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Articles by Sunidhi Bansal
Page 44 of 81
Count the number of pairs (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i] in C++
We are given with an array arr[] of N elements. The goal is to find the count of all valid pairs of indexes (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i] and i!=j.We will do this by traversing the array arr[] using two for loops for each number of pair and check if arr[i]%arr[j]==0 or arr[j]%arr[i]==0 when i!=j. If true increment count of pairs.Let’s understand with examples.Input − Arr[]= { 2, 4, 3, 6 } N=4Output − Count of valid pairs − 3Explanation − Valid pairs are −Arr[0] & Arr[1] → (2, ...
Read MoreCount the number of pairs that have column sum greater than row sum in C++
We are given a matrix of size NXN. The goal is to find the count of all valid pairs of indexes (i, j) such that the sum elements of column j is greater than the sum of elements of row i.We will do this by traversing the matrix and calculate sums of elements of each row and column.Store sum of elements of each in rowsum[N] and sum of elements of each column in colsum[N].Now make pairs of rowsum[i] and colsum[j] and check if colsum[j]>rowsum[i]. If true increment count for such pairs.Let’s understand with examples.Input-: matrix= { { 1, 2, ...
Read MoreCount the number of operations required to reduce the given number in C++
We are given with a positive integer K and an array Ops[] which contains integers. The goal is to find the number of operations required to reduce K such that it becomes less than 0. Operations are −First operation is K + Ops[0], first element added to KAfter 1. Add Ops[i] to K until K
Read MoreMaximum number that can be display on Seven Segment Display using N segments in C++
Given the task is to find the maximum number that can be displayed using N segment on ant number of seven segment display.Let’s now understand what we have to do using an example −Input − N=5Output − 71Explanation − The largest number will be displayed as follows on the seven segment display −Input − N=6Output − 111Approach used in the below program as followsThe following situation can be divided into 3 case −Case 1 −If N is 0 or 1, then it is not possible to display any number.Case 2 −If N is odd. Then the numbers that can be ...
Read MoreCount numbers with same first and last digits in C++
We are given an interval [first, last]. The goal is to find the count of numbers that have the same first and last digit within this interval. For example, 232 has the same first and last digit as 2.We will do this by traversing from i=first to i=last. For each number I compare its first digit with the last digit, if they are the same increment the count.Let’s understand with examples.Input − first=8 last=40Output − Count of numbers with same first and last digits − 5Explanation − Numbers between 8 and 40 with same first and last digit −8, 9, ...
Read MoreCount number of ordered pairs with Even and Odd Product in C++
We are given an array of n positive numbers.The goal is to count the ordered pairs (arr[x], arr[y]) with the product of arr[x] and arr[y] is even or odd. Pair ( arr[i], arr[j] ) and ( arr[j], arr[i] are counted as separate.We will traverse the array using two for loops for each number of pairs. Now calculate product, if it is even increment count by 2 for even products else increment count by 2 for odd products.Let’s understand with examples.Input− Arr[]= { 1, 1, 2, 3 } N=4Output − Count of even product pairs: 6 Count of odd product pairs: ...
Read MoreMaximum number with same digit factorial product in C++
Given the task is to find the maximum number without any leading or trailing zeroes or ones whose product of factorial of its digits is equal to the product of factorial of digits of the given number N.Let’s now understand what we have to do using an example −Input − N = 4912Output − 73332222Explanation − 4! * 9! * 1! * 2! = 7! * 3! * 3! * 3! * 2! * 2! *2! *2! = 17, 418, 240Input − N = 340Output − 3322Approach used in the below program as followsIn order to attain the maximum answer ...
Read MoreCount number of ordered pairs with Even and Odd Sums in C++
We are given an array of n positive numbers.The goal is to count the ordered pairs (arr[x], arr[y]) with the sum of arr[x] and arr[y] is even or odd. Pair ( arr[i], arr[j] ) and ( arr[j], arr[i] are counted as separate.We will traverse the array using two for loops for each number of pairs. Now calculate sum, if it is even increment count by 2 for even sums else increment count by 2 for odd sums.Let’s understand with examples.Input− Arr[]= { 1, 1, 2, 3 } N=4Output− Count of even product sums − 6 Count of odd sum pairs ...
Read MoreCount number of pairs (i, j) such that arr[i] * arr[j] > arr[i] + arr[j] in C++
We are given an array of n positive numbers.The goal is to count the ordered pairs (i,j) such that arr[i]*arr[j] > arr[i]+arr[j] and 0sum.Traverse array using two for loops for each element of the pair.Outer Loop from 0
Read MoreCount of m digit integers that are divisible by an integer n in C++
We are given two integers m and n. The goal is to count m digit numbers that are divisible by n.If m=1, then numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and n=3 then numbers divisible by 3=0, 3, 6, 9 count=4.Let’s understand with examples.Input − m=2, n=9Output − Count of m digit numbers divisible by n − 10Explanation − between 10 and 99 numbers divisible by 9 are −18, 27, 36, 45, 54, 63, 72, 81, 90, 99Input m=3, n=300Output − Count of m digit numbers divisible by n: 3Explanation − between 100 and 999 ...
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