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authorJack Sangdahl <[email protected]>2025-01-11 12:58:39 -0700
committerJack Sangdahl <[email protected]>2025-01-13 13:02:43 -0700
commit32b3c8cc7a72294298aabcedc2033b06f2a7ca5e (patch)
treeff4e5338da75dbe7e024a2d2fd3f0c8b1262e796
parent86fae540ca9b40b8294541593b32600549c5a653 (diff)
downloadleetcode-32b3c8cc7a72294298aabcedc2033b06f2a7ca5e.tar.gz
leetcode-32b3c8cc7a72294298aabcedc2033b06f2a7ca5e.zip
2116. check if a parentheses string can be valid
-rw-r--r--check_parentheses_valid.c50
-rw-r--r--check_parentheses_valid.py49
2 files changed, 99 insertions, 0 deletions
diff --git a/check_parentheses_valid.c b/check_parentheses_valid.c
new file mode 100644
index 0000000..06efb94
--- /dev/null
+++ b/check_parentheses_valid.c
@@ -0,0 +1,50 @@
+// 2116. Check if a Parentheses String Can Be Valid
+#include "leetcode.h"
+#include <math.h>
+#include <string.h>
+
+/*
+ * a parentheses string is a non-empty string consisting only of '(' and ')'. it
+ * is valid if any of the following conditions is true: it is '()', it can be
+ * written as 'AB' where 'A' and 'B' are valid parentheses, it can be written as
+ * '(A)' where 'A' is a valid parentheses string. you are given a parentheses
+ * string 's' and a string 'locked', both of length 'n'. 'locked' is a binary
+ * string consistion only of '0's and '1's. for each indiex 'i' of 'locked'.
+ */
+
+bool canBeValid(char *s, char *locked) {
+ int n = strlen(s), total = 0, open = 0, closed = 0;
+ if (n % 2 == 1)
+ return false;
+ for (int i = n - 1; i >= 0; i--) {
+ if (locked[i] == '0')
+ total++;
+ else if (s[i] == '(')
+ open++;
+ else if (s[i] == ')')
+ closed++;
+ if (total - open + closed < 0)
+ return false;
+ }
+ total = open = closed = 0;
+ for (int i = 0; i < n; i++) {
+ if (locked[i] == '0')
+ total++;
+ else if (s[i] == '(')
+ open++;
+ else if (s[i] == ')')
+ closed++;
+ if (total + open - closed < 0)
+ return false;
+ }
+ return true;
+}
+
+int main() {
+ char *s1 = "))()))", *l1 = "010100";
+ char *s2 = "()()", *l2 = "0000";
+ char *s3 = ")", *l3 = "0";
+ printf("%d\n", canBeValid(s1, l1)); // expect: 1
+ printf("%d\n", canBeValid(s2, l2)); // expect: 1
+ printf("%d\n", canBeValid(s3, l3)); // expect: 0
+}
diff --git a/check_parentheses_valid.py b/check_parentheses_valid.py
new file mode 100644
index 0000000..a6f5c28
--- /dev/null
+++ b/check_parentheses_valid.py
@@ -0,0 +1,49 @@
+# 2116. Check if a Parentheses String Can Be Valid
+
+"""
+a parentheses string is a non-empty string consisting only of '(' and ')'. it
+is valid if any of the following conditions is true: it is '()', it can be
+written as 'AB' where 'A' and 'B' are valid parentheses, it can be written as
+'(A)' where 'A' is a valid parentheses string. you are given a parentheses
+string 's' and a string 'locked', both of length 'n'. 'locked' is a binary
+string consistion only of '0's and '1's. for each indiex 'i' of 'locked'.
+"""
+
+
+class Solution(object):
+ def canBeValid(self, s, locked):
+ """
+ :type s: str
+ :type locked: str
+ :rtype: bool
+ """
+ if len(s) % 2 == 1:
+ return False
+ total = open = closed = 0
+ for i in range(len(s) - 1, -1, -1):
+ if locked[i] == "0":
+ total += 1
+ elif s[i] == "(":
+ open += 1
+ elif s[i] == ")":
+ closed += 1
+ if total - open + closed < 0:
+ return False
+ total = open = closed = 0
+ for i in range(len(s)):
+ if locked[i] == "0":
+ total += 1
+ elif s[i] == "(":
+ open += 1
+ elif s[i] == ")":
+ closed += 1
+ if total + open - closed < 0:
+ return False
+ return True
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.canBeValid(s="))()))", locked="010100"))
+ print(obj.canBeValid(s="()()", locked="0000"))
+ print(obj.canBeValid(s=")", locked="0"))