LeetCode in Kotlin

2569. Handling Sum Queries After Update

Hard

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:

1. For a query of type 1, queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed.
2. For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.
3. For a query of type 3, queries[i] = [3, 0, 0]. Find the sum of the elements in nums2.

Return an array containing all the answers to the third type queries.

Example 1:

Input: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]

Output: [3]

Explanation: After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.

Example 2:

Input: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]

Output: [5]

Explanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.

Constraints:

• 1 <= nums1.length,nums2.length <= 105
• nums1.length = nums2.length
• 1 <= queries.length <= 105
• queries[i].length = 3
• 0 <= l <= r <= nums1.length - 1
• 0 <= p <= 106
• 0 <= nums1[i] <= 1
• 0 <= nums2[i] <= 109

Solution

import java.util.BitSet

class Solution {
    fun handleQuery(nums1: IntArray, nums2: IntArray, queries: Array<IntArray>): LongArray {
        val results: MutableList<Long> = ArrayList()
        val n = nums1.size
        val bs = BitSet(n)
        var sum: Long = 0
        for (i in 0 until n) {
            sum += 1L * nums2[i]
            if (nums1[i] == 1) {
                bs.set(i)
            }
        }
        for (query in queries) {
            when (query[0]) {
                1 -> bs.flip(query[1], query[2] + 1)
                2 -> sum += 1L * query[1] * bs.cardinality()
                else -> results.add(sum)
            }
        }
        val ans = LongArray(results.size)
        for (i in ans.indices) {
            ans[i] = results[i]
        }
        return ans
    }
}

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