Added tasks 3184-3213

Author: javadev

README.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3184\. Count Pairs That Form a Complete Day I
+
+Easy
+
+Given an integer array `hours` representing times in **hours**, return an integer denoting the number of pairs `i`, `j` where `i < j` and `hours[i] + hours[j]` forms a **complete day**.
+
+A **complete day** is defined as a time duration that is an **exact** **multiple** of 24 hours.
+
+For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.
+
+**Example 1:**
+
+**Input:** hours = [12,12,30,24,24]
+
+**Output:** 2
+
+**Explanation:**
+
+The pairs of indices that form a complete day are `(0, 1)` and `(3, 4)`.
+
+**Example 2:**
+
+**Input:** hours = [72,48,24,3]
+
+**Output:** 3
+
+**Explanation:**
+
+The pairs of indices that form a complete day are `(0, 1)`, `(0, 2)`, and `(1, 2)`.
+
+**Constraints:**
+
+*   `1 <= hours.length <= 100`
+*   <code>1 <= hours[i] <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+class Solution {
+    fun countCompleteDayPairs(hours: IntArray): Int {
+        val modular = IntArray(26)
+        var ans = 0
+        for (hour in hours) {
+            val mod = hour % 24
+            ans += modular[24 - mod]
+            if (mod == 0) {
+                modular[24]++
+            } else {
+                modular[mod]++
+            }
+        }
+        return ans
+    }
+}
+```

src/main/kotlin/g3101_3200/s3184_count_pairs_that_form_a_complete_day_i/readme.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3185\. Count Pairs That Form a Complete Day II
+
+Medium
+
+Given an integer array `hours` representing times in **hours**, return an integer denoting the number of pairs `i`, `j` where `i < j` and `hours[i] + hours[j]` forms a **complete day**.
+
+A **complete day** is defined as a time duration that is an **exact** **multiple** of 24 hours.
+
+For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.
+
+**Example 1:**
+
+**Input:** hours = [12,12,30,24,24]
+
+**Output:** 2
+
+**Explanation:** The pairs of indices that form a complete day are `(0, 1)` and `(3, 4)`.
+
+**Example 2:**
+
+**Input:** hours = [72,48,24,3]
+
+**Output:** 3
+
+**Explanation:** The pairs of indices that form a complete day are `(0, 1)`, `(0, 2)`, and `(1, 2)`.
+
+**Constraints:**
+
+*   <code>1 <= hours.length <= 5 * 10<sup>5</sup></code>
+*   <code>1 <= hours[i] <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+class Solution {
+    fun countCompleteDayPairs(hours: IntArray): Long {
+        val hour = LongArray(24)
+        for (j in hours) {
+            hour[j % 24]++
+        }
+        var counter = hour[0] * (hour[0] - 1) / 2
+        counter += hour[12] * (hour[12] - 1) / 2
+        for (i in 1..11) {
+            counter += hour[i] * hour[24 - i]
+        }
+        return counter
+    }
+}
+```

src/main/kotlin/g3101_3200/s3185_count_pairs_that_form_a_complete_day_ii/readme.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3186\. Maximum Total Damage With Spell Casting
+
+Medium
+
+A magician has various spells.
+
+You are given an array `power`, where each element represents the damage of a spell. Multiple spells can have the same damage value.
+
+It is a known fact that if a magician decides to cast a spell with a damage of `power[i]`, they **cannot** cast any spell with a damage of `power[i] - 2`, `power[i] - 1`, `power[i] + 1`, or `power[i] + 2`.
+
+Each spell can be cast **only once**.
+
+Return the **maximum** possible _total damage_ that a magician can cast.
+
+**Example 1:**
+
+**Input:** power = [1,1,3,4]
+
+**Output:** 6
+
+**Explanation:**
+
+The maximum possible damage of 6 is produced by casting spells 0, 1, 3 with damage 1, 1, 4.
+
+**Example 2:**
+
+**Input:** power = [7,1,6,6]
+
+**Output:** 13
+
+**Explanation:**
+
+The maximum possible damage of 13 is produced by casting spells 1, 2, 3 with damage 1, 6, 6.
+
+**Constraints:**
+
+*   <code>1 <= power.length <= 10<sup>5</sup></code>
+*   <code>1 <= power[i] <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun maximumTotalDamage(power: IntArray): Long {
+        var maxPower = 0
+        for (p in power) {
+            if (p > maxPower) {
+                maxPower = p
+            }
+        }
+        return if ((maxPower <= 1000000)) smallPower(power, maxPower) else bigPower(power)
+    }
+
+    private fun smallPower(power: IntArray, maxPower: Int): Long {
+        val counts = IntArray(maxPower + 6)
+        for (p in power) {
+            counts[p]++
+        }
+        val dp = LongArray(maxPower + 6)
+        dp[1] = counts[1].toLong()
+        dp[2] = max((counts[2] * 2L).toDouble(), dp[1].toDouble()).toLong()
+        for (i in 3..maxPower) {
+            dp[i] = max((counts[i] * i + dp[i - 3]).toDouble(), max(dp[i - 1].toDouble(), dp[i - 2].toDouble()))
+                .toLong()
+        }
+        return dp[maxPower]
+    }
+
+    private fun bigPower(power: IntArray): Long {
+        power.sort()
+        val n = power.size
+        val prevs = LongArray(4)
+        var curPower = power[0]
+        var count = 1
+        var result: Long = 0
+        for (i in 1..n) {
+            val p = if ((i == n)) 1000000009 else power[i]
+            if (p == curPower) {
+                count++
+            } else {
+                val curVal = max(
+                    (curPower.toLong() * count + prevs[3]).toDouble(),
+                    max(prevs[1].toDouble(), prevs[2].toDouble())
+                )
+                    .toLong()
+                val diff = min((p - curPower).toDouble(), (prevs.size - 1).toDouble()).toInt()
+                val nextCurVal =
+                    if ((diff == 1)) 0 else max(prevs[3].toDouble(), max(curVal.toDouble(), prevs[2].toDouble()))
+                        .toLong()
+                // Shift the values in prevs[].
+                var k = prevs.size - 1
+                if (diff < prevs.size - 1) {
+                    while (k > diff) {
+                        prevs[k] = prevs[k-- - diff]
+                    }
+                    prevs[k--] = curVal
+                }
+                while (k > 0) {
+                    prevs[k--] = nextCurVal
+                }
+                curPower = p
+                count = 1
+            }
+        }
+        for (v in prevs) {
+            if (v > result) {
+                result = v
+            }
+        }
+        return result
+    }
+}
+```

src/main/kotlin/g3101_3200/s3186_maximum_total_damage_with_spell_casting/readme.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3187\. Peaks in Array
+
+Hard
+
+A **peak** in an array `arr` is an element that is **greater** than its previous and next element in `arr`.
+
+You are given an integer array `nums` and a 2D integer array `queries`.
+
+You have to process queries of two types:
+
+*   <code>queries[i] = [1, l<sub>i</sub>, r<sub>i</sub>]</code>, determine the count of **peak** elements in the subarray <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code>.
+*   <code>queries[i] = [2, index<sub>i</sub>, val<sub>i</sub>]</code>, change <code>nums[index<sub>i</sub>]</code> to <code>val<sub>i</sub></code>.
+
+Return an array `answer` containing the results of the queries of the first type in order.
+
+**Notes:**
+
+*   The **first** and the **last** element of an array or a subarray **cannot** be a peak.
+
+**Example 1:**
+
+**Input:** nums = [3,1,4,2,5], queries = \[\[2,3,4],[1,0,4]]
+
+**Output:** [0]
+
+**Explanation:**
+
+First query: We change `nums[3]` to 4 and `nums` becomes `[3,1,4,4,5]`.
+
+Second query: The number of peaks in the `[3,1,4,4,5]` is 0.
+
+**Example 2:**
+
+**Input:** nums = [4,1,4,2,1,5], queries = \[\[2,2,4],[1,0,2],[1,0,4]]
+
+**Output:** [0,1]
+
+**Explanation:**
+
+First query: `nums[2]` should become 4, but it is already set to 4.
+
+Second query: The number of peaks in the `[4,1,4]` is 0.
+
+Third query: The second 4 is a peak in the `[4,1,4,2,1]`.
+
+**Constraints:**
+
+*   <code>3 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>5</sup></code>
+*   <code>1 <= queries.length <= 10<sup>5</sup></code>
+*   `queries[i][0] == 1` or `queries[i][0] == 2`
+*   For all `i` that:
+    *   `queries[i][0] == 1`: `0 <= queries[i][1] <= queries[i][2] <= nums.length - 1`
+    *   `queries[i][0] == 2`: `0 <= queries[i][1] <= nums.length - 1`, <code>1 <= queries[i][2] <= 10<sup>5</sup></code>
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun countOfPeaks(nums: IntArray, queries: Array<IntArray>): List<Int> {
+        val peaks = BooleanArray(nums.size)
+        val binaryIndexedTree = IntArray(Integer.highestOneBit(peaks.size) * 2 + 1)
+        for (i in 1 until peaks.size - 1) {
+            if (nums[i] > max(nums[i - 1], nums[i + 1])) {
+                peaks[i] = true
+                update(binaryIndexedTree, i + 1, 1)
+            }
+        }
+        val result: MutableList<Int> = ArrayList()
+        for (query in queries) {
+            if (query[0] == 1) {
+                val leftIndex = query[1]
+                val rightIndex = query[2]
+                result.add(computeRangeSum(binaryIndexedTree, leftIndex + 2, rightIndex))
+            } else {
+                val index = query[1]
+                val value = query[2]
+                nums[index] = value
+                for (i in -1..1) {
+                    val affected = index + i
+                    if (affected >= 1 && affected <= nums.size - 2) {
+                        val peak =
+                            nums[affected] > max(nums[affected - 1], nums[affected + 1])
+                        if (peak != peaks[affected]) {
+                            if (peak) {
+                                update(binaryIndexedTree, affected + 1, 1)
+                            } else {
+                                update(binaryIndexedTree, affected + 1, -1)
+                            }
+                            peaks[affected] = peak
+                        }
+                    }
+                }
+            }
+        }
+        return result
+    }
+
+    private fun computeRangeSum(binaryIndexedTree: IntArray, beginIndex: Int, endIndex: Int): Int {
+        return if (beginIndex <= endIndex) query(binaryIndexedTree, endIndex) - query(binaryIndexedTree, beginIndex - 1)
+        else 0
+    }
+
+    private fun query(binaryIndexedTree: IntArray, index: Int): Int {
+        var index = index
+        var result = 0
+        while (index != 0) {
+            result += binaryIndexedTree[index]
+            index -= index and -index
+        }
+
+        return result
+    }
+
+    private fun update(binaryIndexedTree: IntArray, index: Int, delta: Int) {
+        var index = index
+        while (index < binaryIndexedTree.size) {
+            binaryIndexedTree[index] += delta
+            index += index and -index
+        }
+    }
+}
+```