Added tasks 3467-3475

Author: javadev

README.md

@@ -71,44 +71,41 @@ The substring starting at 12 is "thefoobar". It is the concatenation of ["the","
 ```kotlin
 class Solution {
     fun findSubstring(s: String, words: Array<String>): List<Int> {
-        val indices: MutableList<Int> = ArrayList()
-        if (words.size == 0) {
-            return indices
+        val ans: MutableList<Int> = ArrayList<Int>()
+        val n1 = words[0].length
+        val n2 = s.length
+        val map1: MutableMap<String, Int> = HashMap<String, Int>()
+        for (ch in words) {
+            map1.put(ch, map1.getOrDefault(ch, 0) + 1)
         }
-        // Put each word into a HashMap and calculate word frequency
-        val wordMap: MutableMap<String, Int> = HashMap()
-        for (word in words) {
-            wordMap[word] = wordMap.getOrDefault(word, 0) + 1
-        }
-        val wordLength = words[0].length
-        val window = words.size * wordLength
-        for (i in 0 until wordLength) {
-            // move a word's length each time
+        for (i in 0..<n1) {
+            var left = i
             var j = i
-            while (j + window <= s.length) {
-                // get the subStr
-                val subStr = s.substring(j, j + window)
-                val map: MutableMap<String, Int> = HashMap()
-                // start from the last word
-                for (k in words.indices.reversed()) {
-                    // get the word from subStr
-                    val word = subStr.substring(k * wordLength, (k + 1) * wordLength)
-                    val count = map.getOrDefault(word, 0) + 1
-                    // if the num of the word is greater than wordMap's, move (k * wordLength) and
-                    // break
-                    if (count > wordMap.getOrDefault(word, 0)) {
-                        j = j + k * wordLength
-                        break
-                    } else if (k == 0) {
-                        indices.add(j)
-                    } else {
-                        map[word] = count
+            var c = 0
+            val map2: MutableMap<String, Int> = HashMap<String, Int>()
+            while (j + n1 <= n2) {
+                val word1 = s.substring(j, j + n1)
+                j += n1
+                if (map1.containsKey(word1)) {
+                    map2.put(word1, map2.getOrDefault(word1, 0) + 1)
+                    c++
+                    while (map2[word1]!! > map1[word1]!!) {
+                        val word2 = s.substring(left, left + n1)
+                        map2.put(word2, map2[word2]!! - 1)
+                        left += n1
+                        c--
+                    }
+                    if (c == words.size) {
+                        ans.add(left)
                     }
+                } else {
+                    map2.clear()
+                    c = 0
+                    left = j
                 }
-                j = j + wordLength
             }
         }
-        return indices
+        return ans
     }
 }
 ```

src/main/kotlin/g0001_0100/s0030_substring_with_concatenation_of_all_words/readme.md

@@ -0,0 +1,62 @@
+[![](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)
+
+## 3467\. Transform Array by Parity
+
+Easy
+
+You are given an integer array `nums`. Transform `nums` by performing the following operations in the **exact** order specified:
+
+1.  Replace each even number with 0.
+2.  Replace each odd numbers with 1.
+3.  Sort the modified array in **non-decreasing** order.
+
+Return the resulting array after performing these operations.
+
+**Example 1:**
+
+**Input:** nums = [4,3,2,1]
+
+**Output:** [0,0,1,1]
+
+**Explanation:**
+
+*   Replace the even numbers (4 and 2) with 0 and the odd numbers (3 and 1) with 1. Now, `nums = [0, 1, 0, 1]`.
+*   After sorting `nums` in non-descending order, `nums = [0, 0, 1, 1]`.
+
+**Example 2:**
+
+**Input:** nums = [1,5,1,4,2]
+
+**Output:** [0,0,1,1,1]
+
+**Explanation:**
+
+*   Replace the even numbers (4 and 2) with 0 and the odd numbers (1, 5 and 1) with 1. Now, `nums = [1, 1, 1, 0, 0]`.
+*   After sorting `nums` in non-descending order, `nums = [0, 0, 1, 1, 1]`.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 1000`
+
+## Solution
+
+```kotlin
+class Solution {
+    fun transformArray(nums: IntArray): IntArray {
+        val size = nums.size
+        val ans = IntArray(size)
+        var countEven = 0
+        for (i in nums.indices) {
+            if (nums[i] and 1 == 0) {
+                countEven++
+            }
+        }
+        for (i in countEven until size) {
+            ans[i] = 1
+        }
+        return ans
+    }
+}
+```

src/main/kotlin/g3401_3500/s3467_transform_array_by_parity/readme.md

@@ -0,0 +1,83 @@
+[![](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)
+
+## 3468\. Find the Number of Copy Arrays
+
+Medium
+
+You are given an array `original` of length `n` and a 2D array `bounds` of length `n x 2`, where <code>bounds[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>.
+
+You need to find the number of **possible** arrays `copy` of length `n` such that:
+
+1.  `(copy[i] - copy[i - 1]) == (original[i] - original[i - 1])` for `1 <= i <= n - 1`.
+2.  <code>u<sub>i</sub> <= copy[i] <= v<sub>i</sub></code> for `0 <= i <= n - 1`.
+
+Return the number of such arrays.
+
+**Example 1:**
+
+**Input:** original = [1,2,3,4], bounds = \[\[1,2],[2,3],[3,4],[4,5]]
+
+**Output:** 2
+
+**Explanation:**
+
+The possible arrays are:
+
+*   `[1, 2, 3, 4]`
+*   `[2, 3, 4, 5]`
+
+**Example 2:**
+
+**Input:** original = [1,2,3,4], bounds = \[\[1,10],[2,9],[3,8],[4,7]]
+
+**Output:** 4
+
+**Explanation:**
+
+The possible arrays are:
+
+*   `[1, 2, 3, 4]`
+*   `[2, 3, 4, 5]`
+*   `[3, 4, 5, 6]`
+*   `[4, 5, 6, 7]`
+
+**Example 3:**
+
+**Input:** original = [1,2,1,2], bounds = \[\[1,1],[2,3],[3,3],[2,3]]
+
+**Output:** 0
+
+**Explanation:**
+
+No array is possible.
+
+**Constraints:**
+
+*   <code>2 <= n == original.length <= 10<sup>5</sup></code>
+*   <code>1 <= original[i] <= 10<sup>9</sup></code>
+*   `bounds.length == n`
+*   `bounds[i].length == 2`
+*   <code>1 <= bounds[i][0] <= bounds[i][1] <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun countArrays(original: IntArray, bounds: Array<IntArray>): Int {
+        var low = bounds[0][0]
+        var high = bounds[0][1]
+        var ans = high - low + 1
+        for (i in 1..<original.size) {
+            val diff = original[i] - original[i - 1]
+            low = max((low + diff), bounds[i][0])
+            high = min((high + diff), bounds[i][1])
+            ans = min(ans, (high - low + 1)).toInt()
+        }
+        return max(ans, 0)
+    }
+}
+```

src/main/kotlin/g3401_3500/s3468_find_the_number_of_copy_arrays/readme.md

@@ -0,0 +1,90 @@
+[![](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)
+
+## 3469\. Find Minimum Cost to Remove Array Elements
+
+Medium
+
+You are given an integer array `nums`. Your task is to remove **all elements** from the array by performing one of the following operations at each step until `nums` is empty:
+
+*   Choose any two elements from the first three elements of `nums` and remove them. The cost of this operation is the **maximum** of the two elements removed.
+*   If fewer than three elements remain in `nums`, remove all the remaining elements in a single operation. The cost of this operation is the **maximum** of the remaining elements.
+
+Return the **minimum** cost required to remove all the elements.
+
+**Example 1:**
+
+**Input:** nums = [6,2,8,4]
+
+**Output:** 12
+
+**Explanation:**
+
+Initially, `nums = [6, 2, 8, 4]`.
+
+*   In the first operation, remove `nums[0] = 6` and `nums[2] = 8` with a cost of `max(6, 8) = 8`. Now, `nums = [2, 4]`.
+*   In the second operation, remove the remaining elements with a cost of `max(2, 4) = 4`.
+
+The cost to remove all elements is `8 + 4 = 12`. This is the minimum cost to remove all elements in `nums`. Hence, the output is 12.
+
+**Example 2:**
+
+**Input:** nums = [2,1,3,3]
+
+**Output:** 5
+
+**Explanation:**
+
+Initially, `nums = [2, 1, 3, 3]`.
+
+*   In the first operation, remove `nums[0] = 2` and `nums[1] = 1` with a cost of `max(2, 1) = 2`. Now, `nums = [3, 3]`.
+*   In the second operation remove the remaining elements with a cost of `max(3, 3) = 3`.
+
+The cost to remove all elements is `2 + 3 = 5`. This is the minimum cost to remove all elements in `nums`. Hence, the output is 5.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 1000`
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun minCost(nums: IntArray): Int {
+        var nums = nums
+        var n = nums.size
+        if (n % 2 == 0) {
+            nums = nums.copyOf(++n)
+        }
+        val dp = IntArray(n)
+        var j = 1
+        while (j < n - 1) {
+            var cost1: Int = INF
+            var cost2: Int = INF
+            val max = max(nums[j], nums[j + 1])
+            for (i in 0..<j) {
+                cost1 =
+                    min(cost1, dp[i] + max(nums[i], nums[j + 1]))
+                cost2 = min(cost2, dp[i] + max(nums[i], nums[j]))
+                dp[i] += max
+            }
+            dp[j] = cost1
+            dp[j + 1] = cost2
+            j += 2
+        }
+        var result: Int = INF
+        for (i in 0..<n) {
+            result = min(result, dp[i] + nums[i])
+        }
+        return result
+    }
+
+    companion object {
+        private const val INF = 1e9.toInt()
+    }
+}
+```