Added tasks 3375-3382

src/main/java/g3301_3400/s3375_minimum_operations_to_make_array_values_equal_to_k/Solution.java

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+package g3301_3400.s3375_minimum_operations_to_make_array_values_equal_to_k;
+
+// #Easy #Array #Hash_Table #2024_12_10_Time_3_ms_(78.92%)_Space_44.6_MB_(67.39%)
+
+import java.util.HashSet;
+import java.util.Set;
+
+public class Solution {
+    public int minOperations(int[] nums, int k) {
+        Set<Integer> s = new HashSet<>();
+        for (int i : nums) {
+            s.add(i);
+        }
+        int res = 0;
+        for (int i : s) {
+            if (i > k) {
+                res++;
+            } else if (i < k) {
+                return -1;
+            }
+        }
+        return res;
+    }
+}

src/main/java/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/java/g3301_3400/s3376_minimum_time_to_break_locks_i/Solution.java

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+package g3301_3400.s3376_minimum_time_to_break_locks_i;
+
+// #Medium #Array #Dynamic_Programming #Bit_Manipulation #Backtracking #Bitmask
+// #2024_12_10_Time_3_ms_(99.63%)_Space_42_MB_(92.34%)
+
+import java.util.ArrayList;
+import java.util.Collections;
+import java.util.List;
+
+public class Solution {
+    public int findMinimumTime(List<Integer> strength, int k) {
+        List<Integer> strengthLocal = new ArrayList<>(strength);
+        Collections.sort(strengthLocal);
+        int res = strengthLocal.get(0);
+        strengthLocal.remove(0);
+        int x = 1;
+        while (!strengthLocal.isEmpty()) {
+            x += k;
+            int nextTime = (strengthLocal.get(0) - 1) / x + 1;
+            int canBreak = nextTime * x;
+            int indexRemove = findIndex(strengthLocal, canBreak);
+            if (strengthLocal.size() > 1) {
+                int nextTime1 = (strengthLocal.get(1) - 1) / x + 1;
+                int canBreak1 = nextTime1 * x;
+                int indexRemove1 = findIndex(strengthLocal, canBreak1);
+                if (nextTime1 + (strengthLocal.get(0) - 1) / (x + k)
+                        < nextTime + (strengthLocal.get(1) - 1) / (x + k)) {
+                    nextTime = nextTime1;
+                    indexRemove = indexRemove1;
+                }
+            }
+            res += nextTime;
+            strengthLocal.remove(indexRemove);
+        }
+        return res;
+    }
+
+    private int findIndex(List<Integer> strength, int canBreak) {
+        int l = 0;
+        int r = strength.size() - 1;
+        int res = -1;
+        while (l <= r) {
+            int mid = (l + r) / 2;
+            if (strength.get(mid) <= canBreak) {
+                res = mid;
+                l = mid + 1;
+            } else {
+                r = mid - 1;
+            }
+        }
+        return res;
+    }
+}

src/main/java/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/java/g3301_3400/s3377_digit_operations_to_make_two_integers_equal/Solution.java

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+package g3301_3400.s3377_digit_operations_to_make_two_integers_equal;
+
+// #Medium #Math #Heap_Priority_Queue #Graph #Shortest_Path #Number_Theory
+// #2024_12_10_Time_246_ms_(38.59%)_Space_45.2_MB_(73.52%)
+
+import java.util.Arrays;
+import java.util.PriorityQueue;
+
+public class Solution {
+    public int minOperations(int n, int m) {
+        int limit = 100000;
+        boolean[] sieve = new boolean[limit + 1];
+        boolean[] visited = new boolean[limit];
+        Arrays.fill(sieve, true);
+        sieve[0] = false;
+        sieve[1] = false;
+        for (int i = 2; i * i <= limit; i++) {
+            if (sieve[i]) {
+                for (int j = i * i; j <= limit; j += i) {
+                    sieve[j] = false;
+                }
+            }
+        }
+        if (sieve[n]) {
+            return -1;
+        }
+        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
+        visited[n] = true;
+        pq.add(new int[] {n, n});
+        while (!pq.isEmpty()) {
+            int[] current = pq.poll();
+            int cost = current[0];
+            int num = current[1];
+            char[] temp = Integer.toString(num).toCharArray();
+            if (num == m) {
+                return cost;
+            }
+            for (int j = 0; j < temp.length; j++) {
+                char old = temp[j];
+                for (int i = -1; i <= 1; i++) {
+                    int digit = old - '0';
+                    if ((digit == 9 && i == 1) || (digit == 0 && i == -1)) {
+                        continue;
+                    }
+                    temp[j] = (char) (i + digit + '0');
+                    int newnum = Integer.parseInt(new String(temp));
+                    if (!sieve[newnum] && !visited[newnum]) {
+                        visited[newnum] = true;
+                        pq.add(new int[] {cost + newnum, newnum});
+                    }
+                }
+                temp[j] = old;
+            }
+        }
+        return -1;
+    }
+}

src/main/java/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/java/g3301_3400/s3378_count_connected_components_in_lcm_graph/Solution.java

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+package g3301_3400.s3378_count_connected_components_in_lcm_graph;
+
+// #Hard #Array #Hash_Table #Math #Union_Find #Number_Theory
+// #2024_12_10_Time_68_ms_(67.83%)_Space_59.8_MB_(62.24%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class Solution {
+    private static class Unionfind {
+        int[] parent;
+        int[] rank;
+        int totalComponents;
+
+        public Unionfind(int n) {
+            parent = new int[n];
+            rank = new int[n];
+            totalComponents = n;
+            for (int i = 0; i < n; i++) {
+                parent[i] = i;
+            }
+        }
+
+        public int find(int u) {
+            if (parent[u] == u) {
+                return u;
+            }
+            parent[u] = find(parent[u]);
+            return parent[u];
+        }
+
+        public void union(int u, int v) {
+            int parentU = find(u);
+            int parentV = find(v);
+            if (parentU != parentV) {
+                totalComponents--;
+                if (rank[parentU] == rank[parentV]) {
+                    parent[parentV] = parentU;
+                    rank[parentU]++;
+                } else if (rank[parentU] > rank[parentV]) {
+                    parent[parentV] = parentU;
+                } else {
+                    parent[parentU] = parentV;
+                }
+            }
+        }
+    }
+
+    public int countComponents(int[] nums, int threshold) {
+        List<Integer> goodNums = new ArrayList<>();
+        int totalNums = nums.length;
+        for (int num : nums) {
+            if (num <= threshold) {
+                goodNums.add(num);
+            }
+        }
+        if (goodNums.isEmpty()) {
+            return totalNums;
+        }
+        Unionfind uf = new Unionfind(goodNums.size());
+        int[] presentElements = new int[threshold + 1];
+        Arrays.fill(presentElements, -1);
+        for (int i = 0; i < goodNums.size(); i++) {
+            presentElements[goodNums.get(i)] = i;
+        }
+        for (int d : goodNums) {
+            for (int i = d; i <= threshold; i += 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();
+    }
+}