2024 June 7 Problems

[daryl...@gmail.com]

Feel free to work on any of the problems you want; we'll have people present their solutions at 11:30.

Please post your solutions to this thread so others can use this as a reference.

3289. The Two Sneaky Numbers of Digitville Easy 88.4%

3335. Total Characters in String After Transformations I Medium 45.8%

3337. Total Characters in String After Transformations II Hard 58.3%

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Meeting ID: 767 4768 4609

[FloM]

3289. The Two Sneaky Numbers of Digitville Easy 88.4%

Time: O(n) Memory: O(1)

class Solution {

public:

vector<int> getSneakyNumbers(vector<int>& nums) {

unordered_set<int> seen{};

vector<int> twice{};

for(int ii=0; ii<nums.size(); ++ii)

{

if(seen.find(nums[ii]) != seen.end())

{

twice.push_back(nums[ii]);

}

seen.insert(nums[ii]);

}

return twice;

}

};

[Anuj Patnaik]

class Solution:

def getSneakyNumbers(self, nums: List[int]) -> List[int]:

n = len(nums) - 2

expected_sum = n*(n-1)//2

expected_sos_sum = n*(2*n-1)*(n-1)//6

actual_sum = sum(nums)

actual_sos_sum = 0

for i in nums:

actual_sos_sum += i*i

d1 = actual_sum - expected_sum

d2 = actual_sos_sum - expected_sos_sum

#Solving the equation (t - x)(t - y) = 0

xy = (d1*d1 - d2)//2

discriminant = int(sqrt(d1*d1 - 4*xy))

x = (d1 + discriminant) // 2

y= (d1 - discriminant) // 2

return [x,y]

[FloM]

3335. Total Characters in String After Transformations I Medium 45.8%

Had to look at the Editorial Answer. Time: O(t) Mem: O(1)

class Solution {

public:

int lengthAfterTransformations(string s, int t) {

int modulo = 1000000007;

vector<int> charCount(26, 0);

for(const char& x : s)

{

++charCount[x - 'a'];

}

for(int ii=0; ii<t; ++ii)

{

int origNumOfA = charCount[0];

int origNumOfB = charCount[1];

int origNumOfZ = charCount[25];

// charCount for A, B, and C

charCount[0] = origNumOfZ % modulo;

charCount[1] = (origNumOfA + origNumOfZ) % modulo;

for(int ii=25; ii>=3; --ii)

{

charCount[ii] = charCount[ii-1] % modulo;

}

charCount[2] = origNumOfB;

}

int count = 0;

for(int x : charCount)

{

count = (count + x) % modulo;

}

return count;

}

};

[Sharad Bagri]

3289. The Two Sneaky Numbers of Digitville Easy 88.4%

class Solution:

def getSneakyNumbers(self, nums: List[int]) -> List[int]:

sneaky = []

seen = set()

for num in nums:

if num in seen:

sneaky.append(num)

else:

seen.add(num)

return sneaky

3335. Total Characters in String After Transformations I Medium 45.8%

from collections import Counter

import copy

class Solution:

def lengthAfterTransformations(self, s: str, t: int) -> int:

cnt = Counter(s)

a_to_y = list('abdcefghijklmnopqrstuvwxy')

for i in range(t):

new_cnt = Counter()

for a_char in a_to_y:

if a_char in cnt:

new_cnt[chr(ord(a_char) + 1)] = cnt[a_char]

if 'z' in cnt:

new_cnt['a'] += cnt['z']

new_cnt['b'] += cnt['z']

cnt = new_cnt

return cnt.total() % ((10**9) + 7)