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Find area of parallelogram if vectors of two adjacent sides are given using C++.

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 211 Views

Suppose we have two vectors for two adjacent sides of a parallelogram in the form $x\hat{i}+y\hat{j}+z\hat{k}$ Our task is to find the area of parallelogram. The area of parallelogram is magnitude of the cross product of two vectors. (|A × B|)$$\rvert \vec{A}\times\vec{B}\rvert=\sqrt{\lgroup y_{1}*z_{2}-y_{2}*z_{1}\rgroup^{2}+\lgroup x_{1}*z_{2}-x_{2}*z_{1}\rgroup^{2}+\lgroup x_{1}*y_{2}-x_{2}*y_{1}\rgroup^{2}}$$Example#include #include using namespace std; float area(float A[], float B[]) {    float area = sqrt(pow((A[1] * B[2] - B[1] * A[2]), 2) + pow((A[0] * B[2] - B[0] * A[2]), 2) + pow((A[0] * B[1] - B[0] * A[1]), 2));    return area; } int main() {    float A[] = {3, 1, -2};   ...

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Find the longest common prefix between two strings after performing swaps on second string in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 359 Views

Suppose we have two strings str1 and str2. Find the longest common prefix between them after performing zero or more operation on the second string. In each operation, we can swap any two letters. So if str1 = “HERE”, str2 = “THERE”, then output will be 4. The second string can be made “HERET” by just swapping the characters. So the longest prefix is of length 4.As we know that we can only swap on str2. And the length of the matrix should be maximized. So the idea is to traverse str1, and check if the frequency of the current ...

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Find ceil of a/b without using ceil() function in C++.

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 959 Views

Here we will see how to get the ceiling value of a/b without using the ceil() function. If a = 5, b = 4, then (a/b) = 5/4. ceiling(5/4) = 2. To solve this, we can follow this simple formula −$$ceil\lgroup a,b\rgroup=\frac{a+b-1}{b}$$Example#include using namespace std; int ceiling(int a, int b) {    return (a+b-1)/b; } int main() {    cout

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Find the longest sub-array having exactly k odd numbers in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 175 Views

Suppose we have one array with n elements. The problem is to find longest sub-array which has exactly k odd numbers. So if A = [2, 3, 4, 11, 4, 12, 7], and k = 1, then output will be 4, sub-array is [4, 11, 4, 12]We can solve this using sliding window. The task is like below −Initialize max := 0, count := 0, and start := 0for i in range 0 to n – 1, do the followingif arr[i] mod 2 is not 0, then increase count by 1while count > k and start k && start

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Hello World using Perl.

Mohd Mohtashim
Mohd Mohtashim
Updated on 11-Mar-2026 1K+ Views

Perl is a programming language developed by Larry Wall, specially designed for text processing.Just to give you a little excitement about Perl, I'm going to give you a small conventional Perl Hello World program,You can try it using the Demo link.Example#!/usr/bin/perl # This will print "Hello, World" print "Hello, world";

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Find difference between sums of two diagonals in C++.

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 603 Views

Here we will see how to get the difference between the sums of two diagonals of a given matrix. Suppose we have a matrix of order N x N, we have to get the sum of primary and secondary diagonals, then get the difference of them. To get the major diagonal, we know that the row index and column index increases simultaneously. For the second diagonal, row index and column index values are increased by this formula row_index = n – 1 – col_index. After getting the sum, take the difference and return a result.Example#include #include #define MAX 100 using ...

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Find number of diagonals in n sided convex polygon in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 449 Views

Suppose we have a number n, and we have to find the number of diagonals for n sided convex polygon. So if the n = 5, then diagonals count will be 5.As this is n-sided convex polygon, from each vertex we can draw n – 3 diagonals leaving two sided adjacent vertices and itself. So for n vertices, it will be n*(n-3), but as we are considering twice, so it will be n(n – 3)/2.Example#include using namespace std; int diagonalCount(int n) {    return n * (n - 3) / 2; } int main() {    int n = 8;    cout

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Find the minimum number of steps to reach M from N in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 924 Views

Suppose we have two integers N and M. We have to find minimum number of steps to reach M from N, by performing given operations −Multiply the number x by 2, so x will be 2*xSubtract one from the number x, so the number will be x – 1If N = 4 and M = 6, then output will be 2. So if we perform operation number 2 on N, then N becomes 3, then perform operation number one on updated value of N, so it becomes 2 * 3 = 6. So the minimum number of steps will be ...

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Find out the minimum number of coins required to pay total amount in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 352 Views

Suppose we have a number N, and unlimited number of coins worth 1, 10 and 25 currency coins. Find minimum number of coins we need to use to pay exactly amount N. Suppose N is 14, then number of coins will be 5, as one 10 value coin and four 1 value coin.To solve this, we have to use these steps −If N < 10, then return N number of 1 value coinsIf N > 9 and N < 25, then divide the value with 10, and get the result, the remaining will be covered using 1 value coin, add ...

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Find Harmonic mean using Arithmetic mean and Geometric mean using C++.

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 415 Views

Here we will see how to get the Harmonic mean using the arithmetic mean and the geometric mean. The formula for these three means are like below −Arithmetic Mean − (a + b)/2Geometric Mean − $$\sqrt{\lgroup a*b\rgroup}$$Harmonic Mean − 2ab/(a+b)The Harmonic Mean can be expressed using arithmetic mean and geometric mean using this formula −$$HM=\frac{GM^{2}}{AM}$$Example#include #include using namespace std; double getHarmonicMean(int a, int b) {    double AM, GM, HM;    AM = (a + b) / 2;    GM = sqrt(a * b);    HM = (GM * GM) / AM;    return HM; } int main() ...

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