Description
You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.
Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.
Example 1:
Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
Example 2:
Input: nums = [5,6,7,8,9], x = 4
Output: -1
Example 3:
Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4
1 <= x <= 10^9
Solution
Queue
Every time try with leftest and rightest items, use BFS.
Time complexity:
o
(
2
n
)
o(2^n)
o(2n)
Space complexity:
o
(
n
)
o(n)
o(n)
Code
Queue
class Solution:
def minOperations(self, nums: List[int], x: int) -> int:
queue = collections.deque([(0, len(nums) - 1, x, 0)])
visited = set()
while queue:
left, right, r, res = queue.popleft()
if (left, right, r) in visited:
continue
visited.add((left, right, r))
if r == 0:
return res
elif r < 0:
continue
if left > right:
continue
queue.append((left + 1, right, r - nums[left], res + 1))
queue.append((left, right - 1, r - nums[right], res + 1))
return -1
