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authorJack Sangdahl <[email protected]>2025-01-08 10:25:35 +0100
committerJack Sangdahl <[email protected]>2025-01-08 10:25:35 +0100
commitf416afbe5c617eb8aeae025d0b2d73c60145dbd3 (patch)
treec5161adaa26c5f4f2a302ef129de1e89d76d99d5
parent8a2b42fc8d607c983f5ce0f746253441fe7c38c4 (diff)
downloadleetcode-f416afbe5c617eb8aeae025d0b2d73c60145dbd3.tar.gz
leetcode-f416afbe5c617eb8aeae025d0b2d73c60145dbd3.zip
3042. count prefix and suffix pairs 1
-rw-r--r--count_prefix_suffix_pair1.c42
-rw-r--r--count_prefix_suffix_pair1.py34
2 files changed, 76 insertions, 0 deletions
diff --git a/count_prefix_suffix_pair1.c b/count_prefix_suffix_pair1.c
new file mode 100644
index 0000000..e22e030
--- /dev/null
+++ b/count_prefix_suffix_pair1.c
@@ -0,0 +1,42 @@
+// 3042. Count Prefix and Suffix Pairs I
+#include "leetcode.h"
+
+/*
+ * you are given a 0-indexed string array 'words'. let's define a boolean
+ * function 'isPrefixSuffix' that takes two strings 'str1' and 'str2'.
+ * 'isPrefixSuffix()' returns true if 'str1' is both a prefix and suffix of
+ * 'str2', and false otherwise. return an integer denoting the number of index
+ * pairs '(i, j)' such that 'i < j' and 'isPrefixSuffix(words[i], words[j])' is
+ * true.
+ */
+
+bool isPrefixAndSuffix(char *str1, char *str2) {
+ int n = strlen(str1), m = strlen(str2);
+ for (int i = 0, j = 0; str1[i]; i++, j++)
+ if (str1[i] != str2[j])
+ return false;
+ for (int i = n - 1, j = m - 1; j >= 0 && i >= 0; i--, j--)
+ if (str1[i] != str2[j])
+ return false;
+ return true;
+}
+
+int countPrefixSuffixPairs(char **words, int wordsSize) {
+ int cnt = 0;
+ for (int i = 0; i < wordsSize; i++)
+ for (int j = i + 1; j < wordsSize; j++)
+ if (isPrefixAndSuffix(words[i], words[j]))
+ cnt++;
+ return cnt;
+}
+
+int main() {
+ char *w1[] = {"a", "aba", "ababa", "aa"},
+ *w2[] = {"pa", "papa", "ma", "mama"}, *w3[] = {"abab", "ab"};
+ printf("%d\n",
+ countPrefixSuffixPairs((char **)w1, ARRAY_SIZE(w1))); // expect: 4
+ printf("%d\n",
+ countPrefixSuffixPairs((char **)w2, ARRAY_SIZE(w2))); // expect: 2
+ printf("%d\n",
+ countPrefixSuffixPairs((char **)w3, ARRAY_SIZE(w3))); // expect: 0
+}
diff --git a/count_prefix_suffix_pair1.py b/count_prefix_suffix_pair1.py
new file mode 100644
index 0000000..7248d9b
--- /dev/null
+++ b/count_prefix_suffix_pair1.py
@@ -0,0 +1,34 @@
+# 3042. Count Prefix and Suffix Pairs I
+
+"""
+you are given a 0-indexed string array 'words'. let's define a boolean
+function 'isPrefixSuffix' that takes two strings 'str1' and 'str2'.
+'isPrefixSuffix()' returns true if 'str1' is both a prefix and suffix of
+'str2', and false otherwise. return an integer denoting the number of index
+pairs '(i, j)' such that 'i < j' and 'isPrefixSuffix(words[i], words[j])' is
+true.
+"""
+
+
+class Solution(object):
+ def countPrefixSuffixPairs(self, words):
+ """
+ :type words: List[str]
+ :rtype: int
+ """
+ cnt = 0
+ for i in range(len(words)):
+ for j in range(i + 1, len(words)):
+ if (
+ words[j].find(words[i], 0, len(words[i])) == 0
+ and words[j][::-1].find(words[i][::-1], 0, len(words[i])) == 0
+ ):
+ cnt += 1
+ return cnt
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.countPrefixSuffixPairs(words=["a", "aba", "ababa", "aa"]))
+ print(obj.countPrefixSuffixPairs(words=["pa", "papa", "ma", "mama"]))
+ print(obj.countPrefixSuffixPairs(words=["abab", "ab"]))