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Divide 2 integers (leetcode 29) - recursion issue


Divide 2 integers (leetcode 29) - recursion issue

By : Dan
Date : January 11 2021, 03:34 PM
I hope this helps you . I'm having issues when i need to do a count (or aggregation) in recursion problems, for eg. in this problem where i'm trying to divide 2 integers w/o using multiplication, division and mod operator. , One possible problem is related to this logic:
code :
if(dividend > origDivisor):
    self.helper(dividend, origDivisor, origDivisor, count)
else:
    return count
if(dividend > origDivisor):
    return self.helper(dividend, origDivisor, origDivisor, count)

return count
def divide(dividend, divisor):

    quotient = divide_recursive(dividend, divisor)

    return quotient

def divide_recursive(dividend, divisor, count=0):

    if dividend >= divisor:
        return divide_recursive(dividend - divisor, divisor, count + 1)

    return count

print(divide(125, 5))


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Why is my solution on LeetCode not working (recursion, python)?

Why is my solution on LeetCode not working (recursion, python)?


By : lonelysheep
Date : March 29 2020, 07:55 AM
it fixes the issue You've chosen a poor algorithm. It's not that it's recursive, just horribly inefficient. Here's a more efficient recursive approach:
code :
def climbStairs(self, n, res=1, nxt=1):
    if n == 0:
        return res
    return self.climbStairs(n - 1, nxt, res + nxt)
def climbStairs(self, n):  # doesn't use 'self', could be class method
    if 0 <= n <= 1:
        return 1

    a = b = 1

    for _ in range(n - 1):
        a, b = b, a + b

    return b
divide two integers [leetcode] shift error while using unsigned short

divide two integers [leetcode] shift error while using unsigned short


By : user2635606
Date : March 29 2020, 07:55 AM
this one helps. as KamilCuk mentioned, I don't know why he deleted the comment.
left shift signed int is undefined behavior. and shifting unsigned short more that 16 is also undefined, as unsigned short is 16 bits. So I have to dealing these edges.
Recursion implementation of LeetCode Problem

Recursion implementation of LeetCode Problem


By : user3093475
Date : January 02 2021, 04:59 AM
around this issue Here's one answer accepted by LeetCode. I hope the simple logic makes sense:
JavaScript code:
code :
function f(A, i=2, total=0, prevSeqTotal=0){
  if (i >= A.length)
    return total

  // The number of sequences ending here is 1 (length 3)
  // plus the number ending at the previous
  // element since those all just got extended.
  if (A[i] - A[i-1] == A[i-1] - A[i-2])
    return f(A, i + 1, total + 1 + prevSeqTotal, 1 + prevSeqTotal)

  return f(A, i + 1, total, 0)
}

console.log(f([3, 1, -1, -3]))
Why is FloodFill algorithm exceeding Leetcode's maximum recursion limit?

Why is FloodFill algorithm exceeding Leetcode's maximum recursion limit?


By : gip
Date : March 29 2020, 07:55 AM
Hope this helps I believe the code is fine for the case where newColor != og. But my guess is that you are seeing your error because in the case of newColor == og, the stopping condition
code :
if grid[sr][sc]!= og:
        return
From LeetCode Given an array of integers, return indices of the two numbers such that they add up to a specific target

From LeetCode Given an array of integers, return indices of the two numbers such that they add up to a specific target


By : Brandon
Date : March 29 2020, 07:55 AM
hop of those help? This is an exercise on LeetCode. I get an except of , I believe this would work:
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