2026 July 11 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.

1385. Find the Distance Value Between Two Arrays Easy 72.0%

1658. Minimum Operations to Reduce X to Zero Medium 40.7%

1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows Hard 62.5%

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)
Please download and import the following iCalendar (.ics) files to your calendar system.
Weekly: https://us05web.zoom.us/meeting/tZYocuqgqzMuH9IuHUphHIiZ_4T4IndJLToX/ics?icsToken=DJFc88pL6-me_TneLwAALAAAAA70ShjYgLH6RfEnqUm5LK03v2OgtbXujwlKXRMVgIVWCY9aONPJMCycMlkhyPZ3XSrMVVgXDSYlGF_r7DAwMDAwMQ&meetingMasterEventId=MGJmt5lJTd6mPygcS0hiAQ
Join Zoom Meeting
https://us05web.zoom.us/j/82553858456?pwd=fnK5Vb0CcfFpv0lnouILjeEs2z2Khc.1

Meeting ID: 825 5385 8456
Passcode: 182596

[Anuj Patnaik]

class Solution:

def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:

count = 0

arr2.sort()

i = 0

j = 0

while i < len(arr1):

if (-d > arr1[i] - arr2[j] or arr1[i] - arr2[j] > d) and j < len(arr2) - 1:

j += 1

else:

if -d <= arr1[i] - arr2[j] and arr1[i] - arr2[j] <= d:

i += 1

j = 0

else:

if (-d > arr1[i] - arr2[j] or arr1[i] - arr2[j] > d) and j == len(arr2) - 1:

count += 1

j = 0

i += 1

return count

[Jagrut]

1385. Find the Distance Value Between Two Arrays

class Solution:

def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:

clean_count = 0

for a in arr1:

if all(abs(a - b) > d for b in arr2):

clean_count += 1

return clean_count

[Jagrut]

1658. Minimum Operations to Reduce X to Zero

class Solution:

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

target = sum(nums) - x

if target < 0:

return -1

curr_sum = 0

max_window_size = -1

left = 0

for right in range(len(nums)):

# Incorporate incoming number

curr_sum += nums[right]

# Make sure window satisfies criteria

while left <= right and curr_sum > target:

curr_sum -= nums[left]

left += 1

# Check if window sum is exactly equal to target

if curr_sum == target:

max_window_size = max(max_window_size, right - left + 1)

return -1 if max_window_size == -1 else len(nums) - max_window_size