# 11.06.2023 [1146. Snapshot Array]

11.06.2023

1146. Snapshot Array medium
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https://t.me/leetcode_daily_unstoppable/242

Problem TLDR

Implement an array where all elements can be saved into a `snapshot’s.

Intuition

Consider example:

// 0 1 2 3 4 5 6 <-- snapshot id
// 1 . . 2 . . 3 <-- value

When get()(2) called, 1 must be returned. So, we need to keep all the previous values. We can put them into a list combining with the current snapshot id: (1,0), (2, 3), (3, 6). Then we can do a Binary Search and find the highest_id >= id.

Approach

For more robust Binary Search:

• use inclusive lo, hi

• check last condition lo == hi

• always write the result ind = mid

Complexity

• Time complexity:
  O(log(n)) for get

• Space complexity:
  O(n)

Code

class SnapshotArray(length: Int) {
    // 0 1 2 3 4 5 6
    // 1 . . 2 . . 3
    val arr = Array<MutableList<Pair<Int, Int>>>(length) { mutableListOf() }
    var currId = 0

    fun set(index: Int, v: Int) {
        val idVs = arr[index]
        if (idVs.isEmpty() || idVs.last().first != currId) idVs += currId to v
        else idVs[idVs.lastIndex] = currId to v
    }

    fun snap(): Int = currId.also { currId++ }

    fun get(index: Int, id: Int): Int {
        var lo = 0
        var hi = arr[index].lastIndex
        var ind = -1
        while (lo <= hi) {
            val mid = lo + (hi - lo) / 2
            if (arr[index][mid].first <= id) {
                ind = mid
                lo = mid + 1
            } else hi = mid - 1
        }
        return if (ind == -1) 0 else arr[index][ind].second
    }

}