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‎README.md‎

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| 121 | Best Time to Buy and Sell Stock | [Ruby](./algorithms/ruby/0121-best-time-to-buy-and-sell-stock.rb) | Easy |
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| 129 | Sum Root to Leaf Numbers | [Ruby](./algorithms/ruby/0129-sum-root-to-leaf-numbers.rb) | Medium |
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| 142 | Linked List Cycle II | [Ruby](./algorithms/ruby/0142-linked-list-cycle-ii.rb) | Medium |
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| 200 | Number of Islands | [Ruby](./algorithms/ruby/0200-number-of-islands.rb) | Medium |
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| 205 | Isomorphic Strings | [Ruby](./algorithms/ruby/0205-isomorphic-strings.rb) | Easy |
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| 206 | Reverse Linked List | [Ruby](./algorithms/ruby/0206-reverse-linked-list.rb) | Easy |
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| 208 | Implement Trie (Prefix Tree) | [Ruby](./algorithms/ruby/0208-implement-trie-prefix-tree.rb) | Medium |
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| 1319 | Number of Operations to Make Network Connected | [Ruby](./algorithms/ruby/1319-number-of-operations-to-make-network-connected.rb) | Medium |
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| 1345 | Jump Game IV | [Ruby](./algorithms/ruby/1345-jump-game-iv.rb) | Hard |
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| 1402 | Reducing Dishes | [Ruby](./algorithms/ruby/1402-reducing-dishes.rb) | Hard |
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| 1444 | Number of Ways of Cutting a Pizza | [Ruby](./algorithms/ruby/1444-number-of-ways-of-cutting-a-pizza.rb) [Python3](./algorithms/python3/1444-number-of-ways-of-cutting-a-pizza.py) | Hard |
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| 1466 | Reorder Routes to Make All Paths Lead to the City Zero | [Ruby](./algorithms/ruby/1466-eorder-routes-to-make-all-paths-lead-to-the-city-zero.rb) | Medium |
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| 1470 | Shuffle the Array | [Ruby](./algorithms/ruby/1470-shuffle-the-array.rb) | Easy |
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| 1472 | Design Browser History | [Ruby](./algorithms/ruby/1472-design-browser-history.rb) | Medium |

‎algorithms/python3/1444-number-of-ways-of-cutting-a-pizza.py‎

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# 1444. Number of Ways of Cutting a Pizza
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# https://leetcode.com/problems/number-of-ways-of-cutting-a-pizza
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from functools import lru_cache
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class Solution:
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def ways(self, pizza: list[str], K: int) -> int:
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m, n, MOD = len(pizza), len(pizza[0]), 10 ** 9 + 7
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preSum = [[0] * (n + 1) for _ in range(m + 1)]
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for r in range(m - 1, -1, -1):
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for c in range(n - 1, -1, -1):
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preSum[r][c] = preSum[r][c + 1] + preSum[r + 1][c] - preSum[r + 1][c + 1] + (pizza[r][c] == 'A')
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@lru_cache(None)
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def dp(k, r, c):
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if preSum[r][c] == 0: return 0
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if k == 0: return 1
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ans = 0
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for nr in range(r + 1, m):
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if preSum[r][c] - preSum[nr][c] > 0:
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ans = (ans + dp(k - 1, nr, c)) % MOD
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for nc in range(c + 1, n):
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if preSum[r][c] - preSum[r][nc] > 0:
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ans = (ans + dp(k - 1, r, nc)) % MOD
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return ans
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return dp(K - 1, 0, 0)
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# ********************#
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# TEST #
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# ********************#
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import unittest
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class TestStringMethods(unittest.TestCase):
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def test_ways(self):
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self.assertEqual(Solution.ways(self, ["A..", "AAA", "..."], 3), 3)
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self.assertEqual(Solution.ways(self, ["A..", "AA.", "..."], 3), 1)
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self.assertEqual(Solution.ways(self, ["A..", "A..", "..."], 1), 1)
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if __name__ == '__main__':
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unittest.main()

‎algorithms/ruby/0200-number-of-islands.rb‎

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# frozen_string_literal: true
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# 200. Number of Islands
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# https://leetcode.com/problems/number-of-islands
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=begin
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Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
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An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
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### Example 1:
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Input: grid = [
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["1","1","1","1","0"],
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["1","1","0","1","0"],
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["1","1","0","0","0"],
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["0","0","0","0","0"]
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]
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Output: 1
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### Example 2:
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Input: grid = [
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["1","1","0","0","0"],
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["1","1","0","0","0"],
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["0","0","1","0","0"],
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["0","0","0","1","1"]
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]
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Output: 3
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### Constraints:
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* m == grid.length
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* n == grid[i].length
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* 1 <= m, n <= 300
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* grid[i][j] is '0' or '1'.
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=end
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# Runtime: 180 ms
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# Memory: 219.5 MB
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# @param {Character[][]} grid
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# @return {Integer}
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def num_islands(grid)
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count = 0
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grid.size.times do |i|
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grid.first.size.times do |j|
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count += dfs(grid, i, j) if grid[i][j] == "1"
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end
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end
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count
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end
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def dfs(grid, i, j)
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return 0 if i < 0 || j < 0 || i >= grid.size || j >= grid.first.size || grid[i][j] == "0"
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grid[i][j] = "0"
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dfs(grid, i - 1, j)
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dfs(grid, i + 1, j)
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dfs(grid, i, j - 1)
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dfs(grid, i, j + 1)
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1
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end
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# **************** #
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# TEST #
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# **************** #
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require "test/unit"
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class Test_num_islands < Test::Unit::TestCase
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def test_
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assert_equal 1, num_islands([["1", "1", "1", "1", "0"], ["1", "1", "0", "1", "0"], ["1", "1", "0", "0", "0"], ["0", "0", "0", "0", "0"]])
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assert_equal 3, num_islands([["1", "1", "0", "0", "0"], ["1", "1", "0", "0", "0"], ["0", "0", "1", "0", "0"], ["0", "0", "0", "1", "1"] ])
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end
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end

‎algorithms/ruby/0733-flood-fill.rb‎

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# frozen_string_literal: true
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# Problem: 733. Flood Fill
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# URL: https://leetcode.com/problems/flood-fill
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# 733. Flood Fill
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# https://leetcode.com/problems/flood-fill
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=begin
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An image is represented by an m x n integer grid image where image[i][j] represents the pixel value of the image.
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You are also given three integers sr, sc, and color. You should perform a flood fill on the image starting from the pixel image[sr][sc].
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To perform a flood fill, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color), and so on. Replace the color of all of the aforementioned pixels with color.
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Return the modified image after performing the flood fill.
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### Example 1:
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Input: image = [[1,1,1],[1,1,0],[1,0,1]], sr = 1, sc = 1, color = 2
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Output: [[2,2,2],[2,2,0],[2,0,1]]
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Explanation: From the center of the image with position (sr, sc) = (1, 1) (i.e., the red pixel), all pixels connected by a path of the same color as the starting pixel (i.e., the blue pixels) are colored with the new color.
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Note the bottom corner is not colored 2, because it is not 4-directionally connected to the starting pixel.
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### Example 2:
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Input: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, color = 0
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Output: [[0,0,0],[0,0,0]]
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Explanation: The starting pixel is already colored 0, so no changes are made to the image.
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### Constraints:
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* m == image.length
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* n == image[i].length
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* 1 <= m, n <= 50
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* 0 <= image[i][j], color < 216
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* 0 <= sr < m
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* 0 <= sc < n
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=end
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# @param {Integer[][]} image
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# @param {Integer} sr

‎algorithms/ruby/1444-number-of-ways-of-cutting-a-pizza.rb‎

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# frozen_string_literal: true
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# 1444. Number of Ways of Cutting a Pizza
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# https://leetcode.com/problems/number-of-ways-of-cutting-a-pizza
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=begin
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Given a rectangular pizza represented as a rows x cols matrix containing the following characters: 'A' (an apple) and '.' (empty cell) and given the integer k. You have to cut the pizza into k pieces using k-1 cuts.
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For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.
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Return the number of ways of cutting the pizza such that each piece contains at least one apple. Since the answer can be a huge number, return this modulo 10^9 + 7.
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### Example 1:
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Input: pizza = ["A..","AAA","..."], k = 3
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Output: 3
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Explanation: The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.
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### Example 2:
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Input: pizza = ["A..","AA.","..."], k = 3
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Output: 1
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### Example 3:
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Input: pizza = ["A..","A..","..."], k = 1
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Output: 1
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### Constraints:
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* 1 <= rows, cols <= 50
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* rows == pizza.length
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* cols == pizza[i].length
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* 1 <= k <= 10
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* pizza consists of characters 'A' and '.' only.
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=end
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# @param {String[]} pizza
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# @param {Integer} k
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# @return {Integer}
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def ways(pizza, k)
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rows = pizza.size
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cols = pizza[0].size
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apples = Array.new(rows + 1) { Array.new(cols + 1, 0) }
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f = Array.new(rows) { Array.new(cols, 0) }
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row = rows - 1
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while row >= 0
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col = cols - 1
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while col >= 0
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apples[row][col] = ((pizza[row][col] == "A") && 1 || 0) + apples[row + 1][col] + apples[row][col + 1] - apples[row + 1][col + 1]
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f[row][col] = (apples[row][col] > 0) && 1 || 0
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col -= 1
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end
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row -= 1
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end
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mod = 1000000007
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remain = 1
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while remain < k
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g = Array.new(rows) { Array.new(cols, 0) }
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row = 0
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while row < rows
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col = 0
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while col < cols
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next_row = row + 1
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while next_row < rows
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if apples[row][col] - apples[next_row][col] > 0
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g[row][col] += f[next_row][col] %= mod
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end
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next_row += 1
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end
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next_col = col + 1
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while next_col < cols
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if apples[row][col] - apples[row][next_col] > 0
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g[row][col] += f[row][next_col] %= mod
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end
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next_col += 1
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end
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col += 1
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end
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row += 1
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end
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f = g
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remain += 1
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end
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f[0][0]
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end
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# **************** #
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# TEST #
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# **************** #
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# require "test/unit"
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# class Test_ways < Test::Unit::TestCase
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# def test_
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# assert_equal 3, ways(["A..", "AAA", "..."], 3)
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# assert_equal 1, ways(["A..", "AA.", "..."], 3)
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# assert_equal 1, ways(["A..", "A..", "..."], 1)
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# end
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# end
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pizza = ["..A.A.AAA...AAAAAA.AA..A..A.A......A.AAA.AAAAAA.AA", "A.AA.A.....AA..AA.AA.A....AAA.A........AAAAA.A.AA.", "A..AA.AAA..AAAAAAAA..AA...A..A...A..AAA...AAAA..AA", "....A.A.AA.AA.AA...A.AA.AAA...A....AA.......A..AA.", "AAA....AA.A.A.AAA...A..A....A..AAAA...A.A.A.AAAA..", "....AA..A.AA..A.A...A.A..AAAA..AAAA.A.AA..AAA...AA", "A..A.AA.AA.A.A.AA..A.A..A.A.AAA....AAAAA.A.AA..A.A", ".AA.A...AAAAA.A..A....A...A.AAAA.AA..A.AA.AAAA.AA.", "A.AA.AAAA.....AA..AAA..AAAAAAA...AA.A..A.AAAAA.A..", "A.A...A.A...A..A...A.AAAA.A..A....A..AA.AAA.AA.AA.", ".A.A.A....AAA..AAA...A.AA..AAAAAAA.....AA....A....", "..AAAAAA..A..A...AA.A..A.AA......A.AA....A.A.AAAA.", "...A.AA.AAA.AA....A..AAAA...A..AAA.AAAA.A.....AA.A", "A.AAAAA..A...AAAAAAAA.AAA.....A.AAA.AA.A..A.A.A...", "A.A.AA...A.A.AA...A.AA.AA....AA...AA.A..A.AA....AA", "AA.A..A.AA..AAAAA...A..AAAAA.AA..AA.AA.A..AAAAA..A", "...AA....AAAA.A...AA....AAAAA.A.AAAA.A.AA..AA..AAA", "..AAAA..AA..A.AA.A.A.AA...A...AAAAAAA..A.AAA..AA.A", "AA....AA....AA.A......AAA...A...A.AA.A.AA.A.A.AA.A", "A.AAAA..AA..A..AAA.AAA.A....AAA.....A..A.AA.A.A...", "..AA...AAAAA.A.A......AA...A..AAA.AA..A.A.A.AA..A.", ".......AA..AA.AAA.A....A...A.AA..A.A..AAAAAAA.AA.A", ".A.AAA.AA..A.A.A.A.A.AA...AAAA.A.A.AA..A...A.AAA..", "A..AAAAA.A..A..A.A..AA..A...AAA.AA.A.A.AAA..A.AA..", "A.AAA.A.AAAAA....AA..A.AAA.A..AA...AA..A.A.A.AA.AA", ".A..AAAA.A.A.A.A.......AAAA.AA...AA..AAA..A...A.AA", "A.A.A.A..A...AA..A.AAA..AAAAA.AA.A.A.A..AA.A.A....", "A..A..A.A.AA.A....A...A......A.AA.AAA..A.AA...AA..", ".....A..A...A.A...A..A.AA.A...AA..AAA...AA..A.AAA.", "A...AA..A..AA.A.A.AAA..AA..AAA...AAA..AAA.AAAAA...", "AA...AAA.AAA...AAAA..A...A..A...AA...A..AA.A...A..", "A.AA..AAAA.AA.AAA.A.AA.A..AAAAA.A...A.A...A.AA....", "A.......AA....AA..AAA.AAAAAAA.A.AA..A.A.AA....AA..", ".A.A...AA..AA...AA.AAAA.....A..A..A.AA.A.AA...A.AA", "..AA.AA.AA..A...AA.AA.AAAAAA.....A.AA..AA......A..", "AAA..AA...A....A....AA.AA.AA.A.A.A..AA.AA..AAA.AAA", "..AAA.AAA.A.AA.....AAA.A.AA.AAAAA..AA..AA.........", ".AA..A......A.A.AAA.AAAA...A.AAAA...AAA.AAAA.....A", "AAAAAAA.AA..A....AAAA.A..AA.A....AA.A...A.A....A..", ".A.A.AA..A.AA.....A.A...A.A..A...AAA..A..AA..A.AAA", "AAAA....A...A.AA..AAA..A.AAA..AA.........AA.AAA.A.", "......AAAA..A.AAA.A..AAA...AAAAA...A.AA..A.A.AA.A.", "AA......A.AAAAAAAA..A.AAA...A.A....A.AAA.AA.A.AAA.", ".A.A....A.AAA..A..AA........A.AAAA.AAA.AA....A..AA", ".AA.A...AA.AAA.A....A.A...A........A.AAA......A...", "..AAA....A.A...A.AA..AAA.AAAAA....AAAAA..AA.AAAA..", "..A.AAA.AA..A.AA.A...A.AA....AAA.A.....AAA...A...A", ".AA.AA...A....A.AA.A..A..AAA.A.A.AA.......A.A...A.", "...A...A.AA.A..AAAAA...AA..A.A..AAA.AA...AA...A.A.", "..AAA..A.A..A..A..AA..AA...A..AA.AAAAA.A....A..A.A"]
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k = 8
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puts ways(pizza, k) # 641829390

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