Added tasks 3314-3321

src/main/kotlin/g3301_3400/s3314_construct_the_minimum_bitwise_array_i/Solution.kt

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+package g3301_3400.s3314_construct_the_minimum_bitwise_array_i
+
+// #Easy #Array #Bit_Manipulation #2024_10_15_Time_226_ms_(57.14%)_Space_38_MB_(25.71%)
+
+class Solution {
+    fun minBitwiseArray(nums: List<Int>): IntArray {
+        val l = nums.size
+        val r = IntArray(l)
+        for (i in 0 until l) {
+            r[i] = check(nums[i])
+        }
+        return r
+    }
+
+    private fun check(v: Int): Int {
+        if (v % 2 == 0) {
+            return -1
+        }
+        for (j in 1 until v) {
+            if ((j or (j + 1)) == v) {
+                return j
+            }
+        }
+        return -1
+    }
+}

src/main/kotlin/g3301_3400/s3314_construct_the_minimum_bitwise_array_i/readme.md

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+3314\. Construct the Minimum Bitwise Array I
+
+Easy
+
+You are given an array `nums` consisting of `n` prime integers.
+
+You need to construct an array `ans` of length `n`, such that, for each index `i`, the bitwise `OR` of `ans[i]` and `ans[i] + 1` is equal to `nums[i]`, i.e. `ans[i] OR (ans[i] + 1) == nums[i]`.
+
+Additionally, you must **minimize** each value of `ans[i]` in the resulting array.
+
+If it is _not possible_ to find such a value for `ans[i]` that satisfies the **condition**, then set `ans[i] = -1`.
+
+**Example 1:**
+
+**Input:** nums = [2,3,5,7]
+
+**Output:** [-1,1,4,3]
+
+**Explanation:**
+
+*   For `i = 0`, as there is no value for `ans[0]` that satisfies `ans[0] OR (ans[0] + 1) = 2`, so `ans[0] = -1`.
+*   For `i = 1`, the smallest `ans[1]` that satisfies `ans[1] OR (ans[1] + 1) = 3` is `1`, because `1 OR (1 + 1) = 3`.
+*   For `i = 2`, the smallest `ans[2]` that satisfies `ans[2] OR (ans[2] + 1) = 5` is `4`, because `4 OR (4 + 1) = 5`.
+*   For `i = 3`, the smallest `ans[3]` that satisfies `ans[3] OR (ans[3] + 1) = 7` is `3`, because `3 OR (3 + 1) = 7`.
+
+**Example 2:**
+
+**Input:** nums = [11,13,31]
+
+**Output:** [9,12,15]
+
+**Explanation:**
+
+*   For `i = 0`, the smallest `ans[0]` that satisfies `ans[0] OR (ans[0] + 1) = 11` is `9`, because `9 OR (9 + 1) = 11`.
+*   For `i = 1`, the smallest `ans[1]` that satisfies `ans[1] OR (ans[1] + 1) = 13` is `12`, because `12 OR (12 + 1) = 13`.
+*   For `i = 2`, the smallest `ans[2]` that satisfies `ans[2] OR (ans[2] + 1) = 31` is `15`, because `15 OR (15 + 1) = 31`.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `2 <= nums[i] <= 1000`
+*   `nums[i]` is a prime number.

src/main/kotlin/g3301_3400/s3315_construct_the_minimum_bitwise_array_ii/Solution.kt

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+package g3301_3400.s3315_construct_the_minimum_bitwise_array_ii
+
+// #Medium #Array #Bit_Manipulation #2024_10_15_Time_231_ms_(77.27%)_Space_37.1_MB_(93.18%)
+
+class Solution {
+    fun minBitwiseArray(nums: List<Int>): IntArray {
+        val n = nums.size
+        val result = IntArray(n)
+        for (i in 0 until n) {
+            val num: Int = nums[i]
+            result[i] = -1
+            var p = 0
+            while (p < 31) {
+                if (((num shr p) and 1) == 0) {
+                    break
+                }
+                p++
+            }
+            if (p > 0) {
+                result[i] = ((num shr p) shl p) or ((1 shl (p - 1)) - 1)
+            }
+        }
+        return result
+    }
+}

src/main/kotlin/g3301_3400/s3315_construct_the_minimum_bitwise_array_ii/readme.md

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+3315\. Construct the Minimum Bitwise Array II
+
+Medium
+
+You are given an array `nums` consisting of `n` prime integers.
+
+You need to construct an array `ans` of length `n`, such that, for each index `i`, the bitwise `OR` of `ans[i]` and `ans[i] + 1` is equal to `nums[i]`, i.e. `ans[i] OR (ans[i] + 1) == nums[i]`.
+
+Additionally, you must **minimize** each value of `ans[i]` in the resulting array.
+
+If it is _not possible_ to find such a value for `ans[i]` that satisfies the **condition**, then set `ans[i] = -1`.
+
+**Example 1:**
+
+**Input:** nums = [2,3,5,7]
+
+**Output:** [-1,1,4,3]
+
+**Explanation:**
+
+*   For `i = 0`, as there is no value for `ans[0]` that satisfies `ans[0] OR (ans[0] + 1) = 2`, so `ans[0] = -1`.
+*   For `i = 1`, the smallest `ans[1]` that satisfies `ans[1] OR (ans[1] + 1) = 3` is `1`, because `1 OR (1 + 1) = 3`.
+*   For `i = 2`, the smallest `ans[2]` that satisfies `ans[2] OR (ans[2] + 1) = 5` is `4`, because `4 OR (4 + 1) = 5`.
+*   For `i = 3`, the smallest `ans[3]` that satisfies `ans[3] OR (ans[3] + 1) = 7` is `3`, because `3 OR (3 + 1) = 7`.
+
+**Example 2:**
+
+**Input:** nums = [11,13,31]
+
+**Output:** [9,12,15]
+
+**Explanation:**
+
+*   For `i = 0`, the smallest `ans[0]` that satisfies `ans[0] OR (ans[0] + 1) = 11` is `9`, because `9 OR (9 + 1) = 11`.
+*   For `i = 1`, the smallest `ans[1]` that satisfies `ans[1] OR (ans[1] + 1) = 13` is `12`, because `12 OR (12 + 1) = 13`.
+*   For `i = 2`, the smallest `ans[2]` that satisfies `ans[2] OR (ans[2] + 1) = 31` is `15`, because `15 OR (15 + 1) = 31`.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   <code>2 <= nums[i] <= 10<sup>9</sup></code>
+*   `nums[i]` is a prime number.

src/main/kotlin/g3301_3400/s3316_find_maximum_removals_from_source_string/Solution.kt

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+package g3301_3400.s3316_find_maximum_removals_from_source_string
+
+// #Medium #Array #String #Hash_Table #Dynamic_Programming #Two_Pointers
+// #2024_10_15_Time_220_ms_(100.00%)_Space_37.5_MB_(45.45%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maxRemovals(source: String, pattern: String, targetIndices: IntArray): Int {
+        val sChars = source.toCharArray()
+        val sn = sChars.size
+        val pChars = ("$pattern#").toCharArray()
+        val pn = pattern.length
+        var tn = targetIndices.size
+        val maxPat = IntArray(tn + 1)
+        var i = 0
+        var di = 0
+        var nextTI = targetIndices[0]
+        while (i < sn) {
+            val c = sChars[i]
+            if (i == nextTI) {
+                maxPat[di + 1] = maxPat[di]
+                var p = maxPat[di + 1]
+                for (j in di downTo 1) {
+                    val q = maxPat[j - 1]
+                    maxPat[j] = if (c != pChars[p]) q else max((p + 1), q)
+                    p = q
+                }
+                if (c == pChars[p]) {
+                    maxPat[0] = p + 1
+                }
+                nextTI = if (++di < tn) targetIndices[di] else -1
+            } else {
+                for (j in 0..di) {
+                    val p = maxPat[j]
+                    if (c == pChars[p]) {
+                        maxPat[j] = p + 1
+                    }
+                }
+            }
+            i++
+        }
+        while (maxPat[tn] < pn) {
+            tn--
+        }
+        return tn
+    }
+}

src/main/kotlin/g3301_3400/s3316_find_maximum_removals_from_source_string/readme.md

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+3316\. Find Maximum Removals From Source String
+
+Medium
+
+You are given a string `source` of size `n`, a string `pattern` that is a subsequence of `source`, and a **sorted** integer array `targetIndices` that contains **distinct** numbers in the range `[0, n - 1]`.
+
+We define an **operation** as removing a character at an index `idx` from `source` such that:
+
+*   `idx` is an element of `targetIndices`.
+*   `pattern` remains a subsequence of `source` after removing the character.
+
+Performing an operation **does not** change the indices of the other characters in `source`. For example, if you remove `'c'` from `"acb"`, the character at index 2 would still be `'b'`.
+
+Return the **maximum** number of _operations_ that can be performed.
+
+**Example 1:**
+
+**Input:** source = "abbaa", pattern = "aba", targetIndices = [0,1,2]
+
+**Output:** 1
+
+**Explanation:**
+
+We can't remove `source[0]` but we can do either of these two operations:
+
+*   Remove `source[1]`, so that `source` becomes `"a_baa"`.
+*   Remove `source[2]`, so that `source` becomes `"ab_aa"`.
+
+**Example 2:**
+
+**Input:** source = "bcda", pattern = "d", targetIndices = [0,3]
+
+**Output:** 2
+
+**Explanation:**
+
+We can remove `source[0]` and `source[3]` in two operations.
+
+**Example 3:**
+
+**Input:** source = "dda", pattern = "dda", targetIndices = [0,1,2]
+
+**Output:** 0
+
+**Explanation:**
+
+We can't remove any character from `source`.
+
+**Example 4:**
+
+**Input:** source = "yeyeykyded", pattern = "yeyyd", targetIndices = [0,2,3,4]
+
+**Output:** 2
+
+**Explanation:**
+
+We can remove `source[2]` and `source[3]` in two operations.
+
+**Constraints:**
+
+*   <code>1 <= n == source.length <= 3 * 10<sup>3</sup></code>
+*   `1 <= pattern.length <= n`
+*   `1 <= targetIndices.length <= n`
+*   `targetIndices` is sorted in ascending order.
+*   The input is generated such that `targetIndices` contains distinct elements in the range `[0, n - 1]`.
+*   `source` and `pattern` consist only of lowercase English letters.
+*   The input is generated such that `pattern` appears as a subsequence in `source`.

src/main/kotlin/g3301_3400/s3317_find_the_number_of_possible_ways_for_an_event/Solution.kt

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+package g3301_3400.s3317_find_the_number_of_possible_ways_for_an_event
+
+// #Hard #Dynamic_Programming #Math #Combinatorics
+// #2024_10_15_Time_166_ms_(100.00%)_Space_35_MB_(100.00%)
+
+import kotlin.math.min
+
+class Solution {
+    fun numberOfWays(n: Int, x: Int, y: Int): Int {
+        val fact = LongArray(x + 1)
+        fact[0] = 1
+        for (i in 1..x) {
+            fact[i] = fact[i - 1] * i % MOD
+        }
+        val invFact = LongArray(x + 1)
+        invFact[x] = powMod(fact[x], MOD - 2L)
+        for (i in x - 1 downTo 0) {
+            invFact[i] = invFact[i + 1] * (i + 1) % MOD
+        }
+        val powY = LongArray(x + 1)
+        powY[0] = 1
+        for (k in 1..x) {
+            powY[k] = powY[k - 1] * y % MOD
+        }
+        val localArray = LongArray(x + 1)
+        localArray[0] = 1
+        for (i in 1..n) {
+            val kMax: Int = min(i, x)
+            for (k in kMax downTo 1) {
+                localArray[k] = (k * localArray[k] + localArray[k - 1]) % MOD
+            }
+            localArray[0] = 0
+        }
+        var sum: Long = 0
+        val kLimit: Int = min(n, x)
+        for (k in 1..kLimit) {
+            val localValue: Long = fact[x] * invFact[x - k] % MOD
+            var term: Long = localValue * localArray[k] % MOD
+            term = term * powY[k] % MOD
+            sum = (sum + term) % MOD
+        }
+        return sum.toInt()
+    }
+
+    private fun powMod(a: Long, b: Long): Long {
+        var a = a
+        var b = b
+        var res: Long = 1
+        a = a % MOD
+        while (b > 0) {
+            if ((b and 1L) == 1L) {
+                res = res * a % MOD
+            }
+            a = a * a % MOD
+            b = b shr 1
+        }
+        return res
+    }
+
+    companion object {
+        private const val MOD = 1000000007
+    }
+}

src/main/kotlin/g3301_3400/s3317_find_the_number_of_possible_ways_for_an_event/readme.md

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+3317\. Find the Number of Possible Ways for an Event
+
+Hard
+
+You are given three integers `n`, `x`, and `y`.
+
+An event is being held for `n` performers. When a performer arrives, they are **assigned** to one of the `x` stages. All performers assigned to the **same** stage will perform together as a band, though some stages _might_ remain **empty**.
+
+After all performances are completed, the jury will **award** each band a score in the range `[1, y]`.
+
+Return the **total** number of possible ways the event can take place.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Note** that two events are considered to have been held **differently** if **either** of the following conditions is satisfied:
+
+*   **Any** performer is _assigned_ a different stage.
+*   **Any** band is _awarded_ a different score.
+
+**Example 1:**
+
+**Input:** n = 1, x = 2, y = 3
+
+**Output:** 6
+
+**Explanation:**
+
+*   There are 2 ways to assign a stage to the performer.
+*   The jury can award a score of either 1, 2, or 3 to the only band.
+
+**Example 2:**
+
+**Input:** n = 5, x = 2, y = 1
+
+**Output:** 32
+
+**Explanation:**
+
+*   Each performer will be assigned either stage 1 or stage 2.
+*   All bands will be awarded a score of 1.
+
+**Example 3:**
+
+**Input:** n = 3, x = 3, y = 4
+
+**Output:** 684
+
+**Constraints:**
+
+*   `1 <= n, x, y <= 1000`