Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
Articles by Arnab Chakraborty
Page 324 of 377
Find element in a sorted array whose frequency is greater than or equal to n/2 in C++.
Consider we have an array with size n. This array is sorted. There is one element whose frequency is greater than or equal to n/2, where n is the number of elements in the array. So if the array is like [3, 4, 5, 5, 5], then the output will be 5.If we closely observe these type of array, we can easily notice that the number whose frequency is greater than or equal to n/2, will be present at index n/2 also. So the element can be found at position n/2Example Live DemoSource Code: #include using namespace std; int higherFreq(int arr[], ...
Read MoreFind the sum of digits of a number at even and odd places in C++
Suppose, we have an integer N, We have to find the sum of the odd place digits and the even place digits. So if the number is like 153654, then odd_sum = 9, and even_sum = 15.To solve this, we can extract all digits from last digit, if the original number has odd number of digits, then the last digit must be odd positioned, else it will be even positioned. After process a digit, we can invert the state from odd to even and vice versa.Example Live Demo#include using namespace std; bool isOdd(int x){ if(x % 2 == 0) ...
Read MoreFind the Product of first N Prime Numbers in C++
Suppose we have a number n. We have to find the product of prime numbers between 1 to n. So if n = 7, then output will be 210, as 2 * 3 * 5 * 7 = 210.We will use the Sieve of Eratosthenes method to find all primes. Then calculate the product of them.Example Live Demo#include using namespace std; long PrimeProds(int n) { bool prime[n + 1]; for(int i = 0; i
Read MoreFind distance from root to given node in a binary tree in C++
Consider we have a binary tree with few nodes. We have to find the distance between the root and another node u. suppose the tree is like below:Now the distance between (root, 6) = 2, path length is 2, distance between (root, 8) = 3 etc.To solve this problem, we will use a recursive approach to search the node at the left and right subtrees, and also update the lengths for each level.Example Live Demo#include using namespace std; class Node { public: int data; Node *left, *right; }; Node* getNode(int data) { Node* node = ...
Read MoreFind the number of ways to divide number into four parts such that a = c and b = d in C++
Suppose we have a number n. We have to find number of ways to divide a number into parts (a, b, c and d) such that a = c, and b = d. So if the number is 20, then output will be 4. As [1, 1, 9, 9], [2, 2, 8, 8], [3, 3, 7, 7] and [4, 4, 6, 6]So if N is odd, then answer will be 0. If the number is divisible by 4, then answer will be n/4 – 1 otherwise n/4.Example Live Demo#include using namespace std; int countPossiblity(int num) { if (num % 2 == 1) return 0; else if (num % 4 == 0) return num / 4 - 1; else return num / 4; } int main() { int n = 20; cout
Read MoreFind distance between two nodes of a Binary Tree in C++
Consider we have a binary tree with few nodes. We have to find the distance between two nodes u and v. suppose the tree is like below −Now the distance between (4, 6) = 4, path length is 4, length between (5, 8) = 5 etc.To solve this problem, we will find the LCA (Lowest Common Ancestor), then calculate distance from LCA to two nodes.Example Live Demo#include using namespace std; class Node { public: int data; Node *left, *right; }; Node* getNode(int data) { Node* node = new Node; node->data = data; node->left ...
Read MoreFind the number of jumps to reach X in the number line from zero in C++
Suppose we have an integer X. We have to find minimum number of jumps required to reach X from 0. The first jump made can be of length one unit and each successive jump will be exactly one unit longer than the previous jump in length. It is allowed to go either left or right in each jump. So if X = 8, then output is 4. 0 → -1 → 1→ 4 → 8 are possible stages.If we observe carefully, then we can say thatIf you have always jumped in the right direction, then after n jumps you will ...
Read MoreFind count of common nodes in two Doubly Linked Lists in C++
Suppose we have two doubly-linked lists. We have to find the total number of common nodes in both the doubly linked list. So if two lists are like [15, 16, 10, 9, 7, 17], and [15, 16, 40, 6, 9], there are three common nodes.Traverse both lists up to end of the list using two nested loops, for every node in the list, check if it is matched with any node of the second list or not. If a match is found, then increase the counter, and finally return the count.Example Live Demo#include using namespace std; class Node { public: ...
Read MoreFind the number of integers x in range (1,N) for which x and x+1 have same number of divisors in C++
Suppose, we have an integer N, we have to find the number of integers 1 < x < N, for which x and x + 1 has same number of positive divisors. So if N = 3, then output will be 1, as divisor of 1 is 1, divisor of 2 is 1 and 2, and divisor of 3 is 1 and 3.To solve this, we will find the number of divisors of all numbers below N, and store them in an array. Then count number of integers x such that x, such that x + 1 have the same ...
Read MoreFind consecutive 1s of length >= n in binary representation of a number in C++
Suppose, we have two integers x and n, our task is to search for the first consecutive stream of 1s (32-bit binary) which is greater than or equal to the value of n in length and return its position. If no such string exists, then return -1. For example, if x = 35, and n = 2, then result will be 31. The binary representation of 35 in a 32-bit integer is like −00000000000000000000000000100011. So two consecutive 1s are present at index 31, so the answer is 31.To solve this problem, we have to find the number of leading zeros, ...
Read More