How do I return all possible triplets in the 3Sum problem? The key idea is the same as the TwoSum problem When we fix the 1st number, the 2nd and 3rd number can be found following the same reasoning as TwoSum The only difference is that the TwoSum problem of LEETCODE has a unique solution However, in ThreeSum, we have multiple duplicate solutions that can be found Most of the OLE errors happened here because you could've ended up with a solution
python - What is the difference between num - Stack Overflow Whereas in first case you are copying list nums to vals and both vals and nums are pointing out to same heap storage location So changes made to vals will automatically affects to nums
what does this statement do? `nums[i] *= nums[i - 1] or 1` A series of expressions connected by or s evaluates to the leftmost expression that has a "truthy" value - which for numeric types, means "nonzero" So this is equal to nums[I - 1] if that isn't zero, otherwise 1
nums[i++]=i; performs differently in java and C? - Stack Overflow In nums[i ++] = nums[i + 1] If the i++ happens before the nums[i + 1], i + 1 will be out of bounds on the last iteration In that case, Java rightly causes an error, and C, because it doesn't do bound-checking, happily trusts the code you wrote That code is logically invalid You're just unlucky that it didn't do anything obvious in your C program to cause suspicion
How can I solve the 3-sum challenge using arrayLists? I'm currently trying to solve the "three sum" challenge (I'm using java by the way) Here is the challenge description: Given an array nums of n integers, are there elements a, b, c in nums such
Understanding return [0,size-1][nums[0] lt;nums[size-1]] in Python That last row is an obscure way of writing an if then else expression [0, size-1] creates a list of two elements nums[0] < nums[size-1] returns either True or False when used as a list index, this True False is implicitly converted into 1 or 0 and by that, either size-1 or 0 is picked from the list A clearer way to write it is: return size - 1 if nums[0] < nums[size - 1] else 0