Added tasks 73, 74, 75, 76.

Author: ThanhNIT

src/main/kotlin/g0001_0100/s0073_set_matrix_zeroes/Solution.kt

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+package g0001_0100.s0073_set_matrix_zeroes
+
+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Hash_Table #Matrix
+// #Udemy_2D_Arrays/Matrix #2022_08_31_Time_255_ms_(100.00%)_Space_45.7_MB_(91.72%)
+
+class Solution {
+    // Approach: Use first row and first column for storing whether in future
+    //           the entire row or column needs to be marked 0
+    fun setZeroes(matrix: Array<IntArray>) {
+        val m = matrix.size
+        val n: Int = matrix[0].size
+        var row0 = false
+        var col0 = false
+        // Check if 0th col needs to be market all 0s in future
+        for (ints in matrix) {
+            if (ints[0] == 0) {
+                col0 = true
+                break
+            }
+        }
+        // Check if 0th row needs to be market all 0s in future
+        for (i in 0 until n) {
+            if (matrix[0][i] == 0) {
+                row0 = true
+                break
+            }
+        }
+        // Store the signals in 0th row and column
+        for (i in 1 until m) {
+            for (j in 1 until n) {
+                if (matrix[i][j] == 0) {
+                    matrix[i][0] = 0
+                    matrix[0][j] = 0
+                }
+            }
+        }
+        // Mark 0 for all cells based on signal from 0th row and 0th column
+        for (i in 1 until m) {
+            for (j in 1 until n) {
+                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
+                    matrix[i][j] = 0
+                }
+            }
+        }
+        // Set 0th column
+        for (i in 0 until m) {
+            if (col0) {
+                matrix[i][0] = 0
+            }
+        }
+        // Set 0th row
+        for (i in 0 until n) {
+            if (row0) {
+                matrix[0][i] = 0
+            }
+        }
+    }
+}

src/main/kotlin/g0001_0100/s0073_set_matrix_zeroes/readme.md

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+73\. Set Matrix Zeroes
+
+Medium
+
+Given an `m x n` integer matrix `matrix`, if an element is `0`, set its entire row and column to `0`'s.
+
+You must do it [in place](https://en.wikipedia.org/wiki/In-place_algorithm).
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg)
+
+**Input:** matrix = [[1,1,1],[1,0,1],[1,1,1]]
+
+**Output:** [[1,0,1],[0,0,0],[1,0,1]]
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg)
+
+**Input:** matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
+
+**Output:** [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
+
+**Constraints:**
+
+*   `m == matrix.length`
+*   `n == matrix[0].length`
+*   `1 <= m, n <= 200`
+*   <code>-2<sup>31</sup> <= matrix[i][j] <= 2<sup>31</sup> - 1</code>
+
+**Follow up:**
+
+*   A straightforward solution using `O(mn)` space is probably a bad idea.
+*   A simple improvement uses `O(m + n)` space, but still not the best solution.
+*   Could you devise a constant space solution?

src/main/kotlin/g0001_0100/s0074_search_a_2d_matrix/Solution.kt

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+package g0001_0100.s0074_search_a_2d_matrix
+
+// #Medium #Top_100_Liked_Questions #Array #Binary_Search #Matrix #Data_Structure_I_Day_5_Array
+// #Algorithm_II_Day_1_Binary_Search #Binary_Search_I_Day_8 #Level_2_Day_8_Binary_Search
+// #Udemy_2D_Arrays/Matrix #2022_08_31_Time_290_ms_(40.17%)_Space_35.4_MB_(96.48%)
+
+class Solution {
+    fun searchMatrix(matrix: Array<IntArray>, target: Int): Boolean {
+        val endRow = matrix.size
+        val endCol: Int = matrix[0].size
+        var targetRow = 0
+        var result = false
+        for (i in 0 until endRow) {
+            if (matrix[i][endCol - 1] >= target) {
+                targetRow = i
+                break
+            }
+        }
+        for (i in 0 until endCol) {
+            if (matrix[targetRow][i] == target) {
+                result = true
+                break
+            }
+        }
+        return result
+    }
+}

src/main/kotlin/g0001_0100/s0074_search_a_2d_matrix/readme.md

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+74\. Search a 2D Matrix
+
+Medium
+
+Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
+
+*   Integers in each row are sorted from left to right.
+*   The first integer of each row is greater than the last integer of the previous row.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/05/mat.jpg)
+
+**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+
+**Output:** true
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg)
+
+**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
+
+**Output:** false
+
+**Constraints:**
+
+*   `m == matrix.length`
+*   `n == matrix[i].length`
+*   `1 <= m, n <= 100`
+*   <code>-10<sup>4</sup> <= matrix[i][j], target <= 10<sup>4</sup></code>

src/main/kotlin/g0001_0100/s0075_sort_colors/Solution.kt

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+package g0001_0100.s0075_sort_colors
+
+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Sorting #Two_Pointers
+// #Data_Structure_II_Day_2_Array #Udemy_Arrays
+// #2022_08_31_Time_198_ms_(85.66%)_Space_34.8_MB_(84.84%)
+
+class Solution {
+    fun sortColors(nums: IntArray) {
+        var zeroes = 0
+        var ones = 0
+        for (i in nums.indices) {
+            if (nums[i] == 0) {
+                nums[zeroes++] = 0
+            } else if (nums[i] == 1) {
+                ones++
+            }
+        }
+        for (j in zeroes until zeroes + ones) {
+            nums[j] = 1
+        }
+        for (k in zeroes + ones until nums.size) {
+            nums[k] = 2
+        }
+    }
+}

src/main/kotlin/g0001_0100/s0075_sort_colors/readme.md

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+75\. Sort Colors
+
+Medium
+
+Given an array `nums` with `n` objects colored red, white, or blue, sort them **[in-place](https://en.wikipedia.org/wiki/In-place_algorithm)** so that objects of the same color are adjacent, with the colors in the order red, white, and blue.
+
+We will use the integers `0`, `1`, and `2` to represent the color red, white, and blue, respectively.
+
+You must solve this problem without using the library's sort function.
+
+**Example 1:**
+
+**Input:** nums = [2,0,2,1,1,0]
+
+**Output:** [0,0,1,1,2,2]
+
+**Example 2:**
+
+**Input:** nums = [2,0,1]
+
+**Output:** [0,1,2]
+
+**Constraints:**
+
+*   `n == nums.length`
+*   `1 <= n <= 300`
+*   `nums[i]` is either `0`, `1`, or `2`.
+
+**Follow up:** Could you come up with a one-pass algorithm using only constant extra space?

src/main/kotlin/g0001_0100/s0076_minimum_window_substring/Solution.kt

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+package g0001_0100.s0076_minimum_window_substring
+
+// #Hard #Top_100_Liked_Questions #Top_Interview_Questions #String #Hash_Table #Sliding_Window
+// #Level_2_Day_14_Sliding_Window/Two_Pointer
+// #2022_08_31_Time_346_ms_(85.20%)_Space_39.3_MB_(93.88%)
+
+class Solution {
+    fun minWindow(s: String, t: String): String {
+        val map = IntArray(128)
+        for (i in 0 until t.length) {
+            map[t[i] - 'A']++
+        }
+        var count = t.length
+        var begin = 0
+        var end = 0
+        var d = Int.MAX_VALUE
+        var head = 0
+        while (end < s.length) {
+            if (map[s[end++] - 'A']-- > 0) {
+                count--
+            }
+            while (count == 0) {
+                if (end - begin < d) {
+                    d = end - begin
+                    head = begin
+                }
+                if (map[s[begin++] - 'A']++ == 0) {
+                    count++
+                }
+            }
+        }
+        return if (d == Int.MAX_VALUE) "" else s.substring(head, head + d)
+    }
+}

src/main/kotlin/g0001_0100/s0076_minimum_window_substring/readme.md

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+76\. Minimum Window Substring
+
+Hard
+
+Given two strings `s` and `t` of lengths `m` and `n` respectively, return _the **minimum window substring** of_ `s` _such that every character in_ `t` _(**including duplicates**) is included in the window. If there is no such substring__, return the empty string_ `""`_._
+
+The testcases will be generated such that the answer is **unique**.
+
+A **substring** is a contiguous sequence of characters within the string.
+
+**Example 1:**
+
+**Input:** s = "ADOBECODEBANC", t = "ABC"
+
+**Output:** "BANC"
+
+**Explanation:** The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
+
+**Example 2:**
+
+**Input:** s = "a", t = "a"
+
+**Output:** "a"
+
+**Explanation:** The entire string s is the minimum window.
+
+**Example 3:**
+
+**Input:** s = "a", t = "aa"
+
+**Output:** ""
+
+**Explanation:** Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string.
+
+**Constraints:**
+
+*   `m == s.length`
+*   `n == t.length`
+*   <code>1 <= m, n <= 10<sup>5</sup></code>
+*   `s` and `t` consist of uppercase and lowercase English letters.
+
+**Follow up:** Could you find an algorithm that runs in `O(m + n)` time?

src/test/kotlin/g0001_0100/s0073_set_matrix_zeroes/SolutionTest.kt

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+package g0001_0100.s0073_set_matrix_zeroes
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun setZeroes() {
+        var array = arrayOf(intArrayOf(1, 1, 1), intArrayOf(1, 0, 1), intArrayOf(1, 1, 1))
+        var expected = arrayOf(intArrayOf(1, 0, 1), intArrayOf(0, 0, 0), intArrayOf(1, 0, 1))
+        Solution().setZeroes(array)
+        assertThat(array, equalTo(expected))
+    }
+
+    @Test
+    fun setZeroes2() {
+        var array = arrayOf(intArrayOf(0, 1, 2, 0), intArrayOf(3, 4, 5, 2), intArrayOf(1, 3, 1, 5))
+        var expected = arrayOf(intArrayOf(0, 0, 0, 0), intArrayOf(0, 4, 5, 0), intArrayOf(0, 3, 1, 0))
+        Solution().setZeroes(array)
+        assertThat(array, equalTo(expected))
+    }
+}

src/test/kotlin/g0001_0100/s0074_search_a_2d_matrix/SolutionTest.kt

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+package g0001_0100.s0074_search_a_2d_matrix
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun searchMatrix() {
+        assertThat(Solution().searchMatrix(arrayOf(intArrayOf(1, 3, 5, 7), intArrayOf(10, 11, 16, 20), intArrayOf(23, 30, 34, 60)), 3), equalTo(true))
+    }
+
+    @Test
+    fun searchMatrix2() {
+        assertThat(Solution().searchMatrix(arrayOf(intArrayOf(1, 3, 5, 7), intArrayOf(10, 11, 16, 20), intArrayOf(23, 30, 34, 60)), 13), equalTo(false))
+    }
+}

src/test/kotlin/g0001_0100/s0075_sort_colors/SolutionTest.kt

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+package g0001_0100.s0075_sort_colors
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun sortColors() {
+        var array = intArrayOf(2, 0, 2, 1, 1, 0)
+        Solution().sortColors(array)
+        var expected = intArrayOf(0, 0, 1, 1, 2, 2)
+        assertThat(array, equalTo(expected))
+    }
+
+    @Test
+    fun sortColors2() {
+        var array = intArrayOf(2, 0, 1)
+        Solution().sortColors(array)
+        var expected = intArrayOf(0, 1, 2)
+        assertThat(array, equalTo(expected))
+    }
+}

src/test/kotlin/g0001_0100/s0076_minimum_window_substring/SolutionTest.kt

@@ -0,0 +1,22 @@
+package g0001_0100.s0076_minimum_window_substring
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun minWindow() {
+        assertThat(Solution().minWindow("ADOBECODEBANC", "ABC"), equalTo("BANC"))
+    }
+
+    @Test
+    fun minWindow2() {
+        assertThat(Solution().minWindow("a", "a"), equalTo("a"))
+    }
+
+    @Test
+    fun minWindow3() {
+        assertThat(Solution().minWindow("a", "aa"), equalTo(""))
+    }
+}