Question

Google | OA | Online Coding Test Questions and Solutions | Set - 33 | 2022

Answer 1

Solution 1

 

import sys
def solution(N):

	N = str(N)
	r, l = 0, 0

	for i in range(1, len (N)) :
		if int(N[il) > int (N[i - 1]):
			r = i
			
	for i in range (0, len(N)) :
		if int (N[i]) < int (N[r]) :
			l = i
			break

	rev = N[l : r + 1]
	rev = rev[::-1]
	return int(N[:l] + rev + N[r+l:])

Answer 2

A couple of online assesment questions and solutions that are being asked in Google OA (US Based Position) - HackerRank

 

Question 1

 

One fragment of a given integer N can be selected and its digits reversed (replaced with a right to left version of themselves). What is the maximum number that can beobtained this way from integer N?
Write a function:

 

class Solution { public int solution(int N); }

 

that, given an integer 1 <= N <= 1,000,000,000, returns the greatest integer that can be created by reversing a subset of its digits.

 

Examples:
• Given N = 5340, [he answer is 5430. Fragment "34" can be reversed to "43".
• Given N = 2043, the answer is 4023. Fragment "204" can be reversed to "402"
• Given N = 620, the answer is 620. There is no need to reverse any fragment.

 

Assumptions:
• 1 <= N <= 1.000.000.000.