We are given a number N. The goal is to find ordered pairs of positive numbers such that the sum of their cubes is N.Naive ApproachTraverse all numbers from 1 to N and check if it is a perfect square. If floor(sqrt(i))==ceil(sqrt(i)).Then the number is a perfect square.Efficient ApproachPerfect squares below N can be found using formula: floor(sqrt(N)).Let’s understand with examples.Input N=20Output Count of square numbers: 4 Count of non-square numbers: 16Explanation Square numbers are 1, 4, 9 and 16. Rest all are non-squares and less than 20.Input N=40Output Count of square numbers: 6 Count of non-square numbers: 34Explanation Square numbers are 1, 4, 9, 16, ... Read More

We are provided three numbers X, Y and N ( to define range [1, N] ). The goal is to find all the numbers in the range [1, N] that can be constructed using X and Y only any number of times..For example if X=2 and Y=3. Number 6 can be constructed using 2 thrice ( 2+2+2 ) or 3 twice ( 3+3 ). Similarly 7 can be constructed using 2 twice and 3 once ( 2+2+3 ).We will do this by subtracting X or Y from each number from 1 to N. If the final number reduces to 0 ... Read More

We are given two arrays Arr1[] and Arr2[] and a number K. The goal is to find unique pairs of elements of both arrays such that their sum is K. Pairs will be of form ( Arr1[i], Arr2[j] ) where Arr1[i]+Arr2[j]==K.We will traverse using two loops for i and j. If sum (Arr1[i]+Arr2[j])==K. And the pair doesn’t exist in unordered_map. Add it to the map and increment count.Let’s understand with examples.Input Arr1[]={ 1, 3, 2, 4, 3, 2 }; Arr2[]={ 0, 2, 1, 2, 3 }; K=4Output Number of pairs with sum K : 4Explanation Pairs will be ( Arr1[0], Arr2[4] ) → ... Read More

We are given variables N, M, A and B. The goal is to find ordered pairs of positive numbers( i, j ) such that their sum is divisible by both A and B. And 1

We are given a number N. The goal is to find ordered pairs of positive numbers such that the sum of their squares is N.We will do this by finding solutions to the equation a2+ b2 = N. Where a is not more than square root of N and b can be calculated as square root of (N-a2).Let’s understand with examples.Input N=100Output Count of pairs of (a, b) where a^3+b^3=N: 2Explanation Pairs will be (6, 8) and (8, 6). 62+82=36+64=100Input N=11Output Count of pairs of (a, b) where a^3+b^3=N: 0Explanation No such pairs possible.Approach used in the below program is as followsWe take integer N.Function squareSum(int ... Read More

We are given a number N. The goal is to find ordered pairs of positive numbers such that the sum of their cubes is N.We will do this by finding solutions to the equation a3 + b3 = N. Where a is not more than cube root of N and b can be calculated as cube root of (N-a3).Let’s understand with examples.Input N=35Output Count of pairs of (a, b) where a^3+b^3=N: 2Explanation Pairs will be (2, 3) and (3, 2). 23+33=8+27=35Input N=100Output Count of pairs of (a, b) where a^3+b^3=N: 0Explanation No such pairs possible.Approach used in the below program is as followsWe take integer N.Function ... Read More

We are given a number N. The goal is to find ordered pairs of positive numbers such that their product is less than N.We will do this by starting from i=1 to i

We are given two numbers S and K. The goal is to find ordered pairs of positive numbers such that their sum is S and XOR is K.We will do this by starting from i=1 to i

In this article, we have a sorted array of integers. Our task is to find the count of elements of the given sorted array that are less than or equal to the given value K. Example Here are some examples of counting the array elements smaller than the given target element: Input: arr = {6, 12, 16, 23, 32, 45, 48, 50} target = 40 Output: 5 Input: arr = {6, 12, 15, 20, 32, 45, 48, 50} target = 20 Output: 4 Counting of smaller or equal elements in the sorted arrayHere is a list ... Read More

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] which have no digit as 0 and have sum of digits equal to a given number N. Also the numbers are divisible by MWe will do this by traversing numbers from START to END and for each number we will count the sum of its digit using a while loop ( only if all digits are non zero ). If this sum is equal to N and the number is divisible by M, increment ... Read More