Found 2745 Articles for Csharp

Create a function to Find Power which takes the number x and n, where x is the 2 and n is how many times, we have to do the power. If the number is even then we have to do x*x and if the number is odd multiply the result with x*x. Continue the recursive call until the n becomes 0.Suppose if we have a number 2 and 8, then 2*2*2*2*2*2*2*2 =256.Example Live Demousing System; namespace ConsoleApplication{    public class BackTracking{       public int FindPower(int x, int n){          int result;          if ... Read More

Linear scan the 2d grid map, if a node contains a '1', then it is a root node that triggers a Depth First Search. During DFS, every visited node should be set as '0' to mark as visited node. Count the number of root nodes that trigger DFS, this number would be the number of islands since each DFS starting at some root identifies an island.Example Live Demousing System; namespace ConsoleApplication{    public class Matrix{       public int PrintNumberOfIslands(char[, ] grid){          bool[, ] visited = new bool[grid.GetLength(0), grid.GetLength(1)];          int res = ... Read More

To rotate a matrix in spiral order, we need to do following until all the inner matrix and the outer matrix are covered −Step1 − Move elements of top rowStep2 − Move elements of last columnStep3 − Move elements of bottom rowStep4 − Move elements of first columnStep5 − Repeat above steps for inner ring while there is an inner matrixExample Live Demousing System; namespace ConsoleApplication{    public class Matrix{       public void PrintMatrixInSpiralOrder(int m, int n, int[, ] a){          int i, k = 0, l = 0;          while (k < ... Read More

The entire matrix needs to be rotated k number of times. In a matrix there is a total of n/2 squares in n*n matrix and we can process each square one at a time using nested loop. In each square, elements are moving in a cycle of 4 elements then we swap the elements involved in an anticlockwise direction for each cycle.Element at position (n-1-j, i) will go to position(i, j)Element at position (i, j) will go to position(j, n-1-i)Element at position (j, n-1-i) will go to position(n-1-i, n-1-j)Element at position (n-1-i, n-1-j) will go to position(n-1-j, i)Example Live Demousing System; ... Read More

In a Matrix, there is a total of n/2 squares in n*n matrix and we can process each square one at a time using nested loop. In each square elements are moving in a cycle of 4 elements. we swap the elements involved in an anticlockwise direction for each cycleElement at position (n-1-j, i) will go to position(i, j)Element at position (i, j) will go to position(j, n-1-i)Element at position (j, n-1-i) will go to position(n-1-i, n-1-j)Element at position (n-1-i, n-1-j) will go to position(n-1-j, i)Example Live Demousing System; using System.Text; namespace ConsoleApplication{    public class Matrix{       public ... Read More

We can simply start from the first element and repeatedly call for all the elements reachable from first element. The minimum number of jumps to reach end from first can be calculated using minimum number of jumps needed to reach end from the elements reachable from first.Array == {1, 3, 6, 3, 2, 3, 6, 8, 9, 5};Number of steps required is 4Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public int MinJumps(int[] arr, int l, int h){          if (h == l)             return 0;          if (arr[l] == 0)             return int.MaxValue;          int min = int.MaxValue;          for (int i = l + 1; i

To find the missing numberCreate a new array and traverse through the entire array and make the number true in the new array if the number is found Traverse through the entire array and return the first false element as the missing element.To find the repeating elementThe first true element from the new array will be the repeated element.Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public void MissingNumberAndRepeatedNumber(int[] arr){          bool[] tempArray = new bool[arr.Length + 1];          int missingelement = -1;          int repeatingelement = ... Read More

There are three methods as written below −In the first methodUse the formula n(n+1)/2 that counts the number of elements and then need to subtract from the elements in the array.In the second methodCreate a new array and traverse through the entire array and make the number false whichever is found.In the third methodUse Xor operation. which gives the missing number.Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public int MissingNumber1(int[] arr){          int totalcount = 0;          for (int i = 0; i < arr.Length; i++){       ... Read More

Create an empty new array of length 256, traverse through the entire string character by character and increment the value in the new array. At the end traverse the entire array and return the first character that has value 1.Example 1aabccd -→2 1 2 1 → Return the first character which is having count 1. That is b by subtracting with the ascii values.Example 2 Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public char ReturnCharacterOfFirstUniqueCharachter(string s){          int index = -1;          int[] arrayValues = new int[256];       ... Read More

Create an empty new array of length 256, traverse through the entire string character by character and increment the value in the new array. At the end traverse the entire array and return the first character that has value 1.Example 1aabccd -→2 1 2 1 → Return the first character which is having count 1. That is b.Example 2 Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public int ReturnIndexOfFirstUniqueCharachter(string s){          int index = -1;          int[] arrayValues = new int[256];          for (int i = 0; ... Read More