Added tasks 3105-3108

Author: javadev

src/main/kotlin/g3101_3200/s3105_longest_strictly_increasing_or_strictly_decreasing_subarray/Solution.kt

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+package g3101_3200.s3105_longest_strictly_increasing_or_strictly_decreasing_subarray
+
+// #Easy #Array #2024_04_13_Time_159_ms_(94.00%)_Space_36.4_MB_(92.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun longestMonotonicSubarray(nums: IntArray): Int {
+        var inc = 1
+        var dec = 1
+        var res = 1
+        for (i in 1 until nums.size) {
+            if (nums[i] > nums[i - 1]) {
+                inc += 1
+                dec = 1
+            } else if (nums[i] < nums[i - 1]) {
+                dec += 1
+                inc = 1
+            } else {
+                inc = 1
+                dec = 1
+            }
+            res = max(res, max(inc, dec))
+        }
+        return res
+    }
+}

src/main/kotlin/g3101_3200/s3105_longest_strictly_increasing_or_strictly_decreasing_subarray/readme.md

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+3105\. Longest Strictly Increasing or Strictly Decreasing Subarray
+
+Easy
+
+You are given an array of integers `nums`. Return _the length of the **longest** subarray of_ `nums` _which is either **strictly increasing** or **strictly decreasing**_.
+
+**Example 1:**
+
+**Input:** nums = [1,4,3,3,2]
+
+**Output:** 2
+
+**Explanation:**
+
+The strictly increasing subarrays of `nums` are `[1]`, `[2]`, `[3]`, `[3]`, `[4]`, and `[1,4]`.
+
+The strictly decreasing subarrays of `nums` are `[1]`, `[2]`, `[3]`, `[3]`, `[4]`, `[3,2]`, and `[4,3]`.
+
+Hence, we return `2`.
+
+**Example 2:**
+
+**Input:** nums = [3,3,3,3]
+
+**Output:** 1
+
+**Explanation:**
+
+The strictly increasing subarrays of `nums` are `[3]`, `[3]`, `[3]`, and `[3]`.
+
+The strictly decreasing subarrays of `nums` are `[3]`, `[3]`, `[3]`, and `[3]`.
+
+Hence, we return `1`.
+
+**Example 3:**
+
+**Input:** nums = [3,2,1]
+
+**Output:** 3
+
+**Explanation:**
+
+The strictly increasing subarrays of `nums` are `[3]`, `[2]`, and `[1]`.
+
+The strictly decreasing subarrays of `nums` are `[3]`, `[2]`, `[1]`, `[3,2]`, `[2,1]`, and `[3,2,1]`.
+
+Hence, we return `3`.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 50`
+*   `1 <= nums[i] <= 50`

src/main/kotlin/g3101_3200/s3106_lexicographically_smallest_string_after_operations_with_constraint/Solution.kt

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+package g3101_3200.s3106_lexicographically_smallest_string_after_operations_with_constraint
+
+// #Medium #String #Greedy #2024_04_13_Time_162_ms_(74.19%)_Space_36.2_MB_(77.42%)
+
+import kotlin.math.abs
+import kotlin.math.min
+
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun getSmallestString(s: String, k: Int): String {
+        var k = k
+        val sArray = s.toCharArray()
+        for (i in sArray.indices) {
+            val distToA = cyclicDistance(sArray[i], 'a')
+            if (distToA <= k) {
+                sArray[i] = 'a'
+                k -= distToA
+            } else if (k > 0) {
+                sArray[i] = (sArray[i].code - k).toChar()
+                k = 0
+            }
+        }
+        return String(sArray)
+    }
+
+    private fun cyclicDistance(ch1: Char, ch2: Char): Int {
+        val dist = abs(ch1.code - ch2.code)
+        return min(dist, (26 - dist))
+    }
+}

src/main/kotlin/g3101_3200/s3106_lexicographically_smallest_string_after_operations_with_constraint/readme.md

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+3106\. Lexicographically Smallest String After Operations With Constraint
+
+Medium
+
+You are given a string `s` and an integer `k`.
+
+Define a function <code>distance(s<sub>1</sub>, s<sub>2</sub>)</code> between two strings <code>s<sub>1</sub></code> and <code>s<sub>2</sub></code> of the same length `n` as:
+
+*   The **sum** of the **minimum distance** between <code>s<sub>1</sub>[i]</code> and <code>s<sub>2</sub>[i]</code> when the characters from `'a'` to `'z'` are placed in a **cyclic** order, for all `i` in the range `[0, n - 1]`.
+
+For example, `distance("ab", "cd") == 4`, and `distance("a", "z") == 1`.
+
+You can **change** any letter of `s` to **any** other lowercase English letter, **any** number of times.
+
+Return a string denoting the **lexicographically smallest** string `t` you can get after some changes, such that `distance(s, t) <= k`.
+
+**Example 1:**
+
+**Input:** s = "zbbz", k = 3
+
+**Output:** "aaaz"
+
+**Explanation:**
+
+Change `s` to `"aaaz"`. The distance between `"zbbz"` and `"aaaz"` is equal to `k = 3`.
+
+**Example 2:**
+
+**Input:** s = "xaxcd", k = 4
+
+**Output:** "aawcd"
+
+**Explanation:**
+
+The distance between "xaxcd" and "aawcd" is equal to k = 4.
+
+**Example 3:**
+
+**Input:** s = "lol", k = 0
+
+**Output:** "lol"
+
+**Explanation:**
+
+It's impossible to change any character as `k = 0`.
+
+**Constraints:**
+
+*   `1 <= s.length <= 100`
+*   `0 <= k <= 2000`
+*   `s` consists only of lowercase English letters.

src/main/kotlin/g3101_3200/s3107_minimum_operations_to_make_median_of_array_equal_to_k/Solution.kt

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+package g3101_3200.s3107_minimum_operations_to_make_median_of_array_equal_to_k
+
+// #Medium #Array #Sorting #Greedy #2024_04_13_Time_554_ms_(100.00%)_Space_82.2_MB_(68.00%)
+
+import kotlin.math.abs
+
+class Solution {
+    fun minOperationsToMakeMedianK(nums: IntArray, k: Int): Long {
+        nums.sort()
+        val n = nums.size
+        val medianIndex = n / 2
+        var result: Long = 0
+        var totalElements = 0
+        var totalSum: Long = 0
+        var i = medianIndex
+        if (nums[medianIndex] > k) {
+            while (i >= 0 && nums[i] > k) {
+                totalElements += 1
+                totalSum += nums[i].toLong()
+                i -= 1
+            }
+        } else if (nums[medianIndex] < k) {
+            while (i < n && nums[i] < k) {
+                totalElements += 1
+                totalSum += nums[i].toLong()
+                i += 1
+            }
+        }
+        result += abs(totalSum - (totalElements.toLong() * k))
+        return result
+    }
+}

src/main/kotlin/g3101_3200/s3107_minimum_operations_to_make_median_of_array_equal_to_k/readme.md

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+3107\. Minimum Operations to Make Median of Array Equal to K
+
+Medium
+
+You are given an integer array `nums` and a **non-negative** integer `k`. In one operation, you can increase or decrease any element by 1.
+
+Return the **minimum** number of operations needed to make the **median** of `nums` _equal_ to `k`.
+
+The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.
+
+**Example 1:**
+
+**Input:** nums = [2,5,6,8,5], k = 4
+
+**Output:** 2
+
+**Explanation:**
+
+We can subtract one from `nums[1]` and `nums[4]` to obtain `[2, 4, 6, 8, 4]`. The median of the resulting array is equal to `k`.
+
+**Example 2:**
+
+**Input:** nums = [2,5,6,8,5], k = 7
+
+**Output:** 3
+
+**Explanation:**
+
+We can add one to `nums[1]` twice and add one to `nums[2]` once to obtain `[2, 7, 7, 8, 5]`.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3,4,5,6], k = 4
+
+**Output:** 0
+
+**Explanation:**
+
+The median of the array is already equal to `k`.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 2 * 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>1 <= k <= 10<sup>9</sup></code>

src/main/kotlin/g3101_3200/s3108_minimum_cost_walk_in_weighted_graph/Solution.kt

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+package g3101_3200.s3108_minimum_cost_walk_in_weighted_graph
+
+// #Hard #Array #Bit_Manipulation #Graph #Union_Find
+// #2024_04_13_Time_791_ms_(100.00%)_Space_139.3_MB_(26.67%)
+
+@Suppress("NAME_SHADOWING")
+class Solution {
+    fun minimumCost(n: Int, edges: Array<IntArray>, query: Array<IntArray>): IntArray {
+        val parent = IntArray(n)
+        val bitwise = IntArray(n)
+        val size = IntArray(n)
+        var i = 0
+        while (i < n) {
+            parent[i] = i
+            size[i] = 1
+            bitwise[i] = -1
+            i++
+        }
+        val len = edges.size
+        i = 0
+        while (i < len) {
+            val node1 = edges[i][0]
+            val node2 = edges[i][1]
+            val weight = edges[i][2]
+            val parent1 = findParent(node1, parent)
+            val parent2 = findParent(node2, parent)
+            if (parent1 == parent2) {
+                bitwise[parent1] = bitwise[parent1] and weight
+            } else {
+                var bitwiseVal: Int
+                val check1 = bitwise[parent1] == -1
+                val check2 = bitwise[parent2] == -1
+                bitwiseVal = if (check1 && check2) {
+                    weight
+                } else if (check1) {
+                    weight and bitwise[parent2]
+                } else if (check2) {
+                    weight and bitwise[parent1]
+                } else {
+                    weight and bitwise[parent1] and bitwise[parent2]
+                }
+                if (size[parent1] >= size[parent2]) {
+                    parent[parent2] = parent1
+                    size[parent1] += size[parent2]
+                    bitwise[parent1] = bitwiseVal
+                } else {
+                    parent[parent1] = parent2
+                    size[parent2] += size[parent1]
+                    bitwise[parent2] = bitwiseVal
+                }
+            }
+            i++
+        }
+        val queryLen = query.size
+        val result = IntArray(queryLen)
+        i = 0
+        while (i < queryLen) {
+            val start = query[i][0]
+            val end = query[i][1]
+            val parentStart = findParent(start, parent)
+            val parentEnd = findParent(end, parent)
+            if (start == end) {
+                result[i] = 0
+            } else if (parentStart == parentEnd) {
+                result[i] = bitwise[parentStart]
+            } else {
+                result[i] = -1
+            }
+            i++
+        }
+        return result
+    }
+
+    private fun findParent(node: Int, parent: IntArray): Int {
+        var node = node
+        while (parent[node] != node) {
+            node = parent[node]
+        }
+        return node
+    }
+}

src/main/kotlin/g3101_3200/s3108_minimum_cost_walk_in_weighted_graph/readme.md

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+3108\. Minimum Cost Walk in Weighted Graph
+
+Hard
+
+There is an undirected weighted graph with `n` vertices labeled from `0` to `n - 1`.
+
+You are given the integer `n` and an array `edges`, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between vertices <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with a weight of <code>w<sub>i</sub></code>.
+
+A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It's important to note that a walk may visit the same edge or vertex more than once.
+
+The **cost** of a walk starting at node `u` and ending at node `v` is defined as the bitwise `AND` of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is <code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>, then the cost is calculated as <code>w<sub>0</sub> & w<sub>1</sub> & w<sub>2</sub> & ... & w<sub>k</sub></code>, where `&` denotes the bitwise `AND` operator.
+
+You are also given a 2D array `query`, where <code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>. For each query, you need to find the minimum cost of the walk starting at vertex <code>s<sub>i</sub></code> and ending at vertex <code>t<sub>i</sub></code>. If there exists no such walk, the answer is `-1`.
+
+Return _the array_ `answer`_, where_ `answer[i]` _denotes the **minimum** cost of a walk for query_ `i`.
+
+**Example 1:**
+
+**Input:** n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]
+
+**Output:** [1,-1]
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png)
+
+To achieve the cost of 1 in the first query, we need to move on the following edges: `0->1` (weight 7), `1->2` (weight 1), `2->1` (weight 1), `1->3` (weight 7).
+
+In the second query, there is no walk between nodes 3 and 4, so the answer is -1.
+
+**Example 2:**
+
+**Input:** n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]
+
+**Output:** [0]
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png)
+
+To achieve the cost of 0 in the first query, we need to move on the following edges: `1->2` (weight 1), `2->1` (weight 6), `1->2` (weight 1).
+
+**Constraints:**
+
+*   <code>2 <= n <= 10<sup>5</sup></code>
+*   <code>0 <= edges.length <= 10<sup>5</sup></code>
+*   `edges[i].length == 3`
+*   <code>0 <= u<sub>i</sub>, v<sub>i</sub> <= n - 1</code>
+*   <code>u<sub>i</sub> != v<sub>i</sub></code>
+*   <code>0 <= w<sub>i</sub> <= 10<sup>5</sup></code>
+*   <code>1 <= query.length <= 10<sup>5</sup></code>
+*   `query[i].length == 2`
+*   <code>0 <= s<sub>i</sub>, t<sub>i</sub> <= n - 1</code>
+*   <code>s<sub>i</sub> != t<sub>i</sub></code>