Leetcode 1146: Snapshot Array
You need to implement a data structure that captures snapshots of an array and allows retrieving the value at any given index at a specific snapshot.
Problem
Approach
Steps
Complexity
Input: The input consists of several commands that invoke methods on the SnapshotArray class. The array is initialized with a given length and the following methods are used: 'set', 'snap', and 'get'.
Example: ["SnapshotArray", "set", "snap", "get"]
[[4], [2, 10], [], [2, 0]]
Constraints:
• 1 <= length <= 5 * 10^4
• 0 <= index < length
• 0 <= val <= 10^9
• 0 <= snap_id < (the total number of times we call snap())
• At most 5 * 10^4 calls will be made to set, snap, and get.
Output: For each command, the output is specified in the order of the commands: 'null' for methods that do not return anything and the result of 'snap' and 'get' for the respective method calls.
Example: [null, null, 0, 10]
Constraints:
• The output corresponds to the results of the respective methods called in the input.
Goal: Implement the SnapshotArray class with methods set, snap, and get to manipulate an array and capture its state over time.
Steps:
• Initialize the array with zeros.
• Implement set to store the current value at the specified index for the current snapshot.
• Implement snap to take a snapshot and return the current snap_id.
• Implement get to retrieve the value at the specified index at the time of the given snap_id.
Goal: The array length will be at least 1 and at most 50,000. The operations on set, snap, and get are limited to a maximum of 50,000 calls.
Steps:
• 1 <= length <= 5 * 10^4
• 0 <= index < length
• 0 <= val <= 10^9
• 0 <= snap_id < (the total number of times we call snap())
Assumptions:
• All elements in the array are initially set to 0.
• Each snap_id is unique and increments each time snap() is called.
• Input: ["SnapshotArray", "set", "snap", "set", "get"]
[[3], [0, 5], [], [0, 6], [0, 0]]
• Explanation: This example initializes a SnapshotArray with length 3, performs a set operation, takes a snapshot, performs another set operation, and finally retrieves the value of index 0 at snap_id 0.
Approach: The solution involves using a map to track the history of changes for each index in the array, while maintaining the current snapshot ID.
Observations:
• The array needs to store historical changes for each index to support the snapshot functionality.
• Using a map of maps can help track changes at each snap_id.
• To optimize get operations, it's essential to maintain a way to efficiently fetch the value at any given snap_id for an index.
Steps:
• Use a map of maps to store the values at each snap_id for each index.
• Each time set is called, update the value for the current snap_id.
• For snap(), increment the snap_id counter and return the previous value.
• For get(), find the most recent value at the specified snap_id using binary search.
Empty Inputs:
• Handle the case where no set or snap operations are called.
Large Inputs:
• Efficiently handle cases where the length of the array is large, up to 50,000.
Special Values:
• Handle large values of val up to 10^9.
Constraints:
• The solution must ensure that all operations are performed within the provided time limits.
unordered_map<int, map<int, int>> snaps;
int snap_id = 0;
SnapshotArray(int length) {
for(int i = 0; i < length; i++) {
map<int, int> mp;
mp[0] = 0;
snaps[i] = mp;
}
}
void set(int index, int val) {
snaps[index][snap_id] = val;
}
int snap() {
snap_id++;
return snap_id - 1;
}
int get(int index, int snid) {
auto it = snaps[index].upper_bound(snid);
it--;
return it->second;
}
};
/**
* Your SnapshotArray object will be instantiated and called as such:
* SnapshotArray* obj = new SnapshotArray(length);
* obj->set(index,val);
* int param_2 = obj->snap();
* int param_3 = obj->get(index,snap_id);
1 : Variable Initialization
unordered_map<int, map<int, int>> snaps;
A map to store the snapshots, with the key being the index of the array and the value being another map storing the snapshot ids and their corresponding values.
2 : Variable Initialization
int snap_id = 0;
A counter to track the number of snapshots taken.
3 : Constructor
SnapshotArray(int length) {
Constructor definition to initialize a SnapshotArray object.
4 : Looping & Initialization
for(int i = 0; i < length; i++) {
Loop through each index in the array to initialize the map for each.
5 : Map Initialization
map<int, int> mp;
Initialize a map to store the snapshot values for the current index.
6 : Default Value
mp[0] = 0;
Set the default snapshot value to 0 for the initial snapshot.
7 : Storing Snapshots
snaps[i] = mp;
Store the initialized map of snapshots for the current index in the main snapshot map.
8 : Method Definition
void set(int index, int val) {
Define the `set` method, which allows setting the value at a specific index.
9 : Snapshot Storage
snaps[index][snap_id] = val;
Store the value at the given index under the current snapshot id.
10 : Method Definition
int snap() {
Define the `snap` method, which increments the snapshot id and returns the previous one.
11 : Snap Id Increment
snap_id++;
Increment the snapshot id counter.
12 : Return Snapshot Id
return snap_id - 1;
Return the snapshot id before the increment.
13 : Method Definition
int get(int index, int snid) {
Define the `get` method, which returns the value for a given index at a specific snapshot.
14 : Finding Snapshot
auto it = snaps[index].upper_bound(snid);
Find the snapshot at the given snapshot id using upper bound.
15 : Decrement Iterator
it--;
Decrement the iterator to get the correct snapshot value.
16 : Return Value
return it->second;
Return the value corresponding to the found snapshot.
Best Case: O(1) for snap and set operations.
Average Case: O(log k) for get operations, where k is the number of snapshots.
Worst Case: O(log k) for get operations, where k is the number of snapshots.
Description: The time complexity is optimal due to the use of binary search on maps.
Best Case: O(n) where no snapshots are taken.
Worst Case: O(n * k) where n is the number of array elements and k is the number of snapshots.
Description: The space complexity depends on the number of elements in the array and the number of snapshots taken.
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