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

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 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; } int main() {    float A[] = {3, 1, -2}; ... Read More

Consider we have a list of strings. The list has some duplicate strings. We have to check which strings are occurred more than once. Suppose the string list is like [“Hello”, “Kite”, “Hello”, “C++”, “Tom”, “C++”]Here we will use the hashing technique, so create an empty hash table, then traverse each string, and for each string, s is already present in the hash, then display the string, otherwise insert into the hash.Example Live Demo#include #include #include using namespace std; void displayDupliateStrings(vector strings) {    unordered_set s;    bool hasDuplicate = false;    for (int i = 0; i

Here we will see how to get the equal points in a string of brackets. The equal point is the index I, such that the number of opening brackets before it is equal to the number of the closing bracket after it. Suppose a bracket string is like “(()))(()()())))”, if we see closer, we can getSo the number of opening brackets from 0 to 9 is 5, and the number of the closing brackets from 9 to 14 is also 5, so this is the equal point.To solve this problem, we have to follow these few steps −Store the number ... Read More

Suppose we have an array A, it has n elements. Our task is to divide the array A into two subarrays, such that the sum of each subarray will be the same. Suppose the array A = [2, 3, 4, 1, 4, 5], The output is 1, so subarrays before 1 and after 1 are taken. [2, 3, 4], and [4, 5].To solve this problem, we will calculate the whole array except for the first element in right_sum. Consider that is the partitioning element. We will traverse from left to right. Subtracting an element from right_sum and adding an element ... Read More

Consider we have an array A with few elements. We have to find an element from A, such that all elements can be divided by it. Suppose the A is like [15, 21, 69, 33, 3, 72, 81], then the element will be 3, as all numbers can be divisible by 3.To solve this problem, we will take the smallest number in A, then check whether all numbers can be divided by the smallest number or not, if yes, then return the number, otherwise, return false.Example Live Demo#include #include using namespace std; int getNumber(int a[], int n) {    int minNumber ... Read More

Here we will see one problem, where we take a number n, we have to find another value say x, such that x + digit sum of x is same as the given number n. Suppose the value of n is 21. This program will return a number x = 15, as 15 + digit sum of 15, i.e. 15 + 1 + 5 = 21 = n.To solve this problem, we have to follow simple approach. We will iterate through 1 to n, in each iteration, we will see if the sum of the number and its digit sum ... Read More

Suppose we have one 3x3 matrix, whose diagonal elements are empty at first. We have to fill the diagonal such that the sum of a row, column and diagonal will be the same. Suppose a matrix is like −After filling, it will be −Suppose the diagonal elements are x, y, z. The values will be −x = (M[2, 3] + M[3, 2])/ 2z = (M[1, 2] + M[2, 1])/ 2y = (x + z)/2Example Live Demo#include using namespace std; void displayMatrix(int matrix[3][3]) {    for (int i = 0; i < 3; i++) {       for (int j = 0; j < 3; j++)          cout