new file: bytedance/DNA序列编辑距离.py

	modified:   bytedance/test.py
	new file:   bytedance/二进制之和.py
	new file:   bytedance/小E的怪物挑战.py
	new file:   bytedance/数字翻译成字符串的可能性.py
	new file:   bytedance/有限制的楼梯攀登.py
	modified:   bytedance/构造特定数组的逆序拼接.py
	new file:   bytedance/比赛配对问题.py
	new file:   bytedance/简单四则运算解析器.py
	new file:   bytedance/队列.py

Author: sakuralggm

bytedance/DNA序列编辑距离.py

@@ -0,0 +1,15 @@
+def solution(dna1, dna2):
+    dp = [[0 for _ in range(len(dna2) + 1)] for _ in range(len(dna1) + 1)]
+    for i in range(1, len(dna1) + 1):
+        dp[i][0] = i
+    for i in range(1, len(dna2) + 1):
+        dp[0][i] = i
+    for i in range(1, len(dna1) + 1):
+        for j in range(1, len(dna2) + 1):
+            dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + (dna1[i - 1] != dna2[j - 1]))
+    return dp[-1][-1]
+
+if __name__ == "__main__":
+    #  You can add more test cases here
+    print(solution("AGCTTAGC", "AGCTAGCT") == 2 )
+    print(solution("AGCCGAGC", "GCTAGCT") == 4)

bytedance/test.py

@@ -1,11 +1,26 @@
-def solution(numbers):
+def solution(n: int, H: int, A: int, h: list, a: list) -> int:
+    monsters = sorted(zip(h, a), reverse=True)
+    dp = [0 for _ in range(n + 1)]
+    for i in range(n - 1, -1, -1):
+        dp[i] = 1
+        for j in range(n - 1, i, -1):
+            if monsters[j][0] < monsters[i][0] and monsters[j][1] < monsters[i][1]:
+                dp[i] = max(dp[i], dp[j] + 1)
     ans = 0
-    def dfs(u, sum):
-        if u == len(numbers):
-            nonlocal ans
-            if sum % 2 == 0:
-                ans += 1
-        for i in range(0, len(str(numbers[u]))):
-            dfs(u + 1, sum + int(str(numbers[u])[i]))
-    dfs(0, 0)
-    return ans
+    for i in range(n):
+        if monsters[i][0] < H and monsters[i][1] < A:
+            ans = max(ans, dp[i])
+    return ans
+
+# 3 4 5 第一行n H A
+# 1 2 3 第二行h
+# 3 2 1 第三行a
+if __name__ == "__main__":
+    data = input().split()
+    n = int(data[0])
+    H = int(data[1])
+    A = int(data[2])
+    h = list(map(int, input().split()))
+    a = list(map(int, input().split()))
+    
+    print(solution(n, H, A, h, a))

bytedance/二进制之和.py

@@ -0,0 +1,13 @@
+def solution(binary1, binary2):
+    #  Convert binary to decimal
+    decimal1 = int(binary1, 2)
+    decimal2 = int(binary2, 2)
+    #  Add the two decimal numbers and convert the result to binary
+    ans = decimal1 + decimal2
+    return str(ans)
+if __name__ == "__main__":
+    #  You can add more test cases here
+    print(solution("101", "110") == "11")
+    print(solution("111111", "10100") == "83")
+    print(solution("111010101001001011", "100010101001") == "242420")
+    print(solution("111010101001011", "10010101001") == "31220")

bytedance/小E的怪物挑战.py

@@ -0,0 +1,19 @@
+def solution(n: int, H: int, A: int, h: list, a: list) -> int:
+    monsters = sorted(zip(h, a), reverse=True)
+    dp = [0 for _ in range(n + 1)]
+    for i in range(n - 1, -1, -1):
+        dp[i] = 1
+        for j in range(n - 1, i, -1):
+            if monsters[j][0] < monsters[i][0] and monsters[j][1] < monsters[i][1]:
+                dp[i] = max(dp[i], dp[j] + 1)
+    ans = 0
+    for i in range(n):
+        if monsters[i][0] < H and monsters[i][1] < A:
+            ans = max(ans, dp[i])
+    return ans
+
+
+if __name__ == '__main__':
+    print(solution(3, 4, 5, [1, 2, 3], [3, 2, 1]) == 1)
+    print(solution(5, 10, 10, [6, 9, 12, 4, 7], [8, 9, 10, 2, 5]) == 2)
+    print(solution(4, 20, 25, [10, 15, 18, 22], [12, 18, 20, 26]) == 3)

bytedance/数字翻译成字符串的可能性.py

@@ -0,0 +1,19 @@
+def solution(num):
+    num = str(num)
+    ans = 0
+    def dfs(u):
+        nonlocal ans
+        if u >= len(num) - 1:
+            ans += 1
+            return
+        dfs(u + 1)
+        if num[u:u + 2] <= '25' and num[u:u + 2] >= '10':
+            dfs(u + 2)
+    dfs(0)
+    return ans
+    
+if __name__ == "__main__":
+    #  You can add more test cases here
+    print(solution(12258) == 5)
+    print(solution(1400112) == 6)
+    print(solution(2110101) == 10)

bytedance/有限制的楼梯攀登.py

@@ -0,0 +1,15 @@
+def solution(n):
+    dp = [[0 for _ in range(2)] for _ in range(n + 1)]
+    dp[0][0] = 1
+    dp[1][0] = 1
+    dp[1][1] = 0
+    for i in range(2, n + 1):
+        dp[i][0] = dp[i - 1][0] + dp[i - 1][1]
+        dp[i][1] = dp[i - 2][0]
+    return dp[n][0] + dp[n][1]
+if __name__ == "__main__":
+    # Add your test cases here
+
+    print(solution(2) == 2)  # 样例1
+    print(solution(3) == 3)  # 样例2
+    print(solution(4) == 4)  # 样例3

bytedance/构造特定数组的逆序拼接.py

@@ -0,0 +1,8 @@
+def solution(n: int) -> list:
+    ans = [j for i in range(n) for j in range(n, i, -1)]
+    return ans
+    
+if __name__ == '__main__':
+    print(solution(3) == [3, 2, 1, 3, 2, 3])
+    print(solution(4) == [4, 3, 2, 1, 4, 3, 2, 4, 3, 4])
+    print(solution(5) == [5, 4, 3, 2, 1, 5, 4, 3, 2, 5, 4, 3, 5, 4, 5])

bytedance/比赛配对问题.py

@@ -0,0 +1,23 @@
+# 打表找规律，发现结果都是n-1，所以直接返回n-1即可
+# def solution(x):
+#     ans = 0
+#     while x > 1:
+#         if x % 2 == 0:
+#             ans += x // 2
+#             x //= 2
+#         else:
+#             ans += (x - 1) // 2
+#             x = (x + 1) // 2
+#     return ans
+
+# if __name__ == '__main__':
+#     for i in range(1, 100):
+#         print(i, solution(i))
+
+def solution(n: int) -> int:
+    return n - 1
+
+if __name__ == '__main__':
+    print(solution(7) == 6)
+    print(solution(14) == 13)
+    print(solution(1) == 0)

bytedance/简单四则运算解析器.py

@@ -0,0 +1,72 @@
+def solution(expression):
+    def precedence(op):
+        if op == '+' or op == '-':
+            return 1
+        if op == '*' or op == '/':
+            return 2
+        return 0
+
+    def apply_op(a, b, op):
+        if op == '+':
+            return a + b
+        if op == '-':
+            return a - b
+        if op == '*':
+            return a * b
+        if op == '/':
+            return a // b
+
+    def evaluate(tokens):
+        values = []
+        ops = []
+        i = 0
+
+        while i < len(tokens):
+            if tokens[i] == ' ':
+                i += 1
+                continue
+
+            if tokens[i] == '(':
+                ops.append(tokens[i])
+
+            elif tokens[i].isdigit():
+                val = 0
+                while (i < len(tokens) and tokens[i].isdigit()):
+                    val = (val * 10) + int(tokens[i])
+                    i += 1
+                values.append(val)
+                i -= 1
+
+            elif tokens[i] == ')':
+                while len(ops) != 0 and ops[-1] != '(':
+                    val2 = values.pop()
+                    val1 = values.pop()
+                    op = ops.pop()
+                    values.append(apply_op(val1, val2, op))
+                ops.pop()
+
+            else:
+                while (len(ops) != 0 and precedence(ops[-1]) >= precedence(tokens[i])):
+                    val2 = values.pop()
+                    val1 = values.pop()
+                    op = ops.pop()
+                    values.append(apply_op(val1, val2, op))
+                ops.append(tokens[i])
+            i += 1
+
+        while len(ops) != 0:
+            val2 = values.pop()
+            val1 = values.pop()
+            op = ops.pop()
+            values.append(apply_op(val1, val2, op))
+
+        return values[-1]
+
+    return evaluate(expression)
+
+if __name__ == "__main__":
+    #  You can add more test cases here
+    print(solution("1+1") == 2)
+    print(solution("3+4*5/(3+2)") == 7)
+    print(solution("4+2*5-2/1") == 12)
+    print(solution("(1+(4+5+2)-3)+(6+8)") == 23)

bytedance/队列.py

@@ -0,0 +1,16 @@
+def solution(n: int, a: list) -> int:
+    cur = 1
+    ans = []
+    for i in a:
+        if i == cur:
+            cur += 1
+            ans.append(i)
+    if len(ans) == 0:
+        return -1
+    else:
+        return n - len(ans)
+
+if __name__ == '__main__':
+    print(solution(5, [1, 4, 2, 3, 5]) == 2)
+    print(solution(3, [3, 3, 2]) == -1)
+    print(solution(5, [1, 2, 3, 4, 5]) == 0)