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Page 1445 of 2109
Find 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#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 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#include using namespace std; class Node { public: int data; Node *left, *right; }; Node* getNode(int data) { Node* node = new ...
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#include using namespace std; long PrimeProds(int n) { bool prime[n + 1]; for(int i = 0; i
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#include using namespace std; bool isOdd(int x){ if(x % 2 == 0) return ...
Read MoreFind 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/2ExampleSource Code: #include using namespace std; int higherFreq(int arr[], int ...
Read MoreFind four elements a, b, c and d in an array such that a+b = c+d in C++
Suppose we have a list of integers. Our task is to find four distinct integers as two pairs like (a, b) and (c, d), such that a+b = c+d. If there are multiple answers, then print only one. Suppose the array elements are like: A = [7, 5, 9, 3, 6, 4, 2], then pairs can be (7, 3) and (6, 4)Here we will use the hashing technique. We use the sum as key as pair as the value in the hash table. We have to follow these steps to solve this problem.For i in range 0 to n – ...
Read MoreFind frequency of each element in a limited range array in less than O(n) time in C++
Suppose we have an array of integers. The array is A, and the size is n. Our task is to find the frequency of all elements in the array less than O(n) time. The size of the elements must be less than one value say M. Here we will use the binary search approach. Here we will recursively divide the array into two parts if the end elements are different, if both its end elements are the same, it means all elements in the array are the same as the array is already sorted.Example#include #include using namespace std; void calculateFreq(int ...
Read MoreFind gcd(a^n, c) where a, n and c can vary from 1 to 10^9 in C++
We have to find the GCD of two numbers of which one number can be as big as (109 ^ 109), which cannot be stored in some data types like long or any other. So if the numbers are a = 10248585, n = 1000000, b = 12564, then result of GCD(a^n, b) will be 9.As the numbers are very long, we cannot use the Euclidean algorithm. We have to use the modular exponentiation with O(log n) complexity.Example#include #include using namespace std; long long power(long long a, long long n, long long b) { long long res = 1; ...
Read MoreFind if an array contains a string with one mismatch in C++
Suppose we have a string s, and another array of strings A. We have to find whether the array is containing a string with the one-character difference from the current string of different lengths. Suppose the string is like “banana”, and the array looks like [“bana”, “orange”, “banaba”, “banapy”], the result will be true, as there is one string banaba, here only one character is different than a banana.To solve this problem, we will follow some steps −Traverse through given string s, and check for every string in the array, then follow these steps for every string in arr −Check ...
Read MoreFind if an expression has duplicate parenthesis or not in C++
Consider we have an expression exp, and we have to check whether the exp has a duplicate set of parentheses around it or not. An expression will have duplicate parentheses if one sub-expression will be surrounded by more than one parentheses set. For example, if the expression is like −(5+((7−3)))Here the sub-expression (7 – 3) is surrounded by two parentheses pair, so these are duplicate parentheses.To solve this problem, we will use stacks. We will iterate through each character in the exp, and if the character is opening parentheses ‘(’, or any of the operator or operand, then push it ...
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