1146. Snapshot Array
Description
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 <= 5 * 104
0 <= index < length
0 <= val <= 109
0 <= snap_id <
(the total number of times we callsnap()
)- At most
5 * 104
calls will be made toset
,snap
, andget
.
Solutions
Solution 1: Array + Binary Search
We maintain an array of length length
. Each element in the array is a list, which is used to store the value set each time and the corresponding snapshot ID.
When the set
method is called, we add the value and snapshot ID to the list at the corresponding index. The time complexity is $O(1)$.
When the snap
method is called, we first increment the snapshot ID, then return the snapshot ID minus one. The time complexity is $O(1)$.
When the get
method is called, we use binary search to find the first snapshot ID greater than snap_id
at the corresponding position, and then return the previous value. If it cannot be found, return 0. The time complexity is $O(\log n)$.
The space complexity is $O(n)$.
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