Added tasks 3375-3378

Author: javadev

src/main/kotlin/g3301_3400/s3375_minimum_operations_to_make_array_values_equal_to_k/Solution.kt

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+package g3301_3400.s3375_minimum_operations_to_make_array_values_equal_to_k
+
+// #Easy #2024_12_08_Time_191_ms_(100.00%)_Space_39.9_MB_(100.00%)
+
+class Solution {
+    fun minOperations(nums: IntArray, k: Int): Int {
+        val s: MutableSet<Int?> = HashSet<Int?>()
+        for (i in nums) {
+            s.add(i)
+        }
+        var res = 0
+        for (i in s) {
+            if (i!! > k) {
+                res++
+            } else if (i < k) {
+                return -1
+            }
+        }
+        return res
+    }
+}

src/main/kotlin/g3301_3400/s3375_minimum_operations_to_make_array_values_equal_to_k/readme.md

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+3375\. Minimum Operations to Make Array Values Equal to K
+
+Easy
+
+You are given an integer array `nums` and an integer `k`.
+
+An integer `h` is called **valid** if all values in the array that are **strictly greater** than `h` are _identical_.
+
+For example, if `nums = [10, 8, 10, 8]`, a **valid** integer is `h = 9` because all `nums[i] > 9` are equal to 10, but 5 is not a **valid** integer.
+
+You are allowed to perform the following operation on `nums`:
+
+*   Select an integer `h` that is _valid_ for the **current** values in `nums`.
+*   For each index `i` where `nums[i] > h`, set `nums[i]` to `h`.
+
+Return the **minimum** number of operations required to make every element in `nums` **equal** to `k`. If it is impossible to make all elements equal to `k`, return -1.
+
+**Example 1:**
+
+**Input:** nums = [5,2,5,4,5], k = 2
+
+**Output:** 2
+
+**Explanation:**
+
+The operations can be performed in order using valid integers 4 and then 2.
+
+**Example 2:**
+
+**Input:** nums = [2,1,2], k = 2
+
+**Output:** \-1
+
+**Explanation:**
+
+It is impossible to make all the values equal to 2.
+
+**Example 3:**
+
+**Input:** nums = [9,7,5,3], k = 1
+
+**Output:** 4
+
+**Explanation:**
+
+The operations can be performed using valid integers in the order 7, 5, 3, and 1.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`
+*   `1 <= k <= 100`

src/main/kotlin/g3301_3400/s3376_minimum_time_to_break_locks_i/Solution.kt

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+package g3301_3400.s3376_minimum_time_to_break_locks_i
+
+// #Medium #2024_12_08_Time_202_ms_(100.00%)_Space_40_MB_(100.00%)
+
+import kotlin.math.min
+
+class Solution {
+    fun findMinimumTime(strength: List<Int>, k: Int): Int {
+        val perm: MutableList<Int> = ArrayList<Int>(strength)
+        perm.sort()
+        var minTime = Int.Companion.MAX_VALUE
+        do {
+            var time = 0
+            var factor = 1
+            for (required in perm) {
+                val neededTime = (required + factor - 1) / factor
+                time += neededTime
+                factor += k
+            }
+            minTime = min(minTime, time)
+        } while (nextPermutation(perm))
+        return minTime
+    }
+
+    private fun nextPermutation(nums: MutableList<Int>): Boolean {
+        var i = nums.size - 2
+        while (i >= 0 && nums[i] >= nums[i + 1]) {
+            i--
+        }
+        if (i < 0) {
+            return false
+        }
+        var j = nums.size - 1
+        while (nums[j] <= nums[i]) {
+            j--
+        }
+        nums[i] = nums.set(j, nums[i])
+        nums.subList(i + 1, nums.size).reverse()
+        return true
+    }
+}

src/main/kotlin/g3301_3400/s3376_minimum_time_to_break_locks_i/readme.md

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+3376\. Minimum Time to Break Locks I
+
+Medium
+
+Bob is stuck in a dungeon and must break `n` locks, each requiring some amount of **energy** to break. The required energy for each lock is stored in an array called `strength` where `strength[i]` indicates the energy needed to break the <code>i<sup>th</sup></code> lock.
+
+To break a lock, Bob uses a sword with the following characteristics:
+
+*   The initial energy of the sword is 0.
+*   The initial factor `X` by which the energy of the sword increases is 1.
+*   Every minute, the energy of the sword increases by the current factor `X`.
+*   To break the <code>i<sup>th</sup></code> lock, the energy of the sword must reach **at least** `strength[i]`.
+*   After breaking a lock, the energy of the sword resets to 0, and the factor `X` increases by a given value `K`.
+
+Your task is to determine the **minimum** time in minutes required for Bob to break all `n` locks and escape the dungeon.
+
+Return the **minimum** time required for Bob to break all `n` locks.
+
+**Example 1:**
+
+**Input:** strength = [3,4,1], K = 1
+
+**Output:** 4
+
+**Explanation:**
+
+| Time | Energy | X | Action               | Updated X |
+|------|--------|---|----------------------|-----------|
+| 0    | 0      | 1 | Nothing              | 1         |
+| 1    | 1      | 1 | Break 3rd Lock       | 2         |
+| 2    | 2      | 2 | Nothing              | 2         |
+| 3    | 4      | 2 | Break 2nd Lock       | 3         |
+| 4    | 3      | 3 | Break 1st Lock       | 3         |
+
+The locks cannot be broken in less than 4 minutes; thus, the answer is 4.
+
+**Example 2:**
+
+**Input:** strength = [2,5,4], K = 2
+
+**Output:** 5
+
+**Explanation:**
+
+| Time | Energy | X | Action               | Updated X |
+|------|--------|---|----------------------|-----------|
+| 0    | 0      | 1 | Nothing              | 1         |
+| 1    | 1      | 1 | Nothing              | 1         |
+| 2    | 2      | 1 | Break 1st Lock       | 3         |
+| 3    | 3      | 3 | Nothing              | 3         |
+| 4    | 6      | 3 | Break 2nd Lock       | 5         |
+| 5    | 5      | 5 | Break 3rd Lock       | 7         |
+
+The locks cannot be broken in less than 5 minutes; thus, the answer is 5.
+
+**Constraints:**
+
+*   `n == strength.length`
+*   `1 <= n <= 8`
+*   `1 <= K <= 10`
+*   <code>1 <= strength[i] <= 10<sup>6</sup></code>

src/main/kotlin/g3301_3400/s3377_digit_operations_to_make_two_integers_equal/Solution.kt

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+package g3301_3400.s3377_digit_operations_to_make_two_integers_equal
+
+// #Medium #2024_12_08_Time_215_ms_(100.00%)_Space_40.7_MB_(100.00%)
+
+import java.util.PriorityQueue
+
+class Solution {
+    fun minOperations(n: Int, m: Int): Int {
+        val limit = 100000
+        val sieve = BooleanArray(limit + 1)
+        val visited = BooleanArray(limit)
+        sieve.fill(true)
+        sieve[0] = false
+        sieve[1] = false
+        var i = 2
+        while (i * i <= limit) {
+            if (sieve[i]) {
+                var j = i * i
+                while (j <= limit) {
+                    sieve[j] = false
+                    j += i
+                }
+            }
+            i++
+        }
+        if (sieve[n]) {
+            return -1
+        }
+        val pq = PriorityQueue<IntArray>(Comparator { a: IntArray, b: IntArray -> a[0] - b[0] })
+        visited[n] = true
+        pq.add(intArrayOf(n, n))
+        while (!pq.isEmpty()) {
+            val current = pq.poll()
+            val cost = current[0]
+            val num = current[1]
+            val temp = num.toString().toCharArray()
+            if (num == m) {
+                return cost
+            }
+            for (j in temp.indices) {
+                val old = temp[j]
+                for (i in -1..1) {
+                    val digit = old.code - '0'.code
+                    if ((digit == 9 && i == 1) || (digit == 0 && i == -1)) {
+                        continue
+                    }
+                    temp[j] = (i + digit + '0'.code).toChar()
+                    val newNum = String(temp).toInt()
+                    if (!sieve[newNum] && !visited[newNum]) {
+                        visited[newNum] = true
+                        pq.add(intArrayOf(cost + newNum, newNum))
+                    }
+                }
+                temp[j] = old
+            }
+        }
+        return -1
+    }
+}

src/main/kotlin/g3301_3400/s3377_digit_operations_to_make_two_integers_equal/readme.md

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+3377\. Digit Operations to Make Two Integers Equal
+
+Medium
+
+You are given two integers `n` and `m` that consist of the **same** number of digits.
+
+You can perform the following operations **any** number of times:
+
+*   Choose **any** digit from `n` that is not 9 and **increase** it by 1.
+*   Choose **any** digit from `n` that is not 0 and **decrease** it by 1.
+
+The integer `n` must not be a **prime** number at any point, including its original value and after each operation.
+
+The cost of a transformation is the sum of **all** values that `n` takes throughout the operations performed.
+
+Return the **minimum** cost to transform `n` into `m`. If it is impossible, return -1.
+
+A prime number is a natural number greater than 1 with only two factors, 1 and itself.
+
+**Example 1:**
+
+**Input:** n = 10, m = 12
+
+**Output:** 85
+
+**Explanation:**
+
+We perform the following operations:
+
+*   Increase the first digit, now <code>n = <ins>**2**</ins>0</code>.
+*   Increase the second digit, now <code>n = 2**<ins>1</ins>**</code>.
+*   Increase the second digit, now <code>n = 2**<ins>2</ins>**</code>.
+*   Decrease the first digit, now <code>n = **<ins>1</ins>**2</code>.
+
+**Example 2:**
+
+**Input:** n = 4, m = 8
+
+**Output:** \-1
+
+**Explanation:**
+
+It is impossible to make `n` equal to `m`.
+
+**Example 3:**
+
+**Input:** n = 6, m = 2
+
+**Output:** \-1
+
+**Explanation:**
+
+Since 2 is already a prime, we can't make `n` equal to `m`.
+
+**Constraints:**
+
+*   <code>1 <= n, m < 10<sup>4</sup></code>
+*   `n` and `m` consist of the same number of digits.

src/main/kotlin/g3301_3400/s3378_count_connected_components_in_lcm_graph/Solution.kt

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+package g3301_3400.s3378_count_connected_components_in_lcm_graph
+
+// #Hard #2024_12_08_Time_58_ms_(100.00%)_Space_54.4_MB_(100.00%)
+
+class Solution {
+    private class UnionFind(n: Int) {
+        var parent = IntArray(n) { it }
+        var rank = IntArray(n)
+        var totalComponents = n
+
+        fun find(u: Int): Int {
+            if (parent[u] == u) {
+                return u
+            }
+            parent[u] = find(parent[u])
+            return parent[u]
+        }
+
+        fun union(u: Int, v: Int) {
+            val parentU = find(u)
+            val parentV = find(v)
+            if (parentU != parentV) {
+                totalComponents--
+                when {
+                    rank[parentU] == rank[parentV] -> {
+                        parent[parentV] = parentU
+                        rank[parentU]++
+                    }
+                    rank[parentU] > rank[parentV] -> parent[parentV] = parentU
+                    else -> parent[parentU] = parentV
+                }
+            }
+        }
+    }
+
+    fun countComponents(nums: IntArray, threshold: Int): Int {
+        val goodNums = nums.filter { it <= threshold }
+        val totalNums = nums.size
+        if (goodNums.isEmpty()) {
+            return totalNums
+        }
+        val uf = UnionFind(goodNums.size)
+        val presentElements = IntArray(threshold + 1) { -1 }
+        goodNums.forEachIndexed { index, num ->
+            presentElements[num] = index
+        }
+        for (d in goodNums) {
+            for (i in d..threshold step d) {
+                if (presentElements[i] == -1) {
+                    presentElements[i] = presentElements[d]
+                } else if (presentElements[i] != presentElements[d]) {
+                    uf.union(presentElements[i], presentElements[d])
+                }
+            }
+        }
+        return uf.totalComponents + totalNums - goodNums.size
+    }
+}