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Consider we have an integer n. Our task is to find two numbers a and b, where these three conditions will be satisfied.a mod b = 0a * b > na / b < nIf no pair is found, print -1.For an example, if the number n = 10, then a and b can be a = 90, b = 10. This satisfies given rules.To solve this problem, we will follow these steps −Let b = n. a can be found using these three conditionsa mod b = 0 when a is multiple of ba / b < n, so ... Read More
Suppose we have two numbers N and D. We have to find N digit number, that is divisible by D. If N is 3, and D is 5, then the number can be 500. This can be solved easily. If D is 10 and N is 1, then it will be impossible. We can put D, and suppose the D has m number of digits, then attach N – m number of 0s to make it N digit number and divisible by D.Example#include using namespace std; string nDigitDivByD(int n, int d) { string ans = ""; if (d ... Read More
Here we will see if a number is given, then how to find the value of Most Significant Bit value, that is set. The value is power of 2. So if the number is 10, MSB value will be 8.We have to find the position of MSB, then find the value of the number with a set-bit at kth position.Example#include #include using namespace std; int msbBitValue(int n) { int k = (int)(log2(n)); return (int)(pow(2, k)); } int main() { int n = 150; cout
Consider we have a list of numbers; our task is to find the middle of the linked list using recursion. So if the list elements are [12, 14, 18, 36, 96, 25, 62], then the middle element is 36.To solve this problem, we will count total number of nodes in the list in recursive manner and do half of this. Then rolling back through recursion decrement n by 1 in each call, return element where n is zero.Example#include #include using namespace std; class Node{ public: int data; Node *next; }; Node* getNode(int data){ ... Read More
Consider we have a matrix, our task is to find the maximum element of each row of that matrix and print them. This task is simple. For each row, reset the max, and find the max element, and print it. Let us see the code for better understanding.Example#include #define MAX 10 using namespace std; void largestInEachRow(int mat[][MAX], int rows, int cols) { for (int i = 0; i < rows; i++) { int max_row_element = mat[i][0]; for (int j = 1; j < cols; j++) { if (mat[i][j] > max_row_element) max_row_element = mat[i][j]; } cout
When an exception is cached in a catch block, you can re-throw it using the throw keyword (which is used to throw the exception objects).While re-throwing exceptions you can throw the same exception as it is without adjusting it as −try { int result = (arr[a])/(arr[b]); System.out.println("Result of "+arr[a]+"/"+arr[b]+": "+result); } catch(ArithmeticException e) { throw e; }Or, wrap it within a new exception and throw it. When you wrap a cached exception within another exception and throw it, it is known as exception chaining or, exception wrapping, by doing this you can adjust your exception, throwing a ... Read More
Consider we have a matrix, our task is to find the maximum element of each column of that matrix and print them. This task is simple. For each column, reset the max, and find the max element, and print it. Let us see the code for better understanding.Example#include #define MAX 10 using namespace std; void largestInEachCol(int mat[][MAX], int rows, int cols) { for (int i = 0; i < cols; i++) { int max_col_element = mat[0][i]; for (int j = 1; j < rows; j++) { if (mat[j][i] > max_col_element) max_col_element = mat[j][i]; } cout
Suppose we have a string str. We have another character ch. Our task is to find the last index of ch in the string. Suppose the string is “Hello”, and character ch = ‘l’, then the last index will be 3.To solve this, we will traverse the list from right to left, if the character is not same as ‘l’, then decrease index, if it matches, then stop and return result.Example#include using namespace std; int getLastIndex(string& str, char ch) { for (int i = str.length() - 1; i >= 0; i--) if (str[i] == ch) ... Read More
Consider we have one circular array containing integers from 1 to n. Find the last element, that would remain in the list after deleting every second element starting from first element. If the input is 5, then array will be [1, 2, 3, 4, 5]. Start from 1. After deleting each second element will be like −1 0 3 4 5 1 0 3 0 5 0 0 3 0 5 0 0 3 0 0So the element that remains in the list is 3.We will solve this problem using recursion. Suppose n is even. The numbers 2, 4, 6 ... Read More
Suppose we have an array A with some elements. We have to find the largest element in the array A, but the constraint is, we cannot use any conditional operator. So if A = [12, 63, 32, 24, 78, 56, 20], then maximum element will be 78.To solve this problem, we will use the bitwise AND operation. at first we will insert one extra element INT_MAX (where all bit is 1) in array. Then we will try to find maximum AND value of any pair from the array. This obtained max value will contain AND value of INT_MAX and largest ... Read More