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  Count numbers containing 4   Basic  Accuracy:   48.18%  Submissions:   2010  Points:   1 Count the numbers between 1 to  N  containing 4 as a digit.   Example 1: Input: N = 9 Output: 1 Explanation: 4 is the only number between 1 to 9 which contains 4 as a digit. Example 2: Input: N = 14 Output: 2 Explanation: 4 and 14 are the only number between 1 to 14 that contains 4 as a digit.   Your Task: You don't need to read input or print anything. Your task is to complete the function  countNumberswith4()  which takes an Integer N as input and returns the answer.

Easy  Accuracy:   65.11%  Submissions:   761  Points:   2 Given an array of n integers(duplicates allowed). Print “Yes” if it is a set of contiguous integers else print “No”. Example 1: ​ Input : arr[ ] = {5, 2, 3, 6, 4, 4, 6, 6} Output : Yes Explanation: The elements  of array form a contiguous set of integers which is {2, 3, 4, 5, 6} so the output is "Yes". Example 2: Input : arr[ ] = {10, 14, 10, 12, 12,  13, 15} Output : No Your Task: This is a function problem. The input is already taken care of by the driver code. You only need to complete the function  areElementsContiguous()  that takes an array  (arr) , sizeOfArray  (n) , and return the  true  if it is a set of contiguous integers else print  false . The driver code takes care of the printing.

You are given a pointer/ reference to the node which is to be deleted from the linked list of   N  nodes. The task is to delete the node. Pointer/ reference to head node is not given.  Note:  No head reference is given to you. It is guaranteed that the node to be deleted is   not a tail node   in the linked list. Example 1: Input: N = 2 value[] = {1,2} node = 1 Output: 2 Explanation: After deleting 1 from the linked list, we have remaining nodes as 2. Example 2: Input: N = 4 value[] = {10,20,4,30} node = 20 Output: 10 4 30 Explanation: After deleting 20 from the linked list, we have remaining nodes as 10, 4 and 30. Your Task: You only need to complete the  function deleteNode  that takes  reference  to the node that needs to be  deleted . The  printing  is done  automatically  by the  driver code .