break algorithm---dynamic planning(dp-arr)

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声明：
算法基于https://labuladong.github.io/
python语言实现

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300.longest-increasing-subsequence
354. Russian Doll Envelopes
1143. Longest Common Subsequence
1425. Constrained Subsequence Sum

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300.longest-increasing-subsequence

from typing import List


class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        # dp[i]:其值为以nums[i]结尾的LIS值
        # 求出数组中以每个数字结尾的LIS，最终答案为其中最大的
        dp = [1] * len(nums)
        for i in range(0, len(nums), 1):
            for j in range(0, i, 1):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)

        res = 0
        for i in range(len(nums)):
            res = max(res, dp[i])

        return res


def main():
    # Input: nums = [10,9,2,5,3,7,101,18]
    # Output: 4
    # Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
    solution1 = Solution()
    print(solution1.lengthOfLIS(nums=[10, 9, 2, 5, 3, 7, 101, 18]))

    # Input: nums = [0,1,0,3,2,3]
    # Output: 4
    solution2 = Solution()
    print(solution2.lengthOfLIS(nums=[0, 1, 0, 3, 2, 3]))

    # Input: nums = [7,7,7,7,7,7,7]
    # Output: 1
    solution3 = Solution()
    print(solution3.lengthOfLIS(nums=[7, 7, 7, 7, 7, 7, 7]))


if __name__ == '__main__':
    main()

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354. Russian Doll Envelopes

from typing import List
from bisect import *


class SolutionTimeLimitExceeded:
    def maxEnvelopes(self, envelopes: List[List[int]]) -> int:
        w = len(envelopes)
        # 按宽度w升序排列，若宽度一样，则按高度h降序排列：把排序后的h作为一个数组，在这个数组上计算LIS的长度就是最终答案。
        envelopes.sort(key=lambda x: (x[0], -x[1]))

        # 对高度数组寻找LIS
        height = [0] * w
        for i in range(w):
            height[i] = envelopes[i][1]

        return self.lengthOfLIS(height)

    def lengthOfLIS(self, nums: List[int]) -> int:
        # dp[i]:其值为以nums[i]结尾的LIS值
        # 求出数组中以每个数字结尾的LIS，最终答案为其中最大的
        dp = [1] * len(nums)
        for i in range(len(nums)):
            for j in range(0, i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)

        res = 1
        for i in range(len(nums)):
            res = max(res, dp[i])

        return res


class Solution:
    def maxEnvelopes(self, E: List[List[int]]) -> int:
        E.sort(key=lambda x: (x[0], -x[1]))
        dp = []
        for _, height in E:
            left = bisect_left(dp, height)
            if left == len(dp):
                dp.append(height)
            else:
                dp[left] = height
        return len(dp)


def main():
    # Input: envelopes = [[5,4],[6,4],[6,7],[2,3]]
    # Output: 3
    # Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).
    solution1 = Solution()
    print(solution1.maxEnvelopes([[5, 4], [6, 4], [6, 7], [2, 3]]))

    # Input: envelopes = [[1,1],[1,1],[1,1]]
    # Output: 1
    solution2 = Solution()
    print(solution2.maxEnvelopes([[1, 1], [1, 1], [1, 1]]))

    # Input: [[1,15],[7,18],[7,6],[7,100],[2,200],[17,30],[17,45],[3,5],[7,8],[3,6],[3,10],[7,20],[17,3],[17,45]]
    # Output: 6
    # Expected: 3
    solution3 = Solution()
    print(solution3.maxEnvelopes(
        [[1, 15], [7, 18], [7, 6], [7, 100], [2, 200], [17, 30], [17, 45], [3, 5], [7, 8], [3, 6], [3, 10], [7, 20],
         [17, 3], [17, 45]]))
         

if __name__ == '__main__':
    main()       

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1143. Longest Common Subsequence

# dp[i][j] 的含义是：对于 s1[1..i] 和 s2[1..j]，它们的 LCS 长度是 dp[i][j]。
''' NOTE: 对于m行n列的二维数组 java二维数组初始化：int[][] dp = new int[m + 1][n + 1]; <=前行后列 python二维数组初始化：dp = [[0 for col in range(n+1)] for row in range(m+1)] <=前列后行 '''
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m = len(text1)
        n = len(text2)
        # 定义：text1[0..i-1]和text2[0..j-1]的lcs长度为dp[i][j]
        dp = [[0 for col in range(n+1)] for row in range(m+1)]
        # 目标：text1[0..m-1]和text2[0..n-1]的lcs长度，即dp[m][n]
        # base case: dp[0][..]=dp[..][0]=0

        for i in range(1, m + 1, 1):
            for j in range(1, n + 1, 1):
                # 现在i和j从1开始，故需减1
                if text1[i - 1] == text2[j - 1]:
                    # text1[i-1]和text2[j-1]相等，则该字符必在lcs中
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    # text1[i-1]和text2[j-1]不相等，则这两个字符必有一个不在lcs中
                    dp[i][j] = max(dp[i][j - 1], dp[i - 1][j])

        return dp[m][n]


def main():
    # Input: text1 = "abcde", text2 = "ace"
    # Output: 3
    # Explanation: The longest common subsequence is "ace" and its length is 3.
    solution1 = Solution()
    print(solution1.longestCommonSubsequence(text1="abcde", text2="ace"))

    # Input: text1 = "abc", text2 = "abc"
    # Output: 3
    # Explanation: The longest common subsequence is "abc" and its length is 3.
    solution2 = Solution()
    print(solution2.longestCommonSubsequence(text1="abc", text2="abc"))

    # Input: text1 = "abc", text2 = "def"
    # Output: 0
    # Explanation: There is no such common subsequence, so the result is 0.
    solution3 = Solution()
    print(solution3.longestCommonSubsequence(text1="abc", text2="def"))


if __name__ == '__main__':
    main()

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1425. Constrained Subsequence Sum

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