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Copy path503 Next Greater Element II.py
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503 Next Greater Element II.py
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#!/usr/bin/python3
"""
Given a circular array (the next element of the last element is the first
element of the array), print the Next Greater Number for every element. The Next
Greater Number of a number x is the first greater number to its traversing-order
next in the array, which means you could search circularly to find its next
greater number. If it doesn't exist, output -1 for this number.
Example 1:
Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.
Note: The length of given array won't exceed 10000.
Author: Rajeev Ranjan
"""
from bisect import bisect
class Solution:
def nextGreaterElements(self, nums):
"""
scan the nums from right to left, since next largest number, you can
drop certain information about the A[i:]. Use stack to keep a increasing
numbers. if A[i] > any A[i+1: j] but A[i] < A[j], we can safely drop
the numbers A[i+1:j] since they won't be useful.
:type nums: List[int]
:rtype: List[int]
"""
# initalize the stack
stk = []
for n in nums[::-1]:
while stk and stk[-1] <= n:
stk.pop()
stk.append(n)
ret = []
for n in nums[::-1]:
while stk and stk[-1] <= n:
stk.pop()
ret.append(stk[-1] if stk else -1)
stk.append(n)
return ret[::-1]
def nextGreaterElements_error(self, nums):
"""
brute force O(n^2)
bisect O(n lgn) - error cannot binary search
:type nums: List[int]
:rtype: List[int]
"""
A = nums + nums
print(A)
ret = []
for e in nums:
t = bisect(A, e)
if t == len(A):
ret.append(-1)
else:
ret.append(A[t])
print(ret)
return ret
if __name__ == "__main__":
assert Solution().nextGreaterElements([1,2,1]) == [2, -1, 2]