Added tasks 3127-3149

Author: javadev

README.md

@@ -125,7 +125,7 @@ public class Solution {
         solution.printList(result2);
 
         ListNode l5 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9)))))));
-        ListNode l6 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9)))));
+        ListNode l6 = new ListNode(9, new ListNode(9, new ListNode(9, new ListNode(9))));
         ListNode result3 = solution.addTwoNumbers(l5, l6);
         System.out.print("Example 3 Output: ");
         solution.printList(result3);

src/main/java/g0001_0100/s0002_add_two_numbers/readme.md

@@ -53,36 +53,48 @@ import java.util.ArrayList;
 import java.util.List;
 
 public class Solution {
+    private static final int SEG_COUNT = 4;
+    private List<String> result = new ArrayList<>();
+    private int[] segments = new int[SEG_COUNT];
+
     public List<String> restoreIpAddresses(String s) {
-        List<String> results = new ArrayList<>();
-        step(s, 0, new int[4], 0, results);
-        return results;
+        dfs(s, 0, 0);
+        return result;
     }
 
-    void step(String s, int pos, int[] octets, int count, List<String> results) {
-        if (count == 4 && pos == s.length()) {
-            results.add(
-                    String.valueOf(octets[0])
-                            + '.'
-                            + octets[1]
-                            + '.'
-                            + octets[2]
-                            + '.'
-                            + octets[3]);
-        } else if (count < 4 && pos < 12) {
-            int octet = 0;
-            for (int i = 0; i < 3; i++) {
-                if (pos + i < s.length()) {
-                    int digit = s.charAt(pos + i) - '0';
-                    octet = octet * 10 + digit;
-                    if (octet < 256) {
-                        octets[count] = octet;
-                        step(s, pos + i + 1, octets, count + 1, results);
-                    }
-                    if (i == 0 && digit == 0) {
-                        break;
+    public void dfs(String s, int segId, int segStart) {
+        // find 4 segments and get to last index
+        if (segId == SEG_COUNT) {
+            if (segStart == s.length()) {
+                StringBuilder addr = new StringBuilder();
+                for (int i = 0; i < SEG_COUNT; i++) {
+                    addr.append(segments[i]);
+                    if (i != SEG_COUNT - 1) {
+                        addr.append('.');
                     }
                 }
+                result.add(addr.toString());
+            }
+            return;
+        }
+        // last index and no 4 segments
+        if (segStart == s.length()) {
+            return;
+        }
+        // start with a zero
+        if (s.charAt(segStart) == '0') {
+            segments[segId] = 0;
+            dfs(s, segId + 1, segStart + 1);
+            return;
+        }
+        int addr = 0;
+        for (int index = segStart; index < s.length(); index++) {
+            addr = addr * 10 + s.charAt(index) - '0';
+            if (addr >= 0 && addr <= 255) {
+                segments[segId] = addr;
+                dfs(s, segId + 1, index + 1);
+            } else {
+                break;
             }
         }
     }

src/main/java/g0001_0100/s0093_restore_ip_addresses/readme.md

@@ -66,7 +66,7 @@ class Solution {
     }
 
     // Recursive helper method to construct binary tree
-    private TreeNode build(int[] preorder, int[] inorder, int preStart, preEnd, int inStart, int inEnd) {
+    private TreeNode build(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {
         if (preStart > preEnd || inStart > inEnd) return null; // Base case
 
         int rootValue = preorder[preStart]; // Root node value

src/main/java/g0101_0200/s0105_construct_binary_tree_from_preorder_and_inorder_traversal/readme.md

@@ -56,6 +56,7 @@ import java.util.Deque;
 public class MyQueue {
     private Deque<Integer> left;
     private Deque<Integer> right;
+
     // Initialize your data structure here.
     public MyQueue() {
         left = new ArrayDeque<>();

src/main/java/g0201_0300/s0232_implement_queue_using_stacks/readme.md

@@ -72,6 +72,7 @@ public class Solution {
                 bit[i] += v;
             }
         }
+
         // prefix sum query
         private int ps(int j) {
             int ps = 0;

src/main/java/g0301_0400/s0315_count_of_smaller_numbers_after_self/readme.md

@@ -39,59 +39,32 @@ There is no way to assign the variables to satisfy both equations.
 ## Solution
 
 ```java
-import java.util.HashMap;
-
 public class Solution {
-    private int[] par;
+    private int[] parent = new int[26];
 
-    public boolean equationsPossible(String[] equations) {
-        int counter = 0;
-        HashMap<Character, Integer> map = new HashMap<>();
-        for (String str : equations) {
-            char ch = str.charAt(0);
-            if (!map.containsKey(ch)) {
-                map.put(ch, counter);
-                counter++;
-            }
-            ch = str.charAt(3);
-            if (!map.containsKey(ch)) {
-                map.put(ch, counter);
-                counter++;
-            }
+    private int find(int x) {
+        if (parent[x] == x) {
+            return x;
         }
-        par = new int[counter];
-        for (int i = 0; i < par.length; i++) {
-            par[i] = i;
+        parent[x] = find(parent[x]);
+        return parent[x];
+    }
+
+    public boolean equationsPossible(String[] equations) {
+        for (int i = 0; i < 26; i++) {
+            parent[i] = i;
         }
-        for (String str : equations) {
-            String oper = str.substring(1, 3);
-            if (oper.equals("==")) {
-                int px = find(map.get(str.charAt(0)));
-                int py = find(map.get(str.charAt(3)));
-                if (px != py) {
-                    par[px] = py;
-                }
+        for (String e : equations) {
+            if (e.charAt(1) == '=') {
+                parent[find(e.charAt(0) - 'a')] = find(e.charAt(3) - 'a');
             }
         }
-        for (String str : equations) {
-            String oper = str.substring(1, 3);
-            if (oper.equals("!=")) {
-                int px = find(map.get(str.charAt(0)));
-                int py = find(map.get(str.charAt(3)));
-                if (px == py) {
-                    return false;
-                }
+        for (String e : equations) {
+            if (e.charAt(1) == '!' && find(e.charAt(0) - 'a') == find(e.charAt(3) - 'a')) {
+                return false;
             }
         }
         return true;
     }
-
-    private int find(int x) {
-        if (par[x] == x) {
-            return x;
-        }
-        par[x] = find(par[x]);
-        return par[x];
-    }
 }
 ```

src/main/java/g0901_1000/s0990_satisfiability_of_equality_equations/readme.md

@@ -32,29 +32,30 @@ where the maximum is taken over all `0 <= i, j < arr1.length`.
 
 ```java
 public class Solution {
-    public int maxAbsValExpr(int[] arr1, int[] arr2) {
-        if (arr1.length != arr2.length) {
+    private int max(int[] a1, int[] a2, int k1, int k2, int k3) {
+        int result = Integer.MIN_VALUE;
+        for (int i = 0; i < a1.length; i++) {
+            result = Math.max(result, a1[i] * k1 + a2[i] * k2 + i * k3);
+        }
+        return result;
+    }
+
+    private int min(int[] a1, int[] a2, int k1, int k2, int k3) {
+        return -max(a1, a2, -k1, -k2, -k3);
+    }
+
+    public int maxAbsValExpr(int[] a1, int[] a2) {
+        if (a1 == null || a2 == null || a1.length == 0 || a2.length == 0) {
             return 0;
         }
-        int max1 = Integer.MIN_VALUE;
-        int max2 = Integer.MIN_VALUE;
-        int max3 = Integer.MIN_VALUE;
-        int max4 = Integer.MIN_VALUE;
-        int min1 = Integer.MAX_VALUE;
-        int min2 = Integer.MAX_VALUE;
-        int min3 = Integer.MAX_VALUE;
-        int min4 = Integer.MAX_VALUE;
-        for (int i = 0; i < arr1.length; i++) {
-            max1 = Math.max(arr1[i] + arr2[i] + i, max1);
-            min1 = Math.min(arr1[i] + arr2[i] + i, min1);
-            max2 = Math.max(i - arr1[i] - arr2[i], max2);
-            min2 = Math.min(i - arr1[i] - arr2[i], min2);
-            max3 = Math.max(arr1[i] - arr2[i] + i, max3);
-            min3 = Math.min(arr1[i] - arr2[i] + i, min3);
-            max4 = Math.max(arr2[i] - arr1[i] + i, max4);
-            min4 = Math.min(arr2[i] - arr1[i] + i, min4);
+        int result = 0;
+        int[][] ksArray = { {1, 1, 1}, {1, 1, -1}, {1, -1, 1}, {1, -1, -1}};
+        for (int[] ks : ksArray) {
+            int max = max(a1, a2, ks[0], ks[1], ks[2]);
+            int min = min(a1, a2, ks[0], ks[1], ks[2]);
+            result = Math.max(result, max - min);
         }
-        return Math.max(Math.max(max1 - min1, max2 - min2), Math.max(max3 - min3, max4 - min4));
+        return result;
     }
 }
 ```

src/main/java/g1101_1200/s1131_maximum_of_absolute_value_expression/readme.md

@@ -72,6 +72,7 @@ public class Solution {
         { {0, -1}, {-1, 0}},
         { {0, 1}, {-1, 0}}
     };
+
     // the idea is you need to check port direction match, you can go to next cell and check whether
     // you can come back.
     public boolean hasValidPath(int[][] grid) {

src/main/java/g1301_1400/s1391_check_if_there_is_a_valid_path_in_a_grid/readme.md

@@ -67,7 +67,7 @@ public class Solution {
     private int ans = 0;
 
     public int maximumRows(int[][] matrix, int numSelect) {
-        dfs(matrix, /*colIndex=*/ 0, numSelect, /*mask=*/ 0);
+        dfs(matrix, /* colIndex= */ 0, numSelect, /* mask= */ 0);
         return ans;
     }
 

src/main/java/g2301_2400/s2397_maximum_rows_covered_by_columns/readme.md

@@ -0,0 +1,78 @@
+[![](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)
+
+## 3127\. Make a Square with the Same Color
+
+Easy
+
+You are given a 2D matrix `grid` of size `3 x 3` consisting only of characters `'B'` and `'W'`. Character `'W'` represents the white color, and character `'B'` represents the black color.
+
+Your task is to change the color of **at most one** cell so that the matrix has a `2 x 2` square where all cells are of the same color.
+
+Return `true` if it is possible to create a `2 x 2` square of the same color, otherwise, return `false`.
+
+**Example 1:**
+
+**Input:** grid = \[\["B","W","B"],["B","W","W"],["B","W","B"]]
+
+**Output:** true
+
+**Explanation:**
+
+It can be done by changing the color of the `grid[0][2]`.
+
+**Example 2:**
+
+**Input:** grid = \[\["B","W","B"],["W","B","W"],["B","W","B"]]
+
+**Output:** false
+
+**Explanation:**
+
+It cannot be done by changing at most one cell.
+
+**Example 3:**
+
+**Input:** grid = \[\["B","W","B"],["B","W","W"],["B","W","W"]]
+
+**Output:** true
+
+**Explanation:**
+
+The `grid` already contains a `2 x 2` square of the same color.
+
+**Constraints:**
+
+*   `grid.length == 3`
+*   `grid[i].length == 3`
+*   `grid[i][j]` is either `'W'` or `'B'`.
+
+## Solution
+
+```java
+public class Solution {
+    public boolean canMakeSquare(char[][] grid) {
+        int n = grid.length;
+        int m = grid[0].length;
+        for (int i = 0; i < n - 1; i++) {
+            for (int j = 0; j < m - 1; j++) {
+                int countBlack = 0;
+                int countWhite = 0;
+                for (int k = i; k <= i + 1; k++) {
+                    for (int l = j; l <= j + 1; l++) {
+                        if (grid[k][l] == 'W') {
+                            countWhite++;
+                        } else {
+                            countBlack++;
+                        }
+                    }
+                }
+                if (countBlack >= 3 || countWhite >= 3) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+}
+```