Bitwise AND in Python

 

import math

import os

import random

import re

import sys

from collections import defaultdict

#

# Complete the 'countPairs' function below.

#

# The function is expected to return a LONG_INTEGER.

# The function accepts INTEGER_ARRAY arr as parameter.

#

def countPairs(arr):

    po2 = lambda x: x > 0 and not (x & (x - 1))

    d = defaultdict(int)

    for x in arr:

        d[x] += 1

    d = list(d.items())

    ans = 0

    for i in range(len(d)):

        a, a_cnt = d[i]

        for j in range(i, len(d)):

            b, b_cnt = d[j]

            if po2(a & b):

                if a == b:

                    ans += (a_cnt * (a_cnt - 1)) // 2

                else:

                    ans += a_cnt * b_cnt

    return ans          

if __name__ == '__main__':

    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    arr_count = int(input().strip())

    arr = []

    for _ in range(arr_count):

        arr_item = int(input().strip())

        arr.append(arr_item)

    result = countPairs(arr)

    fptr.write(str(result) + '\n')

    fptr.close()

Task of Pairing

 

#!/bin/python3 import math import os import random import re import sys # # Complete the 'taskOfPairing' function below. # # The function is expected to return a LONG_INTEGER. # The function accepts LONG_INTEGER_ARRAY freq as parameter. # def taskOfPairing(freq):     # Initialize the start of the continuous subarray     i_0 = 0     # Initialize the end of the continuous subarray     i_1 = 1     n = len(freq)     total = 0     while i_1 < n:         # While not at the end of the subarray, check if the next value is non-zero         if freq[i_1] == 0:             # If value is zero, take the sum of the continuous subarray and integer divide by 2. That's the             # number of pairs in this segment             total += sum(freq[i_0: i_1]) // 2             # The new start of the continuous array is here             i_0 = i_1         i_1 += 1     # Upon reaching the end, find the number of pairs of the last subarray     total += sum(freq[i_0:i_1]) // 2     return total                   if __name__ == '__main__':     fptr = open(os.environ['OUTPUT_PATH'], 'w')     freq_count = int(input().strip())     freq = []     for _ in range(freq_count):         freq_item = int(input().strip())         freq.append(freq_item)     result = taskOfPairing(freq)     fptr.write(str(result) + '\n')     fptr.close()

Longest Subarray in python

 

#!/bin/python3
	import math
	import os
	import random
	import re
	import sys
	#
	# Complete the 'longestSubarray' function below.
	#
	# The function is expected to return an INTEGER.
	# The function accepts INTEGER_ARRAY arr as parameter.
	#
	def longestSubarray(arr):
	n = len(arr)
	ans = 0
	# O(n^2) is okay because of constraints.
	for i in range(n):
	w = []
	cnt = 0
	for j in range(i, n):
	if arr[j] in w:
	cnt += 1
	continue
	if len(w) == 0:
	w.append(arr[j])
	elif len(w) == 1:
	if abs(w[0] - arr[j]) > 1:
	break
	else:
	w.append(arr[j])
	else:
	break
	cnt += 1
	ans = max(ans, cnt)
	return ans
	if __name__ == '__main__':
	fptr = open(os.environ['OUTPUT_PATH'], 'w')
	arr_count = int(input().strip())
	arr = []
	for _ in range(arr_count):
	arr_item = int(input().strip())
	arr.append(arr_item)
	result = longestSubarray(arr)
	fptr.write(str(result) + '\n')
	fptr.close()

Username Changes in Python

 

                                                      A company has released a new internal system, and each employee has been assigned a username. Employees are allowed to change their usernames but only in a limited way. More specifically, they can choose letters at two different positions and swap them. For example, the username “bigfish” can be changed to “gibfish” (swapping ‘b’ and ‘g’) or “bighisf” (swapping ‘f’ and ‘h’). The manager would like to know which employees can update their usernames so that the new username is smaller in alphabetical order than the original username.

For each username given, return either “YES” or “NO” based on whether that username can be changed (with one swap) to a new one that is smaller in alphabetical order. Usernames changes certification test problem | Hackerrank Solution

Note: For two different strings A and B of the same length, A is smaller than B in alphabetical order when on the first position where A and B differ, A has a smaller letter in alphabetical order than B has.

For example 

let’s say usernames = [“bee”, “superhero”, “ace”]. For the first username, “bee”, it is not possible to make one swap to change it to a smaller one in alphabetical order, so the answer is “NO”. For the second username, “superhero”, it is possible get a new username that is smaller in alphabetical order (for example, by swapping letters ‘s’ and ‘h’ to get “hupersero”), so the answer is “YES”. Finally, for the last username “ace”, it is not possible to make one swap to change it to a smaller one in alphabetical order, so the answer is “NO”. Therefore you would return the array of strings [“NO”. “YES”, “NO”].

Function Description

Complete the function possibleChanges in the editor below.

possible Changes has the following parameter(s): string usernames[n] an array of strings denoting the usernames of the employees Returns:

string[n]: an array of strings containing either “YES” or “NO” based on whether the username can be changed with one swap to a new one that is smaller in alphabetical order

#!/bin/python3
	import math
	import os
	import random
	import re
	import sys
	#
	# Complete the 'possibleChanges' function below.
	#
	# The function is expected to return a STRING_ARRAY.
	# The function accepts STRING_ARRAY usernames as parameter.
	#
	def possibleChanges(usernames):
	ans = []
	for u in usernames:
	if len(u) <= 1:
	ans.append("NO")
	for i in range(len(u) - 1):
	if u[i] > u[i + 1]:
	ans.append("YES")
	break
	else:
	ans.append("NO")
	return ans
	if __name__ == '__main__':
	fptr = open(os.environ['OUTPUT_PATH'], 'w')
	usernames_count = int(input().strip())
	usernames = []
	for _ in range(usernames_count):
	usernames_item = input()
	usernames.append(usernames_item)
	result = possibleChanges(usernames)
	fptr.write('\n'.join(result))
	fptr.write('\n')
	fptr.close()