Added tasks 3127-3134

Author: javadev

src/main/kotlin/g3101_3200/s3127_make_a_square_with_the_same_color/Solution.kt

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+package g3101_3200.s3127_make_a_square_with_the_same_color
+
+// #Easy #Array #Matrix #Enumeration #2024_05_02_Time_149_ms_(80.00%)_Space_35.1_MB_(40.00%)
+
+class Solution {
+    fun canMakeSquare(grid: Array<CharArray>): Boolean {
+        val n = grid.size
+        val m = grid[0].size
+        for (i in 0 until n - 1) {
+            for (j in 0 until m - 1) {
+                var countBlack = 0
+                var countWhite = 0
+                for (k in i..i + 1) {
+                    for (l in j..j + 1) {
+                        if (grid[k][l] == 'W') {
+                            countWhite++
+                        } else {
+                            countBlack++
+                        }
+                    }
+                }
+                if (countBlack >= 3 || countWhite >= 3) {
+                    return true
+                }
+            }
+        }
+        return false
+    }
+}

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

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+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/kotlin/g3101_3200/s3128_right_triangles/Solution.kt

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+package g3101_3200.s3128_right_triangles
+
+// #Medium #Array #Hash_Table #Math #Counting #Combinatorics
+// #2024_05_02_Time_975_ms_(40.63%)_Space_217.6_MB_(56.25%)
+
+class Solution {
+    fun numberOfRightTriangles(grid: Array<IntArray>): Long {
+        val n = grid.size
+        val m = grid[0].size
+        val columns = IntArray(n)
+        val rows = IntArray(m)
+        var sum: Long = 0
+        for (i in 0 until n) {
+            for (j in 0 until m) {
+                columns[i] += grid[i][j]
+                rows[j] += grid[i][j]
+            }
+        }
+        for (i in 0 until n) {
+            for (j in 0 until m) {
+                sum += grid[i][j].toLong() * (rows[j] - 1) * (columns[i] - 1)
+            }
+        }
+        return sum
+    }
+}

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

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+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/kotlin/g3101_3200/s3129_find_all_possible_stable_binary_arrays_i/Solution.kt

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+package g3101_3200.s3129_find_all_possible_stable_binary_arrays_i
+
+// #Medium #Dynamic_Programming #Prefix_Sum #2024_05_02_Time_169_ms_(92.86%)_Space_36.3_MB_(100.00%)
+
+import kotlin.math.abs
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    private fun add(x: Int, y: Int): Int {
+        return (x + y) % MODULUS
+    }
+
+    private fun subtract(x: Int, y: Int): Int {
+        return (x + MODULUS - y) % MODULUS
+    }
+
+    private fun multiply(x: Int, y: Int): Int {
+        return (x.toLong() * y % MODULUS).toInt()
+    }
+
+    fun numberOfStableArrays(zero: Int, one: Int, limit: Int): Int {
+        if (limit == 1) {
+            return max((2 - abs((zero - one))), 0)
+        }
+        val max = max(zero, one)
+        val min = min(zero, one)
+        val lcn = Array(max + 1) { IntArray(max + 1) }
+        var row0 = lcn[0]
+        var row1: IntArray
+        var row2: IntArray
+        row0[0] = 1
+        var s = 1
+        var sLim = s - limit
+        while (s <= max) {
+            row2 = if (sLim > 0) lcn[sLim - 1] else intArrayOf()
+            row1 = row0
+            row0 = lcn[s]
+            var c = (s - 1) / limit + 1
+            while (c <= sLim) {
+                row0[c] = subtract(add(row1[c], row1[c - 1]), row2[c - 1])
+                c++
+            }
+            while (c <= s) {
+                row0[c] = add(row1[c], row1[c - 1])
+                c++
+            }
+            s++
+            sLim++
+        }
+        row1 = lcn[min]
+        var result = 0
+        var s0 = add(if (min < max) row0[min + 1] else 0, row0[min])
+        for (c in min downTo 1) {
+            val s1 = s0
+            s0 = add(row0[c], row0[c - 1])
+            result = add(result, multiply(row1[c], add(s0, s1)))
+        }
+        return result
+    }
+
+    companion object {
+        private const val MODULUS = 1e9.toInt() + 7
+    }
+}

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

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+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/kotlin/g3101_3200/s3130_find_all_possible_stable_binary_arrays_ii/Solution.kt

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+package g3101_3200.s3130_find_all_possible_stable_binary_arrays_ii
+
+// #Hard #Dynamic_Programming #Prefix_Sum #2024_05_02_Time_242_ms_(100.00%)_Space_36.7_MB_(100.00%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    private var factorial: LongArray? = null
+    private lateinit var reverse: LongArray
+
+    fun numberOfStableArrays(zero: Int, one: Int, limit: Int): Int {
+        if (factorial == null) {
+            factorial = LongArray(N + 1)
+            reverse = LongArray(N + 1)
+            factorial!![0] = 1
+            reverse[0] = 1
+            var x: Long = 1
+            for (i in 1..N) {
+                x = (x * i) % MOD
+                factorial!![i] = x.toInt().toLong()
+                reverse[i] = getInverse(x, MOD.toLong())
+            }
+        }
+        var ans: Long = 0
+        val s = LongArray(one + 1)
+        val n = (min(zero, one) + 1).toInt()
+        for (
+            groups0 in (zero + limit - 1) / limit..min(zero, n)
+                .toInt()
+        ) {
+            val s0 = calc(groups0, zero, limit)
+            for (
+                groups1 in max(
+                    groups0 - 1,
+                    (one + limit - 1) / limit
+                )..min((groups0 + 1), one)
+            ) {
+                var s1: Long
+                if (s[groups1] != 0L) {
+                    s1 = s[groups1]
+                } else {
+                    s[groups1] = calc(groups1, one, limit)
+                    s1 = s[groups1]
+                }
+                ans = (ans + s0 * s1 * (if (groups1 == groups0) 2 else 1)) % MOD
+            }
+        }
+        return ((ans + MOD) % MOD).toInt()
+    }
+
+    fun calc(groups: Int, x: Int, limit: Int): Long {
+        var s: Long = 0
+        var sign = 1
+        var k = 0
+        while (k * limit <= x - groups && k <= groups) {
+            s = (s + sign * comb(groups, k) * comb(x - k * limit - 1, groups - 1)) % MOD
+            sign *= -1
+            k++
+        }
+        return s
+    }
+
+    fun comb(n: Int, k: Int): Long {
+        return (factorial!![n] * reverse[k] % MOD) * reverse[n - k] % MOD
+    }
+
+    fun getInverse(n: Long, mod: Long): Long {
+        var n = n
+        var p = mod
+        var x: Long = 1
+        var y: Long = 0
+        while (p > 0) {
+            val quotient = n / p
+            val remainder = n % p
+            val tempY = x - quotient * y
+            x = y
+            y = tempY
+            n = p
+            p = remainder
+        }
+        return ((x % mod) + mod) % mod
+    }
+
+    companion object {
+        private const val MOD = 1e9.toInt() + 7
+        private const val N = 1000
+    }
+}