’s is 4.Input− String[] = “aabbcdeeeeed”Output − bExplanation − In the above string, the longest consecutive sequence is of character ‘e’. Count of consecutive e’s is 5.Approach used in the below program is as followsThe character array string1[] is used to store the string of alphabets.Function maxRepeating(char str[], int n) takes two ...
Read MoreMaximum length of subarray such that sum of the subarray is even in C++
We are given with an array Arr[] of integers. The goal is to find longest length subarray of Arr[] , sum of whose elements is even. That is, the sum of elements of a subarray is even and that subarray is longest in length.Input − Arr[] = { 2, 3, 5, 2, 6, 7 }.Output −Maximum length of subarray − 4Explanation −The maximum length subarray is { 5, 2, 6, 7 }. Sum is 20 which is even.Input − Arr[] = { 5, 7, 7, 3, 4 }.Output − Maximum length of subarray − 4Explanation − The maximum length subarray ...
Read MoreSum of the series 2^0 + 2^1 + 2^2 +...+ 2^n in C++
In this problem, we are given a number n which defines the n-th term of the series 2^0, 2^1, 2^2, …, 2^n. Our task is to create a program to find the sum of the series 2^0 + 2^1 + 2^2 +...+ 2^n.Let’s take an example to understand the problem, Inputn=6Output Explanation sum = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 sum = 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127A simple solution to the problem is by using the loop. Finding the 2^i, for each value from 0 ...
Read MoreMaximum number of dots after throwing a dice N times in C++
Given the task is to calculate the maximum number of dots that can be expected after throwing a dice N times having M faces.The first face of the dice contains 1 dot, the second face has 2 dots and so on. Likewise the M-th face contains M number of dots.The probability of appearance of each face becomes 1/M.Let’s now understand what we have to do using an example −Input − M=2, N=3Output − 1.875Explanation − The dice has 2 sides = {1, 2}If the dice is thrown 3 times then the sample space will be = MN = 23{(1, 1, ...
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