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

Create a method reverse Words that takes char array as an input and for each and every character until the empty space is not reached reverse the word. At the last step reverse the entire string from length 0 to n-1 length. In the first step the string “This is my book” will be turned into “koob ym si siht”. At the end of the second step the string words will be reversed to “book my is This”Time complexity − O(N)Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       static void reverse(char[] str, int start, int end){ ... Read More

The array is already sorted, we can keep two pointers ii and jj, where ii is the slow-runner while jj is the fast-runner. As long as nums[i] = nums[j]nums[i]=nums[j], we increment jj to skip the duplicate.When we encounter nums[j] != nums[i] the duplicate run has ended so we must copy its value to nums[i + 1]nums[i+1]. ii is then incremented and we repeat the same process again until jj reaches the end of array.Create an new array copy all the elements from the filtered array till the index and return the new array.Time complexity − O(N)Example Live Demousing System; namespace ConsoleApplication{ ... Read More

The array is already sorted, we can keep two pointers ii and jj, where ii is the slow-runner while jj is the fast-runner. As long as nums[i] = nums[j]nums[i]=nums[j], we increment jj to skip the duplicate.When we encounter nums[j] != nums[i] the duplicate run has ended so we must copy its value to nums[i + 1]nums[i+1]. ii is then incremented and we repeat the same process again until jj reaches the end of array.Time complexity − O(N)Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public int RemoveDuplicatesFromSortedArrayAndReturnLength(int[] arr){          int index = 1; ... Read More

Given an array and number k, the problem states that we have to rotate the array k times.If the given number is 3 then the array must be rotated 3 times.Create a function reverse which takes the array, start and end as a parameter.In the 1st step call reverse method from 0 to array length.In the 2nd step call the reverse method from 0 to k-1.In the 3rd step call the reverse method from k+1 to array length.Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public void ReverseArrayKTimes(int[] arr, int k){          Reverse(arr, ... Read More

Take two-pointers, low, high. We will use low pointers at the start, and the high pointer will point at the end of the given array.If array [low] =0, then no swapping requiredIf array [low] = 1, then swapping is required. Decrement high pointer once.Time complexity − O(N)Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       public void SwapZerosOnes(int[] arr){          int low = 0;          int high = arr.Length - 1;          while (low < high){             if (arr[low] == 1){     ... Read More

We need to take three-pointers, low, mid, high. We will use low and mid pointers at the start, and the high pointer will point at the end of the given array.If array [mid] =0, then swap array [mid] with array [low] and increment both pointers once.If array [mid] = 1, then no swapping is required. Increment mid pointer once.If array [mid] = 2, then we swap array [mid] with array [high] and decrement the high pointer once.Time complexity − O(N)Example Live Demousing System; namespace ConsoleApplication{    public class Arrays{       private void Swap(int[] arr, int pos1, int pos2){   ... Read More

HasPathsum takes 2 parameters one is the tree node and other is the sum value, initially we check whether the node is null or not, if the node is null then we return false. If the node is not null then we call HasPathSum recursive method, in each and every recursive step we keep on subtracting the sum value from the node value. If the value of the sum reaches 0 then we come to conclusion that the given tree has the path that is equal to sum and return true.Examplepublic class TreesPgm{    public class Node{       ... Read More