Added tasks 3324-3327

src/main/java/g3301_3400/s3324_find_the_sequence_of_strings_appeared_on_the_screen/Solution.java

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+package g3301_3400.s3324_find_the_sequence_of_strings_appeared_on_the_screen;
+
+// #Medium #String #Simulation #2024_10_22_Time_6_ms_(92.04%)_Space_55.7_MB_(44.25%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public List<String> stringSequence(String t) {
+        List<String> ans = new ArrayList<>();
+        int l = t.length();
+        StringBuilder cur = new StringBuilder();
+        for (int i = 0; i < l; i++) {
+            char tCh = t.charAt(i);
+            cur.append('a');
+            ans.add(cur.toString());
+            while (cur.charAt(i) != tCh) {
+                char lastCh = cur.charAt(i);
+                char nextCh = (char) (lastCh == 'z' ? 'a' : lastCh + 1);
+                cur.setCharAt(i, nextCh);
+                ans.add(cur.toString());
+            }
+        }
+        return ans;
+    }
+}

src/main/java/g3301_3400/s3324_find_the_sequence_of_strings_appeared_on_the_screen/readme.md

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+3324\. Find the Sequence of Strings Appeared on the Screen
+
+Medium
+
+You are given a string `target`.
+
+Alice is going to type `target` on her computer using a special keyboard that has **only two** keys:
+
+*   Key 1 appends the character `"a"` to the string on the screen.
+*   Key 2 changes the **last** character of the string on the screen to its **next** character in the English alphabet. For example, `"c"` changes to `"d"` and `"z"` changes to `"a"`.
+
+**Note** that initially there is an _empty_ string `""` on the screen, so she can **only** press key 1.
+
+Return a list of _all_ strings that appear on the screen as Alice types `target`, in the order they appear, using the **minimum** key presses.
+
+**Example 1:**
+
+**Input:** target = "abc"
+
+**Output:** ["a","aa","ab","aba","abb","abc"]
+
+**Explanation:**
+
+The sequence of key presses done by Alice are:
+
+*   Press key 1, and the string on the screen becomes `"a"`.
+*   Press key 1, and the string on the screen becomes `"aa"`.
+*   Press key 2, and the string on the screen becomes `"ab"`.
+*   Press key 1, and the string on the screen becomes `"aba"`.
+*   Press key 2, and the string on the screen becomes `"abb"`.
+*   Press key 2, and the string on the screen becomes `"abc"`.
+
+**Example 2:**
+
+**Input:** target = "he"
+
+**Output:** ["a","b","c","d","e","f","g","h","ha","hb","hc","hd","he"]
+
+**Constraints:**
+
+*   `1 <= target.length <= 400`
+*   `target` consists only of lowercase English letters.

src/main/java/g3301_3400/s3325_count_substrings_with_k_frequency_characters_i/Solution.java

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+package g3301_3400.s3325_count_substrings_with_k_frequency_characters_i;
+
+// #Medium #String #Hash_Table #Sliding_Window #2024_10_22_Time_1_ms_(100.00%)_Space_42_MB_(98.69%)
+
+public class Solution {
+    public int numberOfSubstrings(String s, int k) {
+        int left = 0;
+        int result = 0;
+        int[] count = new int[26];
+        for (int i = 0; i < s.length(); i++) {
+            char ch = s.charAt(i);
+            count[ch - 'a']++;
+
+            while (count[ch - 'a'] == k) {
+                result += s.length() - i;
+                char atLeft = s.charAt(left);
+                count[atLeft - 'a']--;
+                left++;
+            }
+        }
+        return result;
+    }
+}

src/main/java/g3301_3400/s3325_count_substrings_with_k_frequency_characters_i/readme.md

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+3325\. Count Substrings With K-Frequency Characters I
+
+Medium
+
+Given a string `s` and an integer `k`, return the total number of substrings of `s` where **at least one** character appears **at least** `k` times.
+
+**Example 1:**
+
+**Input:** s = "abacb", k = 2
+
+**Output:** 4
+
+**Explanation:**
+
+The valid substrings are:
+
+*   `"aba"` (character `'a'` appears 2 times).
+*   `"abac"` (character `'a'` appears 2 times).
+*   `"abacb"` (character `'a'` appears 2 times).
+*   `"bacb"` (character `'b'` appears 2 times).
+
+**Example 2:**
+
+**Input:** s = "abcde", k = 1
+
+**Output:** 15
+
+**Explanation:**
+
+All substrings are valid because every character appears at least once.
+
+**Constraints:**
+
+*   `1 <= s.length <= 3000`
+*   `1 <= k <= s.length`
+*   `s` consists only of lowercase English letters.

src/main/java/g3301_3400/s3326_minimum_division_operations_to_make_array_non_decreasing/Solution.java

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+package g3301_3400.s3326_minimum_division_operations_to_make_array_non_decreasing;
+
+// #Medium #Array #Math #Greedy #Number_Theory #2024_10_22_Time_20_ms_(97.34%)_Space_73.1_MB_(5.03%)
+
+public class Solution {
+    private static final int MAXI = 1000001;
+    private static final int[] SIEVE = new int[MAXI];
+    private static boolean precompute = false;
+
+    private static void compute() {
+        if (precompute) {
+            return;
+        }
+        for (int i = 2; i < MAXI; i++) {
+            if (i * i > MAXI) {
+                break;
+            }
+            for (int j = i * i; j < MAXI; j += i) {
+                SIEVE[j] = Math.max(SIEVE[j], Math.max(i, j / i));
+            }
+        }
+        precompute = true;
+    }
+
+    public int minOperations(int[] nums) {
+        compute();
+        int op = 0;
+        int n = nums.length;
+        for (int i = n - 2; i >= 0; i--) {
+            while (nums[i] > nums[i + 1]) {
+                if (SIEVE[nums[i]] == 0) {
+                    return -1;
+                }
+                nums[i] /= SIEVE[nums[i]];
+                op++;
+            }
+            if (nums[i] > nums[i + 1]) {
+                return -1;
+            }
+        }
+        return op;
+    }
+}

src/main/java/g3301_3400/s3326_minimum_division_operations_to_make_array_non_decreasing/readme.md

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+3326\. Minimum Division Operations to Make Array Non Decreasing
+
+Medium
+
+You are given an integer array `nums`.
+
+Any **positive** divisor of a natural number `x` that is **strictly less** than `x` is called a **proper divisor** of `x`. For example, 2 is a _proper divisor_ of 4, while 6 is not a _proper divisor_ of 6.
+
+You are allowed to perform an **operation** any number of times on `nums`, where in each **operation** you select any _one_ element from `nums` and divide it by its **greatest** **proper divisor**.
+
+Return the **minimum** number of **operations** required to make the array **non-decreasing**.
+
+If it is **not** possible to make the array _non-decreasing_ using any number of operations, return `-1`.
+
+**Example 1:**
+
+**Input:** nums = [25,7]
+
+**Output:** 1
+
+**Explanation:**
+
+Using a single operation, 25 gets divided by 5 and `nums` becomes `[5, 7]`.
+
+**Example 2:**
+
+**Input:** nums = [7,7,6]
+
+**Output:** \-1
+
+**Example 3:**
+
+**Input:** nums = [1,1,1,1]
+
+**Output:** 0
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>

src/main/java/g3301_3400/s3327_check_if_dfs_strings_are_palindromes/Solution.java

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+package g3301_3400.s3327_check_if_dfs_strings_are_palindromes;
+
+// #Hard #Array #String #Hash_Table #Tree #Hash_Function #Depth_First_Search
+// #2024_10_22_Time_159_ms_(90.40%)_Space_93.9_MB_(80.80%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    private final List<List<Integer>> e = new ArrayList<>();
+    private final StringBuilder stringBuilder = new StringBuilder();
+    private String s;
+    private int now;
+    private int n;
+    private int[] l;
+    private int[] r;
+    private int[] p;
+    private char[] c;
+
+    private void dfs(int x) {
+        l[x] = now + 1;
+        for (int v : e.get(x)) {
+            dfs(v);
+        }
+        stringBuilder.append(s.charAt(x));
+        r[x] = ++now;
+    }
+
+    private void matcher() {
+        c[0] = '~';
+        c[1] = '#';
+        for (int i = 1; i <= n; ++i) {
+            c[2 * i + 1] = '#';
+            c[2 * i] = stringBuilder.charAt(i - 1);
+        }
+        int j = 1;
+        int mid = 0;
+        int localR = 0;
+        while (j <= 2 * n + 1) {
+            if (j <= localR) {
+                p[j] = Math.min(p[(mid << 1) - j], localR - j + 1);
+            }
+            while (c[j - p[j]] == c[j + p[j]]) {
+                ++p[j];
+            }
+            if (p[j] + j > localR) {
+                localR = p[j] + j - 1;
+                mid = j;
+            }
+            ++j;
+        }
+    }
+
+    public boolean[] findAnswer(int[] parent, String s) {
+        n = parent.length;
+        this.s = s;
+        for (int i = 0; i < n; ++i) {
+            e.add(new ArrayList<>());
+        }
+        for (int i = 1; i < n; ++i) {
+            e.get(parent[i]).add(i);
+        }
+        l = new int[n];
+        r = new int[n];
+        dfs(0);
+        c = new char[2 * n + 10];
+        p = new int[2 * n + 10];
+        matcher();
+        boolean[] ans = new boolean[n];
+        for (int i = 0; i < n; ++i) {
+            int mid = (2 * r[i] - 2 * l[i] + 1) / 2 + 2 * l[i];
+            ans[i] = p[mid] - 1 >= r[i] - l[i] + 1;
+        }
+        return ans;
+    }
+}

src/main/java/g3301_3400/s3327_check_if_dfs_strings_are_palindromes/readme.md

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+3327\. Check if DFS Strings Are Palindromes
+
+Hard
+
+You are given a tree rooted at node 0, consisting of `n` nodes numbered from `0` to `n - 1`. The tree is represented by an array `parent` of size `n`, where `parent[i]` is the parent of node `i`. Since node 0 is the root, `parent[0] == -1`.
+
+You are also given a string `s` of length `n`, where `s[i]` is the character assigned to node `i`.
+
+Consider an empty string `dfsStr`, and define a recursive function `dfs(int x)` that takes a node `x` as a parameter and performs the following steps in order:
+
+*   Iterate over each child `y` of `x` **in increasing order of their numbers**, and call `dfs(y)`.
+*   Add the character `s[x]` to the end of the string `dfsStr`.
+
+**Note** that `dfsStr` is shared across all recursive calls of `dfs`.
+
+You need to find a boolean array `answer` of size `n`, where for each index `i` from `0` to `n - 1`, you do the following:
+
+*   Empty the string `dfsStr` and call `dfs(i)`.
+*   If the resulting string `dfsStr` is a **palindrome**, then set `answer[i]` to `true`. Otherwise, set `answer[i]` to `false`.
+
+Return the array `answer`.
+
+A **palindrome** is a string that reads the same forward and backward.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2024/09/01/tree1drawio.png)
+
+**Input:** parent = [-1,0,0,1,1,2], s = "aababa"
+
+**Output:** [true,true,false,true,true,true]
+
+**Explanation:**
+
+*   Calling `dfs(0)` results in the string `dfsStr = "abaaba"`, which is a palindrome.
+*   Calling `dfs(1)` results in the string `dfsStr = "aba"`, which is a palindrome.
+*   Calling `dfs(2)` results in the string `dfsStr = "ab"`, which is **not** a palindrome.
+*   Calling `dfs(3)` results in the string `dfsStr = "a"`, which is a palindrome.
+*   Calling `dfs(4)` results in the string `dfsStr = "b"`, which is a palindrome.
+*   Calling `dfs(5)` results in the string `dfsStr = "a"`, which is a palindrome.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2024/09/01/tree2drawio-1.png)
+
+**Input:** parent = [-1,0,0,0,0], s = "aabcb"
+
+**Output:** [true,true,true,true,true]
+
+**Explanation:**
+
+Every call on `dfs(x)` results in a palindrome string.
+
+**Constraints:**
+
+*   `n == parent.length == s.length`
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   `0 <= parent[i] <= n - 1` for all `i >= 1`.
+*   `parent[0] == -1`
+*   `parent` represents a valid tree.
+*   `s` consists only of lowercase English letters.

src/test/java/g3301_3400/s3324_find_the_sequence_of_strings_appeared_on_the_screen/SolutionTest.java

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+package g3301_3400.s3324_find_the_sequence_of_strings_appeared_on_the_screen;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import java.util.List;
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void stringSequence() {
+        assertThat(
+                new Solution().stringSequence("abc"),
+                equalTo(List.of("a", "aa", "ab", "aba", "abb", "abc")));
+    }
+
+    @Test
+    void stringSequence2() {
+        assertThat(
+                new Solution().stringSequence("he"),
+                equalTo(
+                        List.of(
+                                "a", "b", "c", "d", "e", "f", "g", "h", "ha", "hb", "hc", "hd",
+                                "he")));
+    }
+}