Added tasks 3264-3277

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)
+
+## 3264\. Final Array State After K Multiplication Operations I
+
+Easy
+
+You are given an integer array `nums`, an integer `k`, and an integer `multiplier`.
+
+You need to perform `k` operations on `nums`. In each operation:
+
+*   Find the **minimum** value `x` in `nums`. If there are multiple occurrences of the minimum value, select the one that appears **first**.
+*   Replace the selected minimum value `x` with `x * multiplier`.
+
+Return an integer array denoting the _final state_ of `nums` after performing all `k` operations.
+
+**Example 1:**
+
+**Input:** nums = [2,1,3,5,6], k = 5, multiplier = 2
+
+**Output:** [8,4,6,5,6]
+
+**Explanation:**
+
+| Operation           | Result           |
+|---------------------|------------------|
+| After operation 1   | [2, 2, 3, 5, 6]  |
+| After operation 2   | [4, 2, 3, 5, 6]  |
+| After operation 3   | [4, 4, 3, 5, 6]  |
+| After operation 4   | [4, 4, 6, 5, 6]  |
+| After operation 5   | [8, 4, 6, 5, 6]  |
+
+**Example 2:**
+
+**Input:** nums = [1,2], k = 3, multiplier = 4
+
+**Output:** [16,8]
+
+**Explanation:**
+
+| Operation           | Result      |
+|---------------------|-------------|
+| After operation 1   | [4, 2]      |
+| After operation 2   | [4, 8]      |
+| After operation 3   | [16, 8]     |
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`
+*   `1 <= k <= 10`
+*   `1 <= multiplier <= 5`
+
+## Solution
+
+```kotlin
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun getFinalState(nums: IntArray, k: Int, multiplier: Int): IntArray {
+        var k = k
+        while (k-- > 0) {
+            var min = nums[0]
+            var index = 0
+            for (i in nums.indices) {
+                if (min > nums[i]) {
+                    min = nums[i]
+                    index = i
+                }
+            }
+            nums[index] = nums[index] * multiplier
+        }
+        return nums
+    }
+}
+```

src/main/kotlin/g3201_3300/s3264_final_array_state_after_k_multiplication_operations_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)
+
+## 3265\. Count Almost Equal Pairs I
+
+Medium
+
+You are given an array `nums` consisting of positive integers.
+
+We call two integers `x` and `y` in this problem **almost equal** if both integers can become equal after performing the following operation **at most once**:
+
+*   Choose **either** `x` or `y` and swap any two digits within the chosen number.
+
+Return the number of indices `i` and `j` in `nums` where `i < j` such that `nums[i]` and `nums[j]` are **almost equal**.
+
+**Note** that it is allowed for an integer to have leading zeros after performing an operation.
+
+**Example 1:**
+
+**Input:** nums = [3,12,30,17,21]
+
+**Output:** 2
+
+**Explanation:**
+
+The almost equal pairs of elements are:
+
+*   3 and 30. By swapping 3 and 0 in 30, you get 3.
+*   12 and 21. By swapping 1 and 2 in 12, you get 21.
+
+**Example 2:**
+
+**Input:** nums = [1,1,1,1,1]
+
+**Output:** 10
+
+**Explanation:**
+
+Every two elements in the array are almost equal.
+
+**Example 3:**
+
+**Input:** nums = [123,231]
+
+**Output:** 0
+
+**Explanation:**
+
+We cannot swap any two digits of 123 or 231 to reach the other.
+
+**Constraints:**
+
+*   `2 <= nums.length <= 100`
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>
+
+## Solution
+
+```kotlin
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun countPairs(nums: IntArray): Int {
+        var ans = 0
+        for (i in 0 until nums.size - 1) {
+            for (j in i + 1 until nums.size) {
+                if (nums[i] == nums[j] ||
+                    ((nums[j] - nums[i]) % 9 == 0 && check(nums[i], nums[j]))
+                ) {
+                    ans++
+                }
+            }
+        }
+        return ans
+    }
+
+    private fun check(a: Int, b: Int): Boolean {
+        var a = a
+        var b = b
+        val ca = IntArray(10)
+        val cb = IntArray(10)
+        var d = 0
+        while (a > 0 || b > 0) {
+            if (a % 10 != b % 10) {
+                d++
+                if (d > 2) {
+                    return false
+                }
+            }
+            ca[a % 10]++
+            cb[b % 10]++
+            a /= 10
+            b /= 10
+        }
+        return d == 2 && areEqual(ca, cb)
+    }
+
+    private fun areEqual(a: IntArray, b: IntArray): Boolean {
+        for (i in 0..9) {
+            if (a[i] != b[i]) {
+                return false
+            }
+        }
+        return true
+    }
+}
+```

src/main/kotlin/g3201_3300/s3265_count_almost_equal_pairs_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)
+
+## 3266\. Final Array State After K Multiplication Operations II
+
+Hard
+
+You are given an integer array `nums`, an integer `k`, and an integer `multiplier`.
+
+You need to perform `k` operations on `nums`. In each operation:
+
+*   Find the **minimum** value `x` in `nums`. If there are multiple occurrences of the minimum value, select the one that appears **first**.
+*   Replace the selected minimum value `x` with `x * multiplier`.
+
+After the `k` operations, apply **modulo** <code>10<sup>9</sup> + 7</code> to every value in `nums`.
+
+Return an integer array denoting the _final state_ of `nums` after performing all `k` operations and then applying the modulo.
+
+**Example 1:**
+
+**Input:** nums = [2,1,3,5,6], k = 5, multiplier = 2
+
+**Output:** [8,4,6,5,6]
+
+**Explanation:**
+
+| Operation               | Result           |
+|-------------------------|------------------|
+| After operation 1       | [2, 2, 3, 5, 6]  |
+| After operation 2       | [4, 2, 3, 5, 6]  |
+| After operation 3       | [4, 4, 3, 5, 6]  |
+| After operation 4       | [4, 4, 6, 5, 6]  |
+| After operation 5       | [8, 4, 6, 5, 6]  |
+| After applying modulo   | [8, 4, 6, 5, 6]  |
+
+**Example 2:**
+
+**Input:** nums = [100000,2000], k = 2, multiplier = 1000000
+
+**Output:** [999999307,999999993]
+
+**Explanation:**
+
+| Operation               | Result               |
+|-------------------------|----------------------|
+| After operation 1       | [100000, 2000000000] |
+| After operation 2       | [100000000000, 2000000000] |
+| After applying modulo   | [999999307, 999999993] |
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>4</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>1 <= k <= 10<sup>9</sup></code>
+*   <code>1 <= multiplier <= 10<sup>6</sup></code>
+
+## Solution
+
+```kotlin
+import java.util.PriorityQueue
+import kotlin.math.max
+
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun getFinalState(nums: IntArray, k: Int, multiplier: Int): IntArray {
+        var k = k
+        if (multiplier == 1) {
+            return nums
+        }
+        val n = nums.size
+        var mx = 0
+        for (x in nums) {
+            mx = max(mx, x)
+        }
+        val a = LongArray(n)
+        var left = k
+        var shouldExit = false
+        run {
+            var i = 0
+            while (i < n && !shouldExit) {
+                var x = nums[i].toLong()
+                while (x < mx) {
+                    x *= multiplier.toLong()
+                    if (--left < 0) {
+                        shouldExit = true
+                        break
+                    }
+                }
+                a[i] = x
+                i++
+            }
+        }
+        if (left < 0) {
+            val pq =
+                PriorityQueue { p: LongArray, q: LongArray ->
+                    if (p[0] != q[0]
+                    ) java.lang.Long.compare(p[0], q[0])
+                    else java.lang.Long.compare(p[1], q[1])
+                }
+            for (i in 0 until n) {
+                pq.offer(longArrayOf(nums[i].toLong(), i.toLong()))
+            }
+            while (k-- > 0) {
+                val p = pq.poll()
+                p[0] *= multiplier.toLong()
+                pq.offer(p)
+            }
+            while (pq.isNotEmpty()) {
+                val p = pq.poll()
+                nums[p[1].toInt()] = (p[0] % MOD).toInt()
+            }
+            return nums
+        }
+
+        val ids: Array<Int> = Array(n) { i: Int -> i }
+        ids.sortWith { i: Int?, j: Int? -> java.lang.Long.compare(a[i!!], a[j!!]) }
+        k = left
+        val pow1 = pow(multiplier.toLong(), k / n)
+        val pow2 = pow1 * multiplier % MOD
+        for (i in 0 until n) {
+            val j = ids[i]
+            nums[j] = (a[j] % MOD * (if (i < k % n) pow2 else pow1) % MOD).toInt()
+        }
+        return nums
+    }
+
+    private fun pow(x: Long, n: Int): Long {
+        var x = x
+        var n = n
+        var res: Long = 1
+        while (n > 0) {
+            if (n % 2 > 0) {
+                res = res * x % MOD
+            }
+            x = x * x % MOD
+            n /= 2
+        }
+        return res
+    }
+
+    companion object {
+        private const val MOD = 1000000007
+    }
+}
+```