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Here we will see how to get the last two digits. The unit place digit and the tens place digit of the sum of N factorials. So if N = 4, it will be 1! + 2! + 3! + 4! = 33. so unit place is 3 and ten place is 3. The result will be 33.If we see this clearly, then as the factorials of N > 5, the unit place is 0, so after 5, it will not contribute to change the unit place. And after N > 10, the ten places will remain 0. For N ... Read More
Consider we have an element x, we have to find the largest prime factor of x. If the value of x is 6, then-largest prime factor is 3. To solve this problem, we will just factorize the number by dividing it with the divisor of a number and keep track of the maximum prime factor.Example Live Demo#include #include using namespace std; long long getMaxPrimefactor(long long n) { long long maxPF = -1; while (n % 2 == 0) { maxPF = 2; n /= 2; } for (int i = 3; i 2) maxPF = n; return maxPF; } int main() { long long n = 162378; cout
We know that a rectangle can be represented using two coordinates, the top left corner, and the bottom right corner. Suppose there are two rectangles, we have to check whether these two overlap or not. There are four coordinate points (l1, r1) and (l2, r2).l1 is the top-left corner of first rectangler1 is the bottom-right corner of the first rectanglel2 is the top-left corner of second rectangler2 is the bottom-right corner of the second rectangleWe have assumed that the rectangles are parallel to the coordinate axes. To solve this, we have to check a few conditions.One rectangle is above the ... Read More
Suppose we have the first element of AP, and the differenced. We have to check whether the given number n is a part of AP or not. If the first term is a = 1, differenced = 3, and the term x = 7 will be checked. The answer is yes.To solve this problem, we will follow these steps −If d is 0, and a = x, then return true, otherwise false.Otherwise, if d is not 0, then if x belongs to the sequence x = a + n * d, where n is a non-negative integer, only if (n ... Read More
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 Live Demo#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 ... Read More
Here we will see how to find the floor and ceiling in an unsorted array. The floor value is larger element which is smaller than or equal to x, and the ceiling value is smallest value which is larger than x. If the array A = [5, 6, 8, 9, 6, 5, 5, 6], and x is 7, then the floor value is 6, and the ceiling value is 8.To solve this problem, we will follow the linear search approach. We will traverse the array and track two distances with respect to x.Min distance of element greater than or equal ... Read More
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 Live Demo#include #include #define MAX 100 ... Read More
Suppose we have a matrix of size M x N. We have to find the column, that has a maximum sum. In this program we will not follow some tricky approach, we will traverse the array column-wise, then get the sum of each column, if the sum is the max, then print the sum and the column index.Example#include #define M 5 #define N 5 using namespace std; int colSum(int colIndex, int mat[M][N]){ int sum = 0; for(int i = 0; i maxSum) { maxSum = sum; index = i; } } cout
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 Live Demo#include using namespace std; int ceiling(int a, int b) { return (a+b-1)/b; } int main() { cout
Suppose we have two vectors for two adjacent sides of a triangle in the form $x\hat{i}+y\hat{j}+z\hat{k}$ Our task is to find the area of triangle. The area of triangle is magnitude of the cross product of two vectors. (|A x B|)$$\frac{1}{2}\rvert \vec{A}\times\vec{B}\rvert=\frac{1}{2}\sqrt{\lgroup y_{1}*z_{2}-y_{2}*z_{1}\rgroup^{2}+\lgroup x_{1}*z_{2}-x_{2}*z_{1}\rgroup^{2}+\lgroup x_{1}*y_{2}-x_{2}*y_{1}\rgroup^{2}}$$Example Live Demo#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*0.5; } int main() { float A[] = {3, 1, -2}; ... Read More