# Array

Given an array of integers. Find the maximum number of consecutive integers present in the array. For example, if the array is: int arr[] = { 2, 24, 22, 60, 56, 23, 25}; Then the answer should be 4, because there are 4 consecutive integers present in the array (22, 23, 24, 25). pinkdiamonds.nl

Give the recursive implementation of the Bubble Sort algorithm

Given an array of integers and a number x. check if there exists two elements in the array whose sum = x.

Given an array of unsigned integers, print all the numbers that have more than k bits set in their binary representation. For example, if k = 2, and input array is: int arr[] = {2, 1, 15, 9, 7, 14, 32, 127}; Then the output should be: 15 7 14 127 Because all of these […]

Glasses are arranged in the form of triangle (on top of each other) as shown below: 1 2 3 4 5 6 7 8 9 10 ……………….. …………………. Liquid is poured into 1st glass (Glass no. 1). When it is full, then extra liquid will flow into the glasses 2 and 3 in equal quantities. […]

Given an array of n integers, find the sub-array of length k with maximum sum. For example, Input Array: {6, 4, 3, 5, 1, 9, 2} k:3 Output: 15 (sum of sub-array 5, 1, 9)

Given a positive integer n, print the consecutive numbers whose sum is equal to n. For example, n = 20 Output: 2, 3, 4, 5, 6 (because 2+3+4+5+6+7 = 20) n = 29 Output: 14, 15

A valid mathematical expression can also have duplicate parenthesis as shown below: ((a+b)) (((a+(b)))+(c+d)) Write code to find if an expression has duplicate parenthesis. You may assume that expression does not have any white spaces.

Print Next Greater Element for every element in the array. The Next greater Element of element x is first greater element on its right side. Elements for which no greater element exist, next greater element is -1. Example: – For any array, rightmost element always has next greater element as -1. – All elements of […]

Given an array of numbers find the maximum XOR value of two numbers in the array. Input Array = {12, 15, 5, 1, 7, 9, 8, 6, 10, 13}; Output = 15 (XOR of 5 and 10) 5 = 0101 10 = 1010 ——— XOR = 1111