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Articles by Sunidhi Bansal
Page 31 of 81
Count pairs of natural numbers with GCD equal to given number in C++
We gave three input variables as ‘start’, ‘end’ and ‘number’. The goal is to find pairs of numbers between start and end that have GCD value equal to ‘number’. For example GCD(A, B)=number and both A, B are in range [start, end].Let us understand with examples.Input − start=5 end=20 number=8Output − Count of pairs of natural numbers with GCD equal to given number are − 3Explanation − Pairs between 5 to 20 such that GCD is 8 are − (8, 8), (8, 16), (16, 8)Input − start=5 end=20 number=7Output − Count of pairs of natural numbers with GCD equal to ...
Read MoreCount passing car pairs in C++
We are given an array of length N containing 0’s and 1’s only. The value 1 represents a car going towards west direction and value 0 represents a car going towards east direction.We count passing cars as 1 if a pair of car A and car B is such that 0
Read MoreCount paths with distance equal to Manhattan distance in C++
We are given variables x1, x2, y1, y2 representing two points on a 2D coordinate system as (x1, y1) and (x2, y2). The goal is to find all the paths that will have distance equal to the Manhattan distance between these two points.Manhattan DistanceManhattan Distance between two points (x1, y1) and (x2, y2) is −MD= |x1 – x2| + |y1 – y2|Let’s take A= |x1 – x2| and B= |y1 – y2|All paths with Manhattan distance equal to MD will have edges count as (A+B). A horizontal edge and B vertical edges. So possible combination of (A+B) edges divided into ...
Read MoreCount number of triplets with product equal to given number in C++
We are given an array Arr[] of integers with length n and a number M. The array is only containing positive integers. The goal is to count the triplets of elements of Arr[] which have product equal to M.We will do this by using three for loops. Increment count if arr[x]*arr[y]*arr[z]=M and x!=y!=z. (0
Read MoreCount numbers from range whose prime factors are only 2 and 3 in C++
We are provided two numbers START and END to define a range of numbers. The goal is to find the numbers that have only 2 and 3 as their prime factors and are in the range [START, END].We will do this by traversing numbers from START to END and for each number we will check if the number is divisible by 2 and 3 only. If divisible, divide it and reduce it. If not, break the loop. In the end if the number is reduced to 1 then it has only 2 and 3 as its factors.Let’s understand with examples.Input START=20 ...
Read MoreCount numbers having 0 as a digit in C++
We are provided a number N. The goal is to find the numbers that have 0 as digit and are in the range [1, N].We will do this by traversing numbers from 10 to N ( no need to check from 1 to 9 ) and for each number we will check each digit using a while loop. If any digit is found as zero increment count and move to next number otherwise reduce the number by 10 to check digits until number is >0.Let’s understand with examples.Input N=11Output Numbers from 1 to N with 0 as digit: 1Explanation Starting from i=10 to ...
Read MoreCount numbers in a range that are divisible by all array elements in C++
We are provided two numbers START and END to define a range of numbers. And also an array of positive numbers Arr[]. The goal is to find all the numbers that are divisible by all elements of Arr[] and are in the range [START, END] .Method 1 ( Naive Approach )We will do this by traversing numbers from START to END and for each number we will check if the number is divisible by all elements of the array. If yes increment count.Method 2 ( check divisibility by LCM of array elements )We will find the LCM of all array ...
Read MoreCount numbers in range 1 to N which are divisible by X but not by Y in C++
We are provided a number N. The goal is to find the numbers that are divisible by X and not by Y and are in the range [1, N].Let’s understand with examples.Input N=20 X=5 Y=20Output Numbers from 1 to N divisible by X not Y: 2Explanation Only 5 and 15 are divisible by 5 and not 10.Input N=20 X=4 Y=7Output Numbers from 1 to N divisible by X not Y: 5Explanation Numbers 4, 8, 12, 16 and 20 are divisible by 4 and not 7.Approach used in the below program is as followsWe take an integer N.Function divisibleXY(int x, int y, int n) returns a count ...
Read MoreCount numbers in range that are divisible by all of its non-zero digits in C++
We are provided two numbers START and END to define a range of numbers.The goal is to find all the numbers in the range [START, END] that are divisible by all of its non-zero digits . We will do this by traversing numbers from START to END and for each number we will check if the number is divisible by all non-zero digits using a while loop. If yes increment count.Let’s understand with examples.Input START=10 END=20Output Numbers that are divisible by all its non-zero digits: 14Explanation Numbers 10, 11, 12, 15, 20 are divisible by all their non-zero digits.Input START=100 END=200Output Numbers that are divisible ...
Read MoreCount numbers with difference between number and its digit sum greater than specific value in C++
We are provided two numbers N which defines a range [1, N] and D which is a difference.The goal is to find all the numbers in the range [1, N] such that the [ number - (sum of its digits) ] > D. We will do this by traversing numbers from 1 to N and for each number we will calculate its digit sum using a while loop. Check if the number and calculated digit sum has a difference more than D.Let’s understand with examples.Input N=15 D=5Output Numbers such that difference b/w no. and its digit sum greater than value D: 6Explanation Numbers ...
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