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Which byte occurs most frequently in an array? Write a function that returns the byte that occurs most frequently in an array of bytes Parameters: Array of bytes Count of the number of bytes in the array (if you want it)
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Software Developer Intern

Interviewed at Glitch

3.3★
Oct 6, 2011

Which byte occurs most frequently in an array? Write a function that returns the byte that occurs most frequently in an array of bytes Parameters: Array of bytes Count of the number of bytes in the array (if you want it)

Let's say a triple (a, b, c) is a zigzag if either a < b > c or a > b < c. Given an array of integers numbers, your task is to check all the triples of its consecutive elements for being a zigzag. More formally, your task is to construct an array of length numbers.length - 2, where the ith element of the output array equals 1 if the triple (numbers[i], numbers[i + 1], numbers[i + 2]) is a zigzag, and 0 otherwise. Example • For numbers = [1, 2, 1, 3, 4], the output should be solution(numbers) = [1, 1, 0]. o (numbers[0], numbers[1], numbers[2]) = (1, 2, 1) is a zigzag, because 1 < 2 > 1; o (numbers[1], numbers[2] , numbers[3]) = (2, 1, 3) is a zigzag, because 2 > 1 < 3; o (numbers[2], numbers[3] , numbers[4]) = (1, 3, 4) is not a zigzag, because 1 < 3 < 4; • For numbers = [1, 2, 3, 4], the output should be solution(numbers) = [0, 0]; Since all the elements of numbers are increasing, there are no zigzags. • For numbers = [1000000000, 1000000000, 1000000000], the output should be solution(numbers) = [0]. Since all the elements of numbers are the same, there are no zigzags.
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Software Developer Intern

Interviewed at Robinhood

3.6★
Sep 18, 2024

Let's say a triple (a, b, c) is a zigzag if either a < b > c or a > b < c. Given an array of integers numbers, your task is to check all the triples of its consecutive elements for being a zigzag. More formally, your task is to construct an array of length numbers.length - 2, where the ith element of the output array equals 1 if the triple (numbers[i], numbers[i + 1], numbers[i + 2]) is a zigzag, and 0 otherwise. Example • For numbers = [1, 2, 1, 3, 4], the output should be solution(numbers) = [1, 1, 0]. o (numbers[0], numbers[1], numbers[2]) = (1, 2, 1) is a zigzag, because 1 < 2 > 1; o (numbers[1], numbers[2] , numbers[3]) = (2, 1, 3) is a zigzag, because 2 > 1 < 3; o (numbers[2], numbers[3] , numbers[4]) = (1, 3, 4) is not a zigzag, because 1 < 3 < 4; • For numbers = [1, 2, 3, 4], the output should be solution(numbers) = [0, 0]; Since all the elements of numbers are increasing, there are no zigzags. • For numbers = [1000000000, 1000000000, 1000000000], the output should be solution(numbers) = [0]. Since all the elements of numbers are the same, there are no zigzags.

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