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 givenindexto 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 5Constraints:
1 <= length <= 50000At most
50000calls will be made toset,snap, andget.0 <= index < length0 <= snap_id <(the total number of times we callsnap())0 <= val <= 10^9
// We are only adding data for the values that change over a snap
// TreeMap -> floorKey(id) & ceilingKey(id)
class SnapshotArray {
    // Arrays of maps, where map are between snap_id -> element
    TreeMap<Integer, Integer>[] array;
    int snap_id;
    // O(NlogN)
    public SnapshotArray(int length) {
        this.snap_id = 0;
        array = new TreeMap[length];
        for (int i = 0; i < length; i++) {
            array[i] = new TreeMap<>();
            array[i].put(0, 0);
        }
    }
    // O(logN)
    public void set(int index, int val) {
        array[index].put(this.snap_id, val);
    }
    // O(1)
    public int snap() {
        int snapped = snap_id;
        this.snap_id++;
        return snapped;
    }
    // O(logN)
    public int get(int index, int id) {
        // We will get that value from the snap_id <= id
        // Which ever is available
        Integer lastSnapID = array[index].floorKey(id);
        return array[index].get(lastSnapID);
    }
}PreviousFind Two Non-overlapping Sub-arrays Each With Target SumNextCount pairs in array whose sum is divisible by K
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