Added tasks 3127-3134

src/main/java/g3101_3200/s3127_make_a_square_with_the_same_color/Solution.java

@@ -0,0 +1,29 @@
+package g3101_3200.s3127_make_a_square_with_the_same_color;
+
+// #Easy #Array #Matrix #Enumeration #2024_05_02_Time_0_ms_(100.00%)_Space_41.7_MB_(64.59%)
+
+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;
+    }
+}

src/main/java/g3101_3200/s3127_make_a_square_with_the_same_color/readme.md

@@ -0,0 +1,45 @@
+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'`.

src/main/java/g3101_3200/s3128_right_triangles/Solution.java

@@ -0,0 +1,26 @@
+package g3101_3200.s3128_right_triangles;
+
+// #Medium #Array #Hash_Table #Math #Counting #Combinatorics
+// #2024_05_02_Time_6_ms_(100.00%)_Space_145.9_MB_(90.67%)
+
+public class Solution {
+    public long numberOfRightTriangles(int[][] grid) {
+        int n = grid.length;
+        int m = grid[0].length;
+        int[] columns = new int[n];
+        int[] rows = new int[m];
+        long sum = 0;
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                columns[i] += grid[i][j];
+                rows[j] += grid[i][j];
+            }
+        }
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                sum += (long) grid[i][j] * (rows[j] - 1) * (columns[i] - 1);
+            }
+        }
+        return sum;
+    }
+}

src/main/java/g3101_3200/s3128_right_triangles/readme.md

@@ -0,0 +1,77 @@
+3128\. Right Triangles
+
+Medium
+
+You are given a 2D boolean matrix `grid`.
+
+Return an integer that is the number of **right triangles** that can be made with the 3 elements of `grid` such that **all** of them have a value of 1.
+
+**Note:**
+
+*   A collection of 3 elements of `grid` is a **right triangle** if one of its elements is in the **same row** with another element and in the **same column** with the third element. The 3 elements do not have to be next to each other.
+
+**Example 1:**
+
+0 **1** 0
+
+0 **1 1**
+
+0 1 0
+
+0 1 0
+
+0 **1 1**
+
+0 **1** 0
+
+**Input:** grid = [[0,1,0],[0,1,1],[0,1,0]]
+
+**Output:** 2
+
+**Explanation:**
+
+There are two right triangles.
+
+**Example 2:**
+
+1 0 0 0
+
+0 1 0 1
+
+1 0 0 0
+
+**Input:** grid = [[1,0,0,0],[0,1,0,1],[1,0,0,0]]
+
+**Output:** 0
+
+**Explanation:**
+
+There are no right triangles.
+
+**Example 3:**
+
+**1** 0 **1**
+
+**1** 0 0
+
+1 0 0
+
+**1** 0 **1**
+
+1 0 0
+
+**1** 0 0
+
+**Input:** grid = [[1,0,1],[1,0,0],[1,0,0]]
+
+**Output: **2
+
+**Explanation:**
+
+There are two right triangles.
+
+**Constraints:**
+
+*   `1 <= grid.length <= 1000`
+*   `1 <= grid[i].length <= 1000`
+*   `0 <= grid[i][j] <= 1`

src/main/java/g3101_3200/s3129_find_all_possible_stable_binary_arrays_i/Solution.java

@@ -0,0 +1,53 @@
+package g3101_3200.s3129_find_all_possible_stable_binary_arrays_i;
+
+// #Medium #Dynamic_Programming #Prefix_Sum #2024_05_02_Time_3_ms_(100.00%)_Space_44.1_MB_(98.38%)
+
+public class Solution {
+    private static final int MODULUS = (int) 1e9 + 7;
+
+    private int add(int x, int y) {
+        return (x + y) % MODULUS;
+    }
+
+    private int subtract(int x, int y) {
+        return (x + MODULUS - y) % MODULUS;
+    }
+
+    private int multiply(int x, int y) {
+        return (int) ((long) x * y % MODULUS);
+    }
+
+    public int numberOfStableArrays(int zero, int one, int limit) {
+        if (limit == 1) {
+            return Math.max(2 - Math.abs(zero - one), 0);
+        }
+        int max = Math.max(zero, one);
+        int min = Math.min(zero, one);
+        int[][] lcn = new int[max + 1][max + 1];
+        int[] row0 = lcn[0];
+        int[] row1;
+        int[] row2;
+        row0[0] = 1;
+        for (int s = 1, sLim = s - limit; s <= max; s++, sLim++) {
+            row2 = sLim > 0 ? lcn[sLim - 1] : new int[] {};
+            row1 = row0;
+            row0 = lcn[s];
+            int c;
+            for (c = (s - 1) / limit + 1; c <= sLim; c++) {
+                row0[c] = subtract(add(row1[c], row1[c - 1]), row2[c - 1]);
+            }
+            for (; c <= s; c++) {
+                row0[c] = add(row1[c], row1[c - 1]);
+            }
+        }
+        row1 = lcn[min];
+        int result = 0;
+        int s0 = add(min < max ? row0[min + 1] : 0, row0[min]);
+        for (int c = min; c > 0; c--) {
+            int s1 = s0;
+            s0 = add(row0[c], row0[c - 1]);
+            result = add(result, multiply(row1[c], add(s0, s1)));
+        }
+        return result;
+    }
+}

src/main/java/g3101_3200/s3129_find_all_possible_stable_binary_arrays_i/readme.md

@@ -0,0 +1,51 @@
+3129\. Find All Possible Stable Binary Arrays I
+
+Medium
+
+You are given 3 positive integers `zero`, `one`, and `limit`.
+
+A binary array `arr` is called **stable** if:
+
+*   The number of occurrences of 0 in `arr` is **exactly** `zero`.
+*   The number of occurrences of 1 in `arr` is **exactly** `one`.
+*   Each subarray of `arr` with a size greater than `limit` must contain **both** 0 and 1.
+
+Return the _total_ number of **stable** binary arrays.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** zero = 1, one = 1, limit = 2
+
+**Output:** 2
+
+**Explanation:**
+
+The two possible stable binary arrays are `[1,0]` and `[0,1]`, as both arrays have a single 0 and a single 1, and no subarray has a length greater than 2.
+
+**Example 2:**
+
+**Input:** zero = 1, one = 2, limit = 1
+
+**Output:** 1
+
+**Explanation:**
+
+The only possible stable binary array is `[1,0,1]`.
+
+Note that the binary arrays `[1,1,0]` and `[0,1,1]` have subarrays of length 2 with identical elements, hence, they are not stable.
+
+**Example 3:**
+
+**Input:** zero = 3, one = 3, limit = 2
+
+**Output:** 14
+
+**Explanation:**
+
+All the possible stable binary arrays are `[0,0,1,0,1,1]`, `[0,0,1,1,0,1]`, `[0,1,0,0,1,1]`, `[0,1,0,1,0,1]`, `[0,1,0,1,1,0]`, `[0,1,1,0,0,1]`, `[0,1,1,0,1,0]`, `[1,0,0,1,0,1]`, `[1,0,0,1,1,0]`, `[1,0,1,0,0,1]`, `[1,0,1,0,1,0]`, `[1,0,1,1,0,0]`, `[1,1,0,0,1,0]`, and `[1,1,0,1,0,0]`.
+
+**Constraints:**
+
+*   `1 <= zero, one, limit <= 200`

src/main/java/g3101_3200/s3130_find_all_possible_stable_binary_arrays_ii/Solution.java

@@ -0,0 +1,85 @@
+package g3101_3200.s3130_find_all_possible_stable_binary_arrays_ii;
+
+// #Hard #Dynamic_Programming #Prefix_Sum #2024_05_02_Time_3_ms_(100.00%)_Space_40.6_MB_(100.00%)
+
+public class Solution {
+    private static final int MOD = (int) 1e9 + 7;
+    private static final int N = 1000;
+    private long[] factorial;
+    private long[] reverse;
+
+    public int numberOfStableArrays(int zero, int one, int limit) {
+        if (factorial == null) {
+            factorial = new long[N + 1];
+            reverse = new long[N + 1];
+            factorial[0] = 1;
+            reverse[0] = 1;
+            long x = 1;
+            for (int i = 1; i <= N; ++i) {
+                x = (x * i) % MOD;
+                factorial[i] = (int) x;
+                reverse[i] = getInverse(x, MOD);
+            }
+        }
+        long ans = 0;
+        long[] s = new long[one + 1];
+        int n = Math.min(zero, one) + 1;
+        for (int groups0 = (zero + limit - 1) / limit; groups0 <= Math.min(zero, n); ++groups0) {
+            long s0 = calc(groups0, zero, limit);
+            for (int groups1 = Math.max(groups0 - 1, (one + limit - 1) / limit);
+                    groups1 <= Math.min(groups0 + 1, one);
+                    ++groups1) {
+                long s1;
+                if (s[groups1] != 0) {
+                    s1 = s[groups1];
+                } else {
+                    s1 = s[groups1] = calc(groups1, one, limit);
+                }
+                ans = (ans + s0 * s1 * (groups1 == groups0 ? 2 : 1)) % MOD;
+            }
+        }
+        return (int) ((ans + MOD) % MOD);
+    }
+
+    long calc(int groups, int x, int limit) {
+        long s = 0;
+        int sign = 1;
+        for (int k = 0; k * limit <= x - groups && k <= groups; k++) {
+            s = (s + sign * comb(groups, k) * comb(x - k * limit - 1, groups - 1)) % MOD;
+            sign *= -1;
+        }
+        return s;
+    }
+
+    public long comb(int n, int k) {
+        return (factorial[n] * reverse[k] % MOD) * reverse[n - k] % MOD;
+    }
+
+    public long getInverse(long n, long mod) {
+        long p = mod;
+        long x = 1;
+        long y = 0;
+        while (p > 0) {
+            long quotient = n / p;
+            long remainder = n % p;
+            long tempY = x - quotient * y;
+            x = y;
+            y = tempY;
+            n = p;
+            p = remainder;
+        }
+        return ((x % mod) + mod) % mod;
+    }
+
+    public long quickPower(long base, long power, long p) {
+        long result = 1;
+        while (power > 0) {
+            if ((power & 1) == 1) {
+                result = result * base % p;
+            }
+            power >>= 1;
+            base = base * base % p;
+        }
+        return result;
+    }
+}