Description
https://leetcode.com/problems/snapshot-array/
Implement a SnapshotArray that supports the following interface:
SnapshotArray(int length)
initializes an array-like data structure with the given length. Initially, each element equals 0.void set(index, val)
sets the element at the givenindex
to be equal toval
.int snap()
takes a snapshot of the array and returns thesnap_id
: the total number of times we calledsnap()
minus1
.int get(index, snap_id)
returns the value at the givenindex
, at the time we took the snapshot with the givensnap_id
Example 1:
Input: ["SnapshotArray","set","snap","set","get"] [[3],[0,5],[],[0,6],[0,0]] Output: [null,null,0,null,5] Explanation: SnapshotArray snapshotArr = new SnapshotArray(3); // set the length to be 3 snapshotArr.set(0,5); // Set array[0] = 5 snapshotArr.snap(); // Take a snapshot, return snap_id = 0 snapshotArr.set(0,6); snapshotArr.get(0,0); // Get the value of array[0] with snap_id = 0, return 5
Constraints:
1 <= length <= 50000
- At most
50000
calls will be made toset
,snap
, andget
. 0 <= index < length
0 <= snap_id <
(the total number of times we callsnap()
)0 <= val <= 10^9
Explanation
Use a hashmap to store the current array arrangements and use a list to store snap shots.
Python Solution
class SnapshotArray:
def __init__(self, length: int):
self.snap_shots = []
self.current = {}
def set(self, index: int, val: int) -> None:
self.current[index] = val
def snap(self) -> int:
self.snap_shots.append(dict(self.current))
return len(self.snap_shots) - 1
def get(self, index: int, snap_id: int) -> int:
snap_shot = self.snap_shots[snap_id]
if index in snap_shot:
return snap_shot[index]
return 0
# Your SnapshotArray object will be instantiated and called as such:
# obj = SnapshotArray(length)
# obj.set(index,val)
# param_2 = obj.snap()
# param_3 = obj.get(index,snap_id)
- Time Complexity: O(1).
- Space Complexity: O(N).
where Q is the length of queries and N is the length of colors.
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