2026 August 8 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.

2974. Minimum Number Game Easy 85.7%

2762. Continuous Subarrays Medium 58.0%

Carry over problem:

2030. Smallest K-Length Subsequence With Occurrences of a Letter Hard 39.9%

Here is the meeting url:

https://us05web.zoom.us/j/82553858456?pwd=fnK5Vb0CcfFpv0lnouILjeEs2z2Khc.1 

Full details:

Topic: Leet Code meeting
Time: Mar 28, 2026 10:00 AM Pacific Time (US and Canada)
        Every week on Sat, 110 occurrence(s)
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[Jagrut]

2974. Minimum Number Game

class Solution:

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

n = len(nums)

res = [-1] * n

INC = 2

ar = 0 # Alice read

br = 1 # Bob read

aw = 1 # Alice write

bw = 0 # Bob write

nums_sorted = sorted(nums)

while (br < n):

res[aw] = nums_sorted[ar]

res[bw] = nums_sorted[br]

ar += INC

br += INC

aw += INC

bw += INC

return res

[Anuj Patnaik]

import heapq

class Solution:

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

arr = []

heapq.heapify(nums)

while len(nums) >= 2:

alice_num = heapq.heappop(nums)

bob_num = heapq.heappop(nums)

arr.append(bob_num)

arr.append(alice_num)

return arr

[Bhupathi Kakarlapudi]

class Solution:

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

nums.sort()

arr = []

for i in range(0, len(nums), 2):

arr.append(nums[i + 1])

arr.append(nums[i])

return arr

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[Jagrut]

2762. Continuous Subarrays

class Solution:

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

# indices of max elements in decreasing order in valid window

maxq = deque()

# indices of min elements in increasing order in valid window

minq = deque()

# start of window

i = 0

# result count

res = 0

for j, n in enumerate(nums):

# maxq maintenance

while maxq and nums[maxq[len(maxq) - 1]] < n:

maxq.pop()

maxq.append(j)

# minq maintenance

while minq and nums[minq[len(minq) - 1]] > n:

minq.pop()

minq.append(j)

# make sure window is valid wrt min - max element criteria

while maxq and minq and abs(nums[maxq[0]] - nums[minq[0]]) > 2:

# one of maxq[0] or minq[0] needs to be taken out

# then condition needs to be checked again

# find the one that occurs earlier

# take that out since it will keep the other

# if you take out the one that occurs later,

# it basically means removing both

if maxq[0] < minq[0]:

i = maxq.popleft() + 1

else:

i = minq.popleft() + 1

# now window is valid wrt min - max criteria

# i and j are the boundary indices

res += j - i + 1

return res