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Java-based LeetCode algorithm problem solutions, regularly updated
3245\. Alternating Groups III
Hard
There are some red and blue tiles arranged circularly. You are given an array of integers `colors` and a 2D integers array `queries`.
The color of tile `i` is represented by `colors[i]`:
* `colors[i] == 0` means that tile `i` is **red**.
* `colors[i] == 1` means that tile `i` is **blue**.
An **alternating** group is a contiguous subset of tiles in the circle with **alternating** colors (each tile in the group except the first and last one has a different color from its **adjacent** tiles in the group).
You have to process queries of two types:
* queries[i] = [1, sizei]
, determine the count of **alternating** groups with size sizei
.
* queries[i] = [2, indexi, colori]
, change colors[indexi]
to colori
.
Return an array `answer` containing the results of the queries of the first type _in order_.
**Note** that since `colors` represents a **circle**, the **first** and the **last** tiles are considered to be next to each other.
**Example 1:**
**Input:** colors = [0,1,1,0,1], queries = [[2,1,0],[1,4]]
**Output:** [2]
**Explanation:**
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First query:
Change `colors[1]` to 0.
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Second query:
Count of the alternating groups with size 4:
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**Example 2:**
**Input:** colors = [0,0,1,0,1,1], queries = [[1,3],[2,3,0],[1,5]]
**Output:** [2,0]
**Explanation:**
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First query:
Count of the alternating groups with size 3:
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Second query: `colors` will not change.
Third query: There is no alternating group with size 5.
**Constraints:**
* 4 <= colors.length <= 5 * 104
* `0 <= colors[i] <= 1`
* 1 <= queries.length <= 5 * 104
* `queries[i][0] == 1` or `queries[i][0] == 2`
* For all `i` that:
* `queries[i][0] == 1`: `queries[i].length == 2`, `3 <= queries[i][1] <= colors.length - 1`
* `queries[i][0] == 2`: `queries[i].length == 3`, `0 <= queries[i][1] <= colors.length - 1`, `0 <= queries[i][2] <= 1`
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