Given an array of size N consisting of only 0's and 1's. The array is sorted in such a manner that all the 1's are placed first and then they are followed by all the 0's. Find the count of all the 0's. Example 1: Input: N = 12 Arr[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0} %3D Output: 3 Explanation: There are 3 0's in the given array.

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.   Example 1: Input: nums = [1,2,3,4,5,6,7], k = 3Output: [5,6,7,1,2,3,4]Explanation:rotate 1 steps to the right: [7,1,2,3,4,5,6]rotate 2 steps to the right: [6,7,1,2,3,4,5]rotate 3 steps to the right: [5,6,7,1,2,3,4]Example 2: Input: nums = [-1,-100,3,99], k = 2Output: [3,99,-1,-100]Explanation: rotate 1 steps to the right: [99,-1,-100,3]rotate 2 steps to the right: [3,99,-1,-100]  Constraints: 1 <= nums.length <= 105-231 <= nums[i] <= 231 - 10 <= k <= 105  Follow up: Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.Could you do it in-place with O(1) extra space?

A majority element is an element that makes up more than half of the items inan array. Given a positive integers array, find the majority element. If there is no majority element,return -1. Do this in O(N) time and 0(1) space.Input: 1 2 5 9 5 9 5 5 5Output: 5

Check if Array Elements are Consecutive in Java. Given an array, we need to check if array contains consecutive elements. For example: Input: array[] = {5, 3, 4, 1, 2} Output: true As array contains consecutive elements from 1 to 5 Input: array[] = {47, 43, 45, 44, 46} Output: true As array contains consecutive elements from 43 to 47 Input: array[] = {6, 7, 5, 6} Output: false As array does not contain consecutive elements.

Write a program that creates a 3x3 2d array and populate it by user input. It then creates a secondarray of size 3 which stores the sum of each row in the user input matrix.If the matrix is [ [1,1,1], [2,2,2], [3,3,3] ], sums should be [3, 6, 9].The sums are then printed on the screen.