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C++ Articles
Page 29 of 597
Find number of Positional Elements in C++
In this problem, we are given two dimensional arrays mat[n][m]. Our task is to find the number of positional elements.An element is said to be a positional element if the element is either maximum or minimum element of the row or column.Let’s take an example to understand the problem, Inputmat[][] = {2, 5, 7} {1, 3, 4} {5, 1, 3}Output8ExplanationElements 2, 5, 7, 1, 4, 5, 1, 3 are positional elements.Solution ApproachA simple solution to the problem is by storing the maximum and minimum element of each row and column. And then check for the condition and count the number.Program ...
Read MoreFind number of transformation to make two Matrix Equal in C++
In this problem, we are given two matrices mat1[][] and mat2[][] of the same size. Our task is to find the number of transformations to make two Matrices Equal.The transformation one matrices are −Select any matrix of the two matrices.Select a row or column from the matricesAdd 1 to all elements of the selected row or column.Let’s take an example to understand the problem, Inputmat1[][] = {{1 2} {2 1}} mat1[][] = {{2 3} {4 3}}Output3Explanation1 2 => 2 2 => 2 3 => 2 3 2 1 => 3 1 => 3 2 => 4 3Solution ApproachA simple solution ...
Read MoreFind number of unique triangles among given N triangles in C++
In this problem, we are given three arrays s1[] , s2[] and s3[] of size N, denoting N triangles. Our task is to find the number of unique triangles among given N triangles.For a triangle to be unique, all its side should be unique i.e. no other triangle should have the same sides.Let’s take an example to understand the problem, Inputs1[] = {1, 5, 3} s2[] = {2, 3, 2} s3[] = {4, 2, 5}Output1ExplanationTriangle with sides 1 2 4 is unique.Solution ApproachA simple solution to the problem is by counting the number of triangles which are unique.For this, we ...
Read MoreFind numbers with K odd divisors in a given range in C++
In this problem, we are given three integer values, L, R, and k. Our task is to find numbers with K odd divisors in a given range. We will be finding the count of numbers in the range [L, R] that have exactly k divisors.We will be counting the 1 and the number itself as a divisor.Let’s take an example to understand the problem, Inputa = 3, b = 10, k = 3Output2ExplanationNumbers with exactly 3 divisors within the range 3 to 10 are 4 : divisors = 1, 2, 4 9 : divisors = 1, 3, 9Solution ApproachA simple ...
Read MoreFind orientation of a pattern in a matrix in C++
In this problem, we are given a matrix consisting of character values that make a pattern, we are also given a pattern to be found. Our task is to find the orientation (horizontal or vertical) of a pattern in a matrix.Let’s take an example to understand the problem, Inputmat[][] = { { r, a, m }, {a, m, c}, {w, f, t} } Patern : rawOutputverticalSolution ApproachA simple solution to the problem is by searching the M sized pattern in all the N rows of the matrix. This solution is ok but a more effective solution to ...
Read MoreFind perimeter of shapes formed with 1s in binary matrix in C++
In this problem, we are given a binary matrix bin[][] of size nXm consisting of 0’s and 1’s only. Our task is to Find perimeter of shapes formed with 1s in a binary matrix.The perimeter taken will cover the figure from all sides, i.e.For 1 single value, the perimeter is 4.Let’s take an example to understand the problem, Inputbin[][] = [1, 0] [1, 0]Output6ExplanationThe cells (0, 0) and (1, 0) are connected making a rectangle of sides 2 and 1. The perimeter is 6.Solution ApproachA simple solution to the problem is simply finding all one and their ...
Read MoreFind other two sides and angles of a right angle triangle in C++
In this problem, we are given an integer a denoting one side of a right angle triangle. We need to check whether it is possible to have a right angle triangle with side a. If it is possible, then find the other two sides and angle of a right angle triangle.Let’s take an example to understand the problem, Inputa = 5OutputSides : 5, 12, 13 Angles : 67.38, 22.62, 90ExplanationThe sides of right angle are found as 52 + 122 = 132 And using these sides we can find the angles are, Sin-1 (5/13) and 90 - Sin-1 (5/13).Solution ApproachA ...
Read MoreFind politeness of a number in C++
In this problem, we are given a positive integer N. Our task is to find the politeness of a number.Polite Number is a number which can be expressed as a sum of two or more consecutive numbers.Politeness of a number is defined as the number of ways the number can be expressed as sum of consecutive integers.Let’s take an example to understand the problem, Inputn = 5Output1Explanation2 + 3 = 5, is the only consecutive sum.Solution ApproachA simple solution to the problem is to check all consecutive numbers till N and if their sum is equal to N, increase count ...
Read MoreFind pairs in array whose sums already exist in array in C++
In this problem, we are given an array arr[] consisting of N integer. Our task is to find pairs in an array whose sums already exist in the array. We need to find pairs with sum value = a value in the arrayLet’s take an example to understand the problem, Inputarr[] = {1, 2, 4, 6, 7}Output(1, 6), (2, 4)ExplanationFor pairs (1, 6), the sum of values is 7 which is present in the array.For pairs (2, 4), the sum of values is 6 which is present in the array.Solution ApproachA simple solution to the problem is by finding all ...
Read MoreFind position of an element in a sorted array of infinite numbers in C++
In this problem we are given an array consisting of infinite sorted numbers. Our task is to Find position of an element in a sorted array of infinite numbers.Let’s take an example to understand the problem, Inputarr[] = {2, 4, 6, 8, 9, 12, 14, 17, ….}, ele = 9Output4ExplanationSolution ApproachFor searching elements from a sorted array efficiently, we will be using the binary searching method. Here, single the end point is not known, we will modify the algorithm a bit.We will fix the start pointer to first position, then take the end pointer to second position. After this, we ...
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