Added tasks 3349-3352

src/main/kotlin/g3301_3400/s3349_adjacent_increasing_subarrays_detection_i/Solution.kt

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+package g3301_3400.s3349_adjacent_increasing_subarrays_detection_i
+
+// #Easy #Array #2024_11_15_Time_179_ms_(97.92%)_Space_37.3_MB_(91.67%)
+
+class Solution {
+    fun hasIncreasingSubarrays(nums: List<Int>, k: Int): Boolean {
+        val l = nums.size
+        if (l < k * 2) {
+            return false
+        }
+        for (i in 0.rangeUntil(l - 2 * k + 1)) {
+            if (check(i, k, nums) && check(i + k, k, nums)) {
+                return true
+            }
+        }
+        return false
+    }
+
+    private fun check(p: Int, k: Int, nums: List<Int>): Boolean {
+        for (i in p.rangeUntil(p + k - 1)) {
+            if (nums[i] >= nums[i + 1]) {
+                return false
+            }
+        }
+        return true
+    }
+}

src/main/kotlin/g3301_3400/s3349_adjacent_increasing_subarrays_detection_i/readme.md

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+3349\. Adjacent Increasing Subarrays Detection I
+
+Easy
+
+Given an array `nums` of `n` integers and an integer `k`, determine whether there exist **two** **adjacent** subarrays of length `k` such that both subarrays are **strictly** **increasing**. Specifically, check if there are **two** subarrays starting at indices `a` and `b` (`a < b`), where:
+
+*   Both subarrays `nums[a..a + k - 1]` and `nums[b..b + k - 1]` are **strictly increasing**.
+*   The subarrays must be **adjacent**, meaning `b = a + k`.
+
+Return `true` if it is _possible_ to find **two** such subarrays, and `false` otherwise.
+
+**Example 1:**
+
+**Input:** nums = [2,5,7,8,9,2,3,4,3,1], k = 3
+
+**Output:** true
+
+**Explanation:**
+
+*   The subarray starting at index `2` is `[7, 8, 9]`, which is strictly increasing.
+*   The subarray starting at index `5` is `[2, 3, 4]`, which is also strictly increasing.
+*   These two subarrays are adjacent, so the result is `true`.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4,4,4,4,5,6,7], k = 5
+
+**Output:** false
+
+**Constraints:**
+
+*   `2 <= nums.length <= 100`
+*   `1 < 2 * k <= nums.length`
+*   `-1000 <= nums[i] <= 1000`

src/main/kotlin/g3301_3400/s3350_adjacent_increasing_subarrays_detection_ii/Solution.kt

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+package g3301_3400.s3350_adjacent_increasing_subarrays_detection_ii
+
+// #Medium #Array #Binary_Search #2024_11_15_Time_947_ms_(48.57%)_Space_87.4_MB_(51.43%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun maxIncreasingSubarrays(nums: List<Int>): Int {
+        val n = nums.size
+        val a = IntArray(n)
+        for (i in 0.rangeUntil(n)) {
+            a[i] = nums[i]
+        }
+        var ans = 1
+        var previousLen = Int.Companion.MAX_VALUE
+        var i = 0
+        while (i < n) {
+            var j = i + 1
+            while (j < n && a[j - 1] < a[j]) {
+                ++j
+            }
+            val len = j - i
+            ans = max(ans, (len / 2))
+            if (previousLen != Int.Companion.MAX_VALUE) {
+                ans = max(ans, min(previousLen, len))
+            }
+            previousLen = len
+            i = j
+        }
+        return ans
+    }
+}

src/main/kotlin/g3301_3400/s3350_adjacent_increasing_subarrays_detection_ii/readme.md

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+3350\. Adjacent Increasing Subarrays Detection II
+
+Medium
+
+Given an array `nums` of `n` integers, your task is to find the **maximum** value of `k` for which there exist **two** adjacent subarrays of length `k` each, such that both subarrays are **strictly** **increasing**. Specifically, check if there are **two** subarrays of length `k` starting at indices `a` and `b` (`a < b`), where:
+
+*   Both subarrays `nums[a..a + k - 1]` and `nums[b..b + k - 1]` are **strictly increasing**.
+*   The subarrays must be **adjacent**, meaning `b = a + k`.
+
+Return the **maximum** _possible_ value of `k`.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [2,5,7,8,9,2,3,4,3,1]
+
+**Output:** 3
+
+**Explanation:**
+
+*   The subarray starting at index 2 is `[7, 8, 9]`, which is strictly increasing.
+*   The subarray starting at index 5 is `[2, 3, 4]`, which is also strictly increasing.
+*   These two subarrays are adjacent, and 3 is the **maximum** possible value of `k` for which two such adjacent strictly increasing subarrays exist.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4,4,4,4,5,6,7]
+
+**Output:** 2
+
+**Explanation:**
+
+*   The subarray starting at index 0 is `[1, 2]`, which is strictly increasing.
+*   The subarray starting at index 2 is `[3, 4]`, which is also strictly increasing.
+*   These two subarrays are adjacent, and 2 is the **maximum** possible value of `k` for which two such adjacent strictly increasing subarrays exist.
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 2 * 10<sup>5</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3301_3400/s3351_sum_of_good_subsequences/Solution.kt

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+package g3301_3400.s3351_sum_of_good_subsequences
+
+// #Hard #Array #Hash_Table #Dynamic_Programming
+// #2024_11_15_Time_16_ms_(100.00%)_Space_61.2_MB_(80.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun sumOfGoodSubsequences(nums: IntArray): Int {
+        var max = 0
+        for (x in nums) {
+            max = max(x, max)
+        }
+        val count = LongArray(max + 3)
+        val total = LongArray(max + 3)
+        val mod = (1e9 + 7).toInt().toLong()
+        var res: Long = 0
+        for (a in nums) {
+            count[a + 1] = (count[a] + count[a + 1] + count[a + 2] + 1) % mod
+            val cur = total[a] + total[a + 2] + a * (count[a] + count[a + 2] + 1)
+            total[a + 1] = (total[a + 1] + cur) % mod
+            res = (res + cur) % mod
+        }
+        return res.toInt()
+    }
+}

src/main/kotlin/g3301_3400/s3351_sum_of_good_subsequences/readme.md

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+3351\. Sum of Good Subsequences
+
+Hard
+
+You are given an integer array `nums`. A **good** subsequence is defined as a subsequence of `nums` where the absolute difference between any **two** consecutive elements in the subsequence is **exactly** 1.
+
+Return the **sum** of all _possible_ **good subsequences** of `nums`.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Note** that a subsequence of size 1 is considered good by definition.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1]
+
+**Output:** 14
+
+**Explanation:**
+
+*   Good subsequences are: `[1]`, `[2]`, `[1]`, `[1,2]`, `[2,1]`, `[1,2,1]`.
+*   The sum of elements in these subsequences is 14.
+
+**Example 2:**
+
+**Input:** nums = [3,4,5]
+
+**Output:** 40
+
+**Explanation:**
+
+*   Good subsequences are: `[3]`, `[4]`, `[5]`, `[3,4]`, `[4,5]`, `[3,4,5]`.
+*   The sum of elements in these subsequences is 40.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums[i] <= 10<sup>5</sup></code>

src/main/kotlin/g3301_3400/s3352_count_k_reducible_numbers_less_than_n/Solution.kt

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+package g3301_3400.s3352_count_k_reducible_numbers_less_than_n
+
+// #Hard #String #Dynamic_Programming #Math #Combinatorics
+// #2024_11_15_Time_170_ms_(100.00%)_Space_34.9_MB_(100.00%)
+
+class Solution {
+    fun countKReducibleNumbers(s: String, k: Int): Int {
+        val n = s.length
+        val reducible = IntArray(n + 1)
+        for (i in 2.rangeUntil(reducible.size)) {
+            reducible[i] = 1 + reducible[Integer.bitCount(i)]
+        }
+        val dp = LongArray(n + 1)
+        var curr = 0
+        for (i in 0.rangeUntil(n)) {
+            for (j in i - 1 downTo 0) {
+                dp[j + 1] += dp[j]
+                dp[j + 1] %= MOD.toLong()
+            }
+            if (s[i] == '1') {
+                dp[curr]++
+                dp[curr] %= MOD.toLong()
+                curr++
+            }
+        }
+        var result: Long = 0
+        for (i in 1..s.length) {
+            if (reducible[i] < k) {
+                result += dp[i]
+                result %= MOD.toLong()
+            }
+        }
+        return (result % MOD).toInt()
+    }
+
+    companion object {
+        private val MOD = (1e9 + 7).toInt()
+    }
+}

src/main/kotlin/g3301_3400/s3352_count_k_reducible_numbers_less_than_n/readme.md

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+3352\. Count K-Reducible Numbers Less Than N
+
+Hard
+
+You are given a **binary** string `s` representing a number `n` in its binary form.
+
+You are also given an integer `k`.
+
+An integer `x` is called **k-reducible** if performing the following operation **at most** `k` times reduces it to 1:
+
+*   Replace `x` with the **count** of set bits in its binary representation.
+
+For example, the binary representation of 6 is `"110"`. Applying the operation once reduces it to 2 (since `"110"` has two set bits). Applying the operation again to 2 (binary `"10"`) reduces it to 1 (since `"10"` has one set bit).
+
+Return an integer denoting the number of positive integers **less** than `n` that are **k-reducible**.
+
+Since the answer may be too large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** s = "111", k = 1
+
+**Output:** 3
+
+**Explanation:**
+
+`n = 7`. The 1-reducible integers less than 7 are 1, 2, and 4.
+
+**Example 2:**
+
+**Input:** s = "1000", k = 2
+
+**Output:** 6
+
+**Explanation:**
+
+`n = 8`. The 2-reducible integers less than 8 are 1, 2, 3, 4, 5, and 6.
+
+**Example 3:**
+
+**Input:** s = "1", k = 3
+
+**Output:** 0
+
+**Explanation:**
+
+There are no positive integers less than `n = 1`, so the answer is 0.
+
+**Constraints:**
+
+*   `1 <= s.length <= 800`
+*   `s` has no leading zeros.
+*   `s` consists only of the characters `'0'` and `'1'`.
+*   `1 <= k <= 5`

src/test/kotlin/g3301_3400/s3349_adjacent_increasing_subarrays_detection_i/SolutionTest.kt

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+package g3301_3400.s3349_adjacent_increasing_subarrays_detection_i
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun hasIncreasingSubarrays() {
+        assertThat<Boolean>(
+            Solution().hasIncreasingSubarrays(listOf<Int>(2, 5, 7, 8, 9, 2, 3, 4, 3, 1), 3),
+            equalTo<Boolean>(true)
+        )
+    }
+
+    @Test
+    fun hasIncreasingSubarrays2() {
+        assertThat<Boolean>(
+            Solution().hasIncreasingSubarrays(listOf<Int>(1, 2, 3, 4, 4, 4, 4, 5, 6, 7), 5),
+            equalTo<Boolean>(false)
+        )
+    }
+}

src/test/kotlin/g3301_3400/s3350_adjacent_increasing_subarrays_detection_ii/SolutionTest.kt

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+package g3301_3400.s3350_adjacent_increasing_subarrays_detection_ii
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun maxIncreasingSubarrays() {
+        assertThat<Int>(
+            Solution().maxIncreasingSubarrays(listOf<Int>(2, 5, 7, 8, 9, 2, 3, 4, 3, 1)),
+            equalTo<Int>(3)
+        )
+    }
+
+    @Test
+    fun maxIncreasingSubarrays2() {
+        assertThat<Int>(
+            Solution().maxIncreasingSubarrays(listOf<Int>(1, 2, 3, 4, 4, 4, 4, 5, 6, 7)),
+            equalTo<Int>(2)
+        )
+    }
+}

src/test/kotlin/g3301_3400/s3351_sum_of_good_subsequences/SolutionTest.kt

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+package g3301_3400.s3351_sum_of_good_subsequences
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun sumOfGoodSubsequences() {
+        assertThat<Int>(
+            Solution().sumOfGoodSubsequences(intArrayOf(1, 2, 1)),
+            equalTo<Int>(14)
+        )
+    }
+
+    @Test
+    fun sumOfGoodSubsequences2() {
+        assertThat<Int>(
+            Solution().sumOfGoodSubsequences(intArrayOf(3, 4, 5)),
+            equalTo<Int>(40)
+        )
+    }
+}

src/test/kotlin/g3301_3400/s3352_count_k_reducible_numbers_less_than_n/SolutionTest.kt

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+package g3301_3400.s3352_count_k_reducible_numbers_less_than_n
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun countKReducibleNumbers() {
+        assertThat<Int>(Solution().countKReducibleNumbers("111", 1), equalTo<Int>(3))
+    }
+
+    @Test
+    fun countKReducibleNumbers2() {
+        assertThat<Int>(Solution().countKReducibleNumbers("1000", 2), equalTo<Int>(6))
+    }
+
+    @Test
+    fun countKReducibleNumbers3() {
+        assertThat<Int>(Solution().countKReducibleNumbers("1", 3), equalTo<Int>(0))
+    }
+}