Server Side Programming Articles - Page 1997 of 2650

Suppose we have three sorted arrays A, B and C, and three elements i, j and k from A, B and C respectively such that max(|A[i] – B[i]|, |B[j] – C[k]|, |C[k] – A[i]|) is minimized. So if A = [1, 4, 10], B = [2, 15, 20], and C = [10, 12], then output elements are 10, 15, 10, these three from A, B and C.Suppose the size of A, B and C are p, q and r respectively. Now follow these steps to solve this −i := 0, j := 0 and k := 0Now do the following ... Read More

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 More

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 More

Suppose we have a list of numbers from 0 to n-1. A number can be repeated as many as a possible number of times. We have to find the repeating numbers without taking any extra space. If the value of n = 7, and list is like [5, 2, 3, 5, 1, 6, 2, 3, 4, 5]. The answer will be 5, 2, 3.To solve this, we have to follow these steps −for each element e in the list, do the following steps −sign := A[absolute value of e]if the sign is positive, then make it negativeOtherwise, it is a ... Read More

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

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 More

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

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 More

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 More

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 More