Added tasks 3345-3348

src/main/kotlin/g3301_3400/s3345_smallest_divisible_digit_product_i/Solution.kt

@@ -0,0 +1,28 @@
+package g3301_3400.s3345_smallest_divisible_digit_product_i
+
+// #Easy #Math #Enumeration #2024_11_14_Time_1_ms_(100.00%)_Space_33.7_MB_(100.00%)
+
+class Solution {
+    fun smallestNumber(n: Int, t: Int): Int {
+        var num = -1
+        var check = n
+        while (num == -1) {
+            val product = findProduct(check)
+            if (product % t == 0) {
+                num = check
+            }
+            check += 1
+        }
+        return num
+    }
+
+    private fun findProduct(check: Int): Int {
+        var check = check
+        var res = 1
+        while (check > 0) {
+            res *= check % 10
+            check = check / 10
+        }
+        return res
+    }
+}

src/main/kotlin/g3301_3400/s3345_smallest_divisible_digit_product_i/readme.md

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+3345\. Smallest Divisible Digit Product I
+
+Easy
+
+You are given two integers `n` and `t`. Return the **smallest** number greater than or equal to `n` such that the **product of its digits** is divisible by `t`.
+
+**Example 1:**
+
+**Input:** n = 10, t = 2
+
+**Output:** 10
+
+**Explanation:**
+
+The digit product of 10 is 0, which is divisible by 2, making it the smallest number greater than or equal to 10 that satisfies the condition.
+
+**Example 2:**
+
+**Input:** n = 15, t = 3
+
+**Output:** 16
+
+**Explanation:**
+
+The digit product of 16 is 6, which is divisible by 3, making it the smallest number greater than or equal to 15 that satisfies the condition.
+
+**Constraints:**
+
+*   `1 <= n <= 100`
+*   `1 <= t <= 10`

src/main/kotlin/g3301_3400/s3346_maximum_frequency_of_an_element_after_performing_operations_i/Solution.kt

@@ -0,0 +1,42 @@
+package g3301_3400.s3346_maximum_frequency_of_an_element_after_performing_operations_i
+
+// #Medium #Array #Sorting #Binary_Search #Prefix_Sum #Sliding_Window
+// #2024_11_14_Time_12_ms_(100.00%)_Space_64.1_MB_(80.00%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    private fun getMax(nums: IntArray): Int {
+        var max = nums[0]
+        for (num in nums) {
+            max = max(num, max)
+        }
+        return max
+    }
+
+    fun maxFrequency(nums: IntArray, k: Int, numOperations: Int): Int {
+        val maxNum = getMax(nums)
+        val n = maxNum + k + 2
+        val freq = IntArray(n)
+        for (num in nums) {
+            freq[num]++
+        }
+        val pref = IntArray(n)
+        pref[0] = freq[0]
+        for (i in 1 until n) {
+            pref[i] = pref[i - 1] + freq[i]
+        }
+        var res = 0
+        for (i in 0 until n) {
+            val left: Int = max(0, (i - k))
+            val right: Int = min((n - 1), (i + k))
+            var tot = pref[right]
+            if (left > 0) {
+                tot -= pref[left - 1]
+            }
+            res = max(res, (freq[i] + min(numOperations, (tot - freq[i]))))
+        }
+        return res
+    }
+}

src/main/kotlin/g3301_3400/s3346_maximum_frequency_of_an_element_after_performing_operations_i/readme.md

@@ -0,0 +1,44 @@
+3346\. Maximum Frequency of an Element After Performing Operations I
+
+Medium
+
+You are given an integer array `nums` and two integers `k` and `numOperations`.
+
+You must perform an **operation** `numOperations` times on `nums`, where in each operation you:
+
+*   Select an index `i` that was **not** selected in any previous operations.
+*   Add an integer in the range `[-k, k]` to `nums[i]`.
+
+Return the **maximum** possible frequency of any element in `nums` after performing the **operations**.
+
+**Example 1:**
+
+**Input:** nums = [1,4,5], k = 1, numOperations = 2
+
+**Output:** 2
+
+**Explanation:**
+
+We can achieve a maximum frequency of two by:
+
+*   Adding 0 to `nums[1]`. `nums` becomes `[1, 4, 5]`.
+*   Adding -1 to `nums[2]`. `nums` becomes `[1, 4, 4]`.
+
+**Example 2:**
+
+**Input:** nums = [5,11,20,20], k = 5, numOperations = 1
+
+**Output:** 2
+
+**Explanation:**
+
+We can achieve a maximum frequency of two by:
+
+*   Adding 0 to `nums[1]`.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>5</sup></code>
+*   <code>0 <= k <= 10<sup>5</sup></code>
+*   `0 <= numOperations <= nums.length`

src/main/kotlin/g3301_3400/s3347_maximum_frequency_of_an_element_after_performing_operations_ii/Solution.kt

@@ -0,0 +1,42 @@
+package g3301_3400.s3347_maximum_frequency_of_an_element_after_performing_operations_ii
+
+// #Hard #Array #Sorting #Binary_Search #Prefix_Sum #Sliding_Window
+// #2024_11_14_Time_48_ms_(100.00%)_Space_67.8_MB_(93.33%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun maxFrequency(nums: IntArray, k: Int, numOperations: Int): Int {
+        nums.sort()
+        val n = nums.size
+        var l = 0
+        var r = 0
+        var i = 0
+        var j = 0
+        var res = 0
+        while (i < n) {
+            while (j < n && nums[j] == nums[i]) {
+                j++
+            }
+            while (l < i && nums[i] - nums[l] > k) {
+                l++
+            }
+            while (r < n && nums[r] - nums[i] <= k) {
+                r++
+            }
+            res = max(res, (min((i - l + r - j), numOperations) + j - i))
+            i = j
+        }
+        i = 0
+        j = 0
+        while (i < n && j < n) {
+            while (j < n && j - i < numOperations && nums[j] - nums[i] <= k * 2) {
+                j++
+            }
+            res = max(res, (j - i))
+            i++
+        }
+        return res
+    }
+}

src/main/kotlin/g3301_3400/s3347_maximum_frequency_of_an_element_after_performing_operations_ii/readme.md

@@ -0,0 +1,44 @@
+3347\. Maximum Frequency of an Element After Performing Operations II
+
+Hard
+
+You are given an integer array `nums` and two integers `k` and `numOperations`.
+
+You must perform an **operation** `numOperations` times on `nums`, where in each operation you:
+
+*   Select an index `i` that was **not** selected in any previous operations.
+*   Add an integer in the range `[-k, k]` to `nums[i]`.
+
+Return the **maximum** possible frequency of any element in `nums` after performing the **operations**.
+
+**Example 1:**
+
+**Input:** nums = [1,4,5], k = 1, numOperations = 2
+
+**Output:** 2
+
+**Explanation:**
+
+We can achieve a maximum frequency of two by:
+
+*   Adding 0 to `nums[1]`, after which `nums` becomes `[1, 4, 5]`.
+*   Adding -1 to `nums[2]`, after which `nums` becomes `[1, 4, 4]`.
+
+**Example 2:**
+
+**Input:** nums = [5,11,20,20], k = 5, numOperations = 1
+
+**Output:** 2
+
+**Explanation:**
+
+We can achieve a maximum frequency of two by:
+
+*   Adding 0 to `nums[1]`.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>0 <= k <= 10<sup>9</sup></code>
+*   `0 <= numOperations <= nums.length`

src/main/kotlin/g3301_3400/s3348_smallest_divisible_digit_product_ii/Solution.kt

@@ -0,0 +1,77 @@
+package g3301_3400.s3348_smallest_divisible_digit_product_ii
+
+// #Hard #String #Math #Greedy #Backtracking #Number_Theory
+// #2024_11_14_Time_46_ms_(100.00%)_Space_48.2_MB_(100.00%)
+
+class Solution {
+    fun smallestNumber(num: String, t: Long): String {
+        var t = t
+        var tmp = t
+        for (i in 9 downTo 2) {
+            while (tmp % i == 0L) {
+                tmp /= i.toLong()
+            }
+        }
+        if (tmp > 1) {
+            return "-1"
+        }
+        val s = num.toCharArray()
+        val n = s.size
+        val leftT = LongArray(n + 1)
+        leftT[0] = t
+        var i0 = n - 1
+        for (i in 0 until n) {
+            if (s[i] == '0') {
+                i0 = i
+                break
+            }
+            leftT[i + 1] = leftT[i] / gcd(leftT[i], s[i].code.toLong() - '0'.code.toLong())
+        }
+        if (leftT[n] == 1L) {
+            return num
+        }
+        for (i in i0 downTo 0) {
+            while (++s[i] <= '9') {
+                var tt = leftT[i] / gcd(leftT[i], s[i].code.toLong() - '0'.code.toLong())
+                for (j in n - 1 downTo i + 1) {
+                    if (tt == 1L) {
+                        s[j] = '1'
+                        continue
+                    }
+                    for (k in 9 downTo 2) {
+                        if (tt % k == 0L) {
+                            s[j] = ('0'.code + k).toChar()
+                            tt /= k.toLong()
+                            break
+                        }
+                    }
+                }
+                if (tt == 1L) {
+                    return String(s)
+                }
+            }
+        }
+        val ans = StringBuilder()
+        for (i in 9 downTo 2) {
+            while (t % i == 0L) {
+                ans.append(('0'.code + i).toChar())
+                t /= i.toLong()
+            }
+        }
+        while (ans.length <= n) {
+            ans.append('1')
+        }
+        return ans.reverse().toString()
+    }
+
+    private fun gcd(a: Long, b: Long): Long {
+        var a = a
+        var b = b
+        while (a != 0L) {
+            val tmp = a
+            a = b % a
+            b = tmp
+        }
+        return b
+    }
+}

src/main/kotlin/g3301_3400/s3348_smallest_divisible_digit_product_ii/readme.md

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+3348\. Smallest Divisible Digit Product II
+
+Hard
+
+You are given a string `num` which represents a **positive** integer, and an integer `t`.
+
+A number is called **zero-free** if _none_ of its digits are 0.
+
+Return a string representing the **smallest** **zero-free** number greater than or equal to `num` such that the **product of its digits** is divisible by `t`. If no such number exists, return `"-1"`.
+
+**Example 1:**
+
+**Input:** num = "1234", t = 256
+
+**Output:** "1488"
+
+**Explanation:**
+
+The smallest zero-free number that is greater than 1234 and has the product of its digits divisible by 256 is 1488, with the product of its digits equal to 256.
+
+**Example 2:**
+
+**Input:** num = "12355", t = 50
+
+**Output:** "12355"
+
+**Explanation:**
+
+12355 is already zero-free and has the product of its digits divisible by 50, with the product of its digits equal to 150.
+
+**Example 3:**
+
+**Input:** num = "11111", t = 26
+
+**Output:** "-1"
+
+**Explanation:**
+
+No number greater than 11111 has the product of its digits divisible by 26.
+
+**Constraints:**
+
+*   <code>2 <= num.length <= 2 * 10<sup>5</sup></code>
+*   `num` consists only of digits in the range `['0', '9']`.
+*   `num` does not contain leading zeros.
+*   <code>1 <= t <= 10<sup>14</sup></code>

src/test/kotlin/g1101_1200/s1195_fizz_buzz_multithreaded/FizzBuzzTest.kt

@@ -39,7 +39,7 @@ internal class FizzBuzzTest {
             }
         }
             .start()
-        TimeUnit.MILLISECONDS.sleep(1600)
+        TimeUnit.MILLISECONDS.sleep(2000)
         assertThat(fizz[0] > 0, equalTo(true))
     }
 

src/test/kotlin/g3301_3400/s3345_smallest_divisible_digit_product_i/SolutionTest.kt

@@ -0,0 +1,17 @@
+package g3301_3400.s3345_smallest_divisible_digit_product_i
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun smallestNumber() {
+        assertThat<Int>(Solution().smallestNumber(10, 2), equalTo<Int>(10))
+    }
+
+    @Test
+    fun smallestNumber2() {
+        assertThat<Int>(Solution().smallestNumber(15, 3), equalTo<Int>(16))
+    }
+}