Example: Input: { 3, 4, -7, 3, 1, 3, 1, -4, -2, -2 } Output: True The subarrays with a sum of 0 are: { 3, 4, -7 } { 4, -7, 3 } etc.

Input: { 3, 4, 2, 6, -2 } Output: False

* Solution Idea:

• Let a series, [some x ints] , 1 , (-2) , 2 , (-1) , [other y ints]. 1,-2, 2,-1 sums to 0. Let, SUM([some x ints]) = x0.
  

x0 + 1 = x1</n x1 + (-2) = x2 x2 + 2 = x3 x3 + (-1) = x4 ———————— x0 + x1 + x2 + x3 = x1 + x2 + x3 + x4 or x0 = x4

• It seems after summing up to ([some x ints] , 1 , (-2) , 2 , (-1)) sum returns to x0 ( where x0 = SUM([some x ints])).
• If no 0-sum subarray exist, x0 = x4 this condition can never be hold. This condition only holds if at least one 0-sum subarray exists.

Time Complexity: O(n) Space Complexity: O(n)

* Python Solution:

from typing import List


def hasZeroSumSubarray(numsL: List[int]):
prevSums = {0}  # if array contains a 0
totalSum = 0
for i in nums:
totalSum += i
if totalSum in prevSums:
return True
prevSums.add (totalSum)
return False


if __name__ == '__main__':
nums = [4, 6, 3, -1, 2, -4, 89, 4, 2, 7, 1, 1]
# nums = [4, 6, 37, 1, 1]
print (hasZeroSumSubarray (nums))