leetcode

commit 8704ee5ba635a64f25efe6088971568bfff6df7c
parent e202f40f8763153b4c10fbd3921402ae95e1bb87
Author: Jack Sangdahl <jack@pngu.org>
Date:   Mon, 29 Jun 2026 09:17:34 -0600

1967. number of strings that appear as substrings in word

Diffstat:
A
num_str_appear_substr_word.c
 | 
31
+++++++++++++++++++++++++++++++
A
num_str_appear_substr_word.py
 | 
23
+++++++++++++++++++++++

2 files changed, 54 insertions(+), 0 deletions(-)
diff --git a/num_str_appear_substr_word.c b/num_str_appear_substr_word.c
@@ -0,0 +1,31 @@
+// 1967. Number of Strings That Appear as Substrings in Word
+#include "leetcode.h"
+
+/*
+ * given an array of strings 'patterns' and a string 'word', return the number
+ * of strings in 'patterns' that exist as a substring in 'word'. a substring is
+ * a contiguous sequence of characters within a string.
+ */
+
+int numOfStrings(char **patterns, int patternsSize, char *word) {
+  int ans = 0;
+  for (int i = 0; i < patternsSize; i++)
+    if (strstr(word, patterns[i]))
+      ans++;
+  return ans;
+}
+
+int main() {
+  char *p1[] = {"a", "abc", "bc", "d"}, *w1 = "abc";
+  char *p2[] = {"a", "b", "c"}, *w2 = "aaaaabbbbb";
+  char *p3[] = {"a", "a", "a"}, *w3 = "ab";
+  int r1 = numOfStrings(p1, ARRAY_SIZE(p1), w1);
+  int r2 = numOfStrings(p2, ARRAY_SIZE(p2), w2);
+  int r3 = numOfStrings(p3, ARRAY_SIZE(p3), w3);
+  printf("%d\n", r1);
+  assert(r1 == 3);
+  printf("%d\n", r2);
+  assert(r2 == 2);
+  printf("%d\n", r3);
+  assert(r3 == 3);
+}
diff --git a/num_str_appear_substr_word.py b/num_str_appear_substr_word.py
@@ -0,0 +1,23 @@
+# 1967. Number of Strings That Appear as Substrings in Word
+
+"""
+given an array of strings 'patterns' and a string 'word', return the number
+of strings in 'patterns' that exist as a substring in 'word'. a substring is
+a contiguous sequence of characters within a string.
+"""
+
+
+class Solution:
+    def numOfStrings(self, patterns: List[str], word: str) -> int:
+        ans = 0
+        for i in patterns:
+            if i in word:
+                ans += 1
+        return ans
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.numOfStrings(patterns=["a", "abc", "bc", "d"], word="abc"))
+    print(obj.numOfStrings(patterns=["a", "b", "c"], word="aaaaabbbbb"))
+    print(obj.numOfStrings(patterns=["a", "a", "a"], word="ab"))