Added tasks 3527-3534

src/main/java/g3501_3600/s3527_find_the_most_common_response/Solution.java

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+package g3501_3600.s3527_find_the_most_common_response;
+
+// #Medium #Array #String #Hash_Table #Counting
+// #2025_04_28_Time_94_ms_(100.00%)_Space_211.70_MB_(22.07%)
+
+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/s3527_find_the_most_common_response/readme.md

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+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

src/main/java/g3501_3600/s3528_unit_conversion_i/Solution.java

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+package g3501_3600.s3528_unit_conversion_i;
+
+// #Medium #Depth_First_Search #Breadth_First_Search #Graph
+// #2025_04_28_Time_3_ms_(99.90%)_Space_127.84_MB_(26.65%)
+
+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/s3528_unit_conversion_i/readme.md

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+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.

src/main/java/g3501_3600/s3529_count_cells_in_overlapping_horizontal_and_vertical_substrings/Solution.java

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+package g3501_3600.s3529_count_cells_in_overlapping_horizontal_and_vertical_substrings;
+
+// #Medium #Array #String #Matrix #Hash_Function #String_Matching #Rolling_Hash
+// #2025_04_28_Time_33_ms_(100.00%)_Space_62.71_MB_(100.00%)
+
+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;
+    }
+}

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

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+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.

src/main/java/g3501_3600/s3530_maximum_profit_from_valid_topological_order_in_dag/Solution.java

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+package g3501_3600.s3530_maximum_profit_from_valid_topological_order_in_dag;
+
+// #Hard #Array #Dynamic_Programming #Bit_Manipulation #Graph #Bitmask #Topological_Sort
+// #2025_04_28_Time_1927_ms_(100.00%)_Space_66.86_MB_(100.00%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class Solution {
+    private int helper(
+            int mask,
+            int pos,
+            int[] inDegree,
+            List<List<Integer>> adj,
+            int[] score,
+            int[] dp,
+            int n) {
+        if (mask == (1 << n) - 1) {
+            return 0;
+        }
+        if (dp[mask] != -1) {
+            return dp[mask];
+        }
+        int res = 0;
+        for (int i = 0; i < n; i++) {
+            if ((mask & (1 << i)) == 0 && inDegree[i] == 0) {
+                for (int ng : adj.get(i)) {
+                    inDegree[ng]--;
+                }
+                int val =
+                        pos * score[i]
+                                + helper(mask | (1 << i), pos + 1, inDegree, adj, score, dp, n);
+                res = Math.max(res, val);
+                for (int ng : adj.get(i)) {
+                    inDegree[ng]++;
+                }
+            }
+        }
+        dp[mask] = res;
+        return res;
+    }
+
+    public int maxProfit(int n, int[][] edges, int[] score) {
+        List<List<Integer>> adj = new ArrayList<>();
+        for (int i = 0; i < n; i++) {
+            adj.add(new ArrayList<>());
+        }
+        int[] inDegree = new int[n];
+        for (int[] e : edges) {
+            adj.get(e[0]).add(e[1]);
+            inDegree[e[1]]++;
+        }
+        int[] dp = new int[1 << n];
+        Arrays.fill(dp, -1);
+        return helper(0, 1, inDegree, adj, score, dp, n);
+    }
+}