Added tasks 3527-3534

Author: javadev

README.md

@@ -2088,6 +2088,14 @@
 
 | #    |      Title     | Difficulty  | Tag         | Time, ms | Time, %
 |------|----------------|-------------|-------------|----------|--------
+| 3534 |[Path Existence Queries in a Graph II](src/main/java/g3501_3600/s3534_path_existence_queries_in_a_graph_ii)| Hard | Array, Sorting, Greedy, Binary_Search, Graph | 84 | 100.00
+| 3533 |[Concatenated Divisibility](src/main/java/g3501_3600/s3533_concatenated_divisibility)| Hard | Array, Dynamic_Programming, Bit_Manipulation, Bitmask | 14 | 100.00
+| 3532 |[Path Existence Queries in a Graph I](src/main/java/g3501_3600/s3532_path_existence_queries_in_a_graph_i)| Medium | Array, Binary_Search, Graph, Union_Find | 3 | 100.00
+| 3531 |[Count Covered Buildings](src/main/java/g3501_3600/s3531_count_covered_buildings)| Medium | Array, Hash_Table, Sorting | 12 | 100.00
+| 3530 |[Maximum Profit from Valid Topological Order in DAG](src/main/java/g3501_3600/s3530_maximum_profit_from_valid_topological_order_in_dag)| Hard | Array, Dynamic_Programming, Bit_Manipulation, Graph, Bitmask, Topological_Sort | 1927 | 100.00
+| 3529 |[Count Cells in Overlapping Horizontal and Vertical Substrings](src/main/java/g3501_3600/s3529_count_cells_in_overlapping_horizontal_and_vertical_substrings)| Medium | Array, String, Matrix, Hash_Function, String_Matching, Rolling_Hash | 33 | 100.00
+| 3528 |[Unit Conversion I](src/main/java/g3501_3600/s3528_unit_conversion_i)| Medium | Depth_First_Search, Breadth_First_Search, Graph | 3 | 99.90
+| 3527 |[Find the Most Common Response](src/main/java/g3501_3600/s3527_find_the_most_common_response)| Medium | Array, String, Hash_Table, Counting | 94 | 100.00
 | 3525 |[Find X Value of Array II](src/main/java/g3501_3600/s3525_find_x_value_of_array_ii)| Hard | Array, Math, Segment_Tree | 177 | 79.87
 | 3524 |[Find X Value of Array I](src/main/java/g3501_3600/s3524_find_x_value_of_array_i)| Medium | Array, Dynamic_Programming, Math | 12 | 95.54
 | 3523 |[Make Array Non-decreasing](src/main/java/g3501_3600/s3523_make_array_non_decreasing)| Medium | Array, Greedy, Stack, Monotonic_Stack | 3 | 63.29

src/main/java/g3501_3600/s3527_find_the_most_common_response/readme.md

@@ -0,0 +1,93 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Java?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Java)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Java?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Java/fork)
+
+## 3527\. Find the Most Common Response
+
+Medium
+
+You are given a 2D string array `responses` where each `responses[i]` is an array of strings representing survey responses from the <code>i<sup>th</sup></code> day.
+
+Return the **most common** response across all days after removing **duplicate** responses within each `responses[i]`. If there is a tie, return the _lexicographically smallest_ response.
+
+**Example 1:**
+
+**Input:** responses = \[\["good","ok","good","ok"],["ok","bad","good","ok","ok"],["good"],["bad"]]
+
+**Output:** "good"
+
+**Explanation:**
+
+*   After removing duplicates within each list, `responses = \[\["good", "ok"], ["ok", "bad", "good"], ["good"], ["bad"]]`.
+*   `"good"` appears 3 times, `"ok"` appears 2 times, and `"bad"` appears 2 times.
+*   Return `"good"` because it has the highest frequency.
+
+**Example 2:**
+
+**Input:** responses = \[\["good","ok","good"],["ok","bad"],["bad","notsure"],["great","good"]]
+
+**Output:** "bad"
+
+**Explanation:**
+
+*   After removing duplicates within each list we have `responses = \[\["good", "ok"], ["ok", "bad"], ["bad", "notsure"], ["great", "good"]]`.
+*   `"bad"`, `"good"`, and `"ok"` each occur 2 times.
+*   The output is `"bad"` because it is the lexicographically smallest amongst the words with the highest frequency.
+
+**Constraints:**
+
+*   `1 <= responses.length <= 1000`
+*   `1 <= responses[i].length <= 1000`
+*   `1 <= responses[i][j].length <= 10`
+*   `responses[i][j]` consists of only lowercase English letters
+
+## Solution
+
+```java
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+public class Solution {
+    private boolean compareStrings(String str1, String str2) {
+        int n = str1.length();
+        int m = str2.length();
+        int i = 0;
+        int j = 0;
+        while (i < n && j < m) {
+            if (str1.charAt(i) < str2.charAt(j)) {
+                return true;
+            } else if (str1.charAt(i) > str2.charAt(j)) {
+                return false;
+            }
+            i++;
+            j++;
+        }
+        return n < m;
+    }
+
+    public String findCommonResponse(List<List<String>> responses) {
+        int n = responses.size();
+        Map<String, int[]> mp = new HashMap<>();
+        String ans = responses.get(0).get(0);
+        int maxFreq = 0;
+        for (int row = 0; row < n; row++) {
+            int m = responses.get(row).size();
+            for (int col = 0; col < m; col++) {
+                String resp = responses.get(row).get(col);
+                int[] arr = mp.getOrDefault(resp, new int[] {0, -1});
+                if (arr[1] != row) {
+                    arr[0]++;
+                    arr[1] = row;
+                    mp.put(resp, arr);
+                }
+                if (arr[0] > maxFreq
+                        || !ans.equals(resp) && arr[0] == maxFreq && compareStrings(resp, ans)) {
+                    ans = resp;
+                    maxFreq = arr[0];
+                }
+            }
+        }
+        return ans;
+    }
+}
+```

src/main/java/g3501_3600/s3528_unit_conversion_i/readme.md

@@ -0,0 +1,63 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Java?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Java)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Java?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Java/fork)
+
+## 3528\. Unit Conversion I
+
+Medium
+
+There are `n` types of units indexed from `0` to `n - 1`. You are given a 2D integer array `conversions` of length `n - 1`, where <code>conversions[i] = [sourceUnit<sub>i</sub>, targetUnit<sub>i</sub>, conversionFactor<sub>i</sub>]</code>. This indicates that a single unit of type <code>sourceUnit<sub>i</sub></code> is equivalent to <code>conversionFactor<sub>i</sub></code> units of type <code>targetUnit<sub>i</sub></code>.
+
+Return an array `baseUnitConversion` of length `n`, where `baseUnitConversion[i]` is the number of units of type `i` equivalent to a single unit of type 0. Since the answer may be large, return each `baseUnitConversion[i]` **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** conversions = \[\[0,1,2],[1,2,3]]
+
+**Output:** [1,2,6]
+
+**Explanation:**
+
+*   Convert a single unit of type 0 into 2 units of type 1 using `conversions[0]`.
+*   Convert a single unit of type 0 into 6 units of type 2 using `conversions[0]`, then `conversions[1]`.
+
+![](https://assets.leetcode.com/uploads/2025/03/12/example1.png)
+
+**Example 2:**
+
+**Input:** conversions = \[\[0,1,2],[0,2,3],[1,3,4],[1,4,5],[2,5,2],[4,6,3],[5,7,4]]
+
+**Output:** [1,2,3,8,10,6,30,24]
+
+**Explanation:**
+
+*   Convert a single unit of type 0 into 2 units of type 1 using `conversions[0]`.
+*   Convert a single unit of type 0 into 3 units of type 2 using `conversions[1]`.
+*   Convert a single unit of type 0 into 8 units of type 3 using `conversions[0]`, then `conversions[2]`.
+*   Convert a single unit of type 0 into 10 units of type 4 using `conversions[0]`, then `conversions[3]`.
+*   Convert a single unit of type 0 into 6 units of type 5 using `conversions[1]`, then `conversions[4]`.
+*   Convert a single unit of type 0 into 30 units of type 6 using `conversions[0]`, `conversions[3]`, then `conversions[5]`.
+*   Convert a single unit of type 0 into 24 units of type 7 using `conversions[1]`, `conversions[4]`, then `conversions[6]`.
+
+**Constraints:**
+
+*   <code>2 <= n <= 10<sup>5</sup></code>
+*   `conversions.length == n - 1`
+*   <code>0 <= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> < n</code>
+*   <code>1 <= conversionFactor<sub>i</sub> <= 10<sup>9</sup></code>
+*   It is guaranteed that unit 0 can be converted into any other unit through a **unique** combination of conversions without using any conversions in the opposite direction.
+
+## Solution
+
+```java
+public class Solution {
+    public int[] baseUnitConversions(int[][] conversions) {
+        int[] arr = new int[conversions.length + 1];
+        arr[0] = 1;
+        for (int[] conversion : conversions) {
+            long val = ((long) arr[conversion[0]] * conversion[2]) % 1000000007;
+            arr[conversion[1]] = (int) val;
+        }
+        return arr;
+    }
+}
+```

src/main/java/g3501_3600/s3529_count_cells_in_overlapping_horizontal_and_vertical_substrings/readme.md

@@ -0,0 +1,142 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Java?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Java)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Java?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Java/fork)
+
+## 3529\. Count Cells in Overlapping Horizontal and Vertical Substrings
+
+Medium
+
+You are given an `m x n` matrix `grid` consisting of characters and a string `pattern`.
+
+A **horizontal substring** is a contiguous sequence of characters read from left to right. If the end of a row is reached before the substring is complete, it wraps to the first column of the next row and continues as needed. You do **not** wrap from the bottom row back to the top.
+
+A **vertical substring** is a contiguous sequence of characters read from top to bottom. If the bottom of a column is reached before the substring is complete, it wraps to the first row of the next column and continues as needed. You do **not** wrap from the last column back to the first.
+
+Count the number of cells in the matrix that satisfy the following condition:
+
+*   The cell must be part of **at least** one horizontal substring and **at least** one vertical substring, where **both** substrings are equal to the given `pattern`.
+
+Return the count of these cells.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2025/03/03/gridtwosubstringsdrawio.png)
+
+**Input:** grid = \[\["a","a","c","c"],["b","b","b","c"],["a","a","b","a"],["c","a","a","c"],["a","a","c","c"]], pattern = "abaca"
+
+**Output:** 1
+
+**Explanation:**
+
+The pattern `"abaca"` appears once as a horizontal substring (colored blue) and once as a vertical substring (colored red), intersecting at one cell (colored purple).
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2025/03/03/gridexample2fixeddrawio.png)
+
+**Input:** grid = \[\["c","a","a","a"],["a","a","b","a"],["b","b","a","a"],["a","a","b","a"]], pattern = "aba"
+
+**Output:** 4
+
+**Explanation:**
+
+The cells colored above are all part of at least one horizontal and one vertical substring matching the pattern `"aba"`.
+
+**Example 3:**
+
+**Input:** grid = \[\["a"]], pattern = "a"
+
+**Output:** 1
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   `1 <= m, n <= 1000`
+*   <code>1 <= m * n <= 10<sup>5</sup></code>
+*   `1 <= pattern.length <= m * n`
+*   `grid` and `pattern` consist of only lowercase English letters.
+
+## Solution
+
+```java
+public class Solution {
+    public int countCells(char[][] grid, String pattern) {
+        int k = pattern.length();
+        int[] lps = makeLps(pattern);
+        int m = grid.length;
+        int n = grid[0].length;
+        int[][] horiPats = new int[m][n];
+        int[][] vertPats = new int[m][n];
+        int i = 0;
+        int j = 0;
+        while (i < m * n) {
+            if (grid[i / n][i % n] == pattern.charAt(j)) {
+                i++;
+                if (++j == k) {
+                    int d = i - j;
+                    horiPats[d / n][d % n]++;
+                    if (i < m * n) {
+                        horiPats[i / n][i % n]--;
+                    }
+                    j = lps[j - 1];
+                }
+            } else if (j != 0) {
+                j = lps[j - 1];
+            } else {
+                i++;
+            }
+        }
+        i = 0;
+        j = 0;
+        // now do vert pattern, use i = 0 to m*n -1 but instead index as grid[i % m][i/m]
+        while (i < m * n) {
+            if (grid[i % m][i / m] == pattern.charAt(j)) {
+                i++;
+                if (++j == k) {
+                    int d = i - j;
+                    vertPats[d % m][d / m]++;
+                    if (i < m * n) {
+                        vertPats[i % m][i / m]--;
+                    }
+                    j = lps[j - 1];
+                }
+            } else if (j != 0) {
+                j = lps[j - 1];
+            } else {
+                i++;
+            }
+        }
+        for (i = 1; i < m * n; i++) {
+            vertPats[i % m][i / m] += vertPats[(i - 1) % m][(i - 1) / m];
+            horiPats[i / n][i % n] += horiPats[(i - 1) / n][(i - 1) % n];
+        }
+        int res = 0;
+        for (i = 0; i < m; i++) {
+            for (j = 0; j < n; j++) {
+                if (horiPats[i][j] > 0 && vertPats[i][j] > 0) {
+                    res++;
+                }
+            }
+        }
+        return res;
+    }
+
+    private int[] makeLps(String pattern) {
+        int n = pattern.length();
+        int[] lps = new int[n];
+        int len = 0;
+        int i = 1;
+        lps[0] = 0;
+        while (i < n) {
+            if (pattern.charAt(i) == pattern.charAt(len)) {
+                lps[i++] = ++len;
+            } else if (len != 0) {
+                len = lps[len - 1];
+            } else {
+                lps[i++] = 0;
+            }
+        }
+        return lps;
+    }
+}
+```