leetcode

commit 6604ace851b84f3a386b787d88b7db9124d4dc2f
parent 9bb4ffd1c46a3106d23e8fbc73b62fe2d6158b47
Author: Jack Sangdahl <jack@pngu.org>
Date:   Tue, 21 Apr 2026 20:18:50 -0700

2452. words within two edits of dictionary

Diffstat:
A
words_within_edits_dict.c
 | 
61
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
A
words_within_edits_dict.py
 | 
39
+++++++++++++++++++++++++++++++++++++++

2 files changed, 100 insertions(+), 0 deletions(-)
diff --git a/words_within_edits_dict.c b/words_within_edits_dict.c
@@ -0,0 +1,61 @@
+// 2452. Words Within Two Edits of Dictionary
+#include "leetcode.h"
+
+/*
+ * you are given two string arrays 'queries' and 'dictionary'. all words in each
+ * array comprise of lowercase english letters and have the same length. in one
+ * edit, you can take a word from 'queries' and change any letter in it to any
+ * other letter. find all words from 'queries' that, after a maximum of two
+ * edits, equal some word from 'dictionary'. return a list of all words from
+ * 'queries' that match with some word from 'dictionary' after a maximum of two
+ * edits. return the words in the same order that they appear in 'queries'.
+ */
+
+char **twoEditWords(char **queries, int queriesSize, char **dictionary,
+                    int dictionarySize, int *returnSize) {
+  int n = strlen(queries[0]), cnt = 0;
+  char **ans = (char **)malloc(queriesSize * sizeof(int));
+  for (int i = 0; i < queriesSize; i++) {
+    char *q = queries[i];
+    for (int k = 0; k < dictionarySize; k++) {
+      char *d = dictionary[k];
+      int diff = 0;
+      for (int j = 0; j < n; j++) {
+        if (q[j] != d[j])
+          diff++;
+        if (diff > 2)
+          break;
+      }
+      if (diff <= 2) {
+        ans[cnt++] = q;
+        break;
+      }
+    }
+  }
+  *returnSize = cnt;
+  return ans;
+}
+
+int main() {
+  char *q1[] = {"word", "note", "ants", "wood"};
+  char *d1[] = {"wood", "joke", "moat"};
+  char *q2[] = {"yes"};
+  char *d2[] = {"not"};
+  char *r1[] = {"word", "note", "wood"};
+  char *r2[] = {};
+  int rs1, rs2;
+  char **tew1 = twoEditWords(q1, ARRAY_SIZE(q1), d1, ARRAY_SIZE(d1), &rs1);
+  char **tew2 = twoEditWords(q2, ARRAY_SIZE(q2), d2, ARRAY_SIZE(d2), &rs2);
+  for (int i = 0; i < rs1; i++) {
+    printf("%s ", tew1[i]);
+    assert(!strcmp(tew1[i], r1[i]));
+  }
+  printf("\n");
+  for (int i = 0; i < rs2; i++) {
+    printf("%s ", tew2[i]);
+    assert(!strcmp(tew2[i], r2[i]));
+  }
+  printf("\n");
+  free(tew1);
+  free(tew2);
+}
diff --git a/words_within_edits_dict.py b/words_within_edits_dict.py
@@ -0,0 +1,39 @@
+# 2452. Words Within Two Edits of Dictionary
+
+"""
+you are given two string arrays 'queries' and 'dictionary'. all words in each
+array comprise of lowercase english letters and have the same length. in one
+edit, you can take a word from 'queries' and change any letter in it to any
+other letter. find all words from 'queries' that, after a maximum of two
+edits, equal some word from 'dictionary'. return a list of all words from
+'queries' that match with some word from 'dictionary' after a maximum of two
+edits. return the words in the same order that they appear in 'queries'.
+"""
+
+
+class Solution:
+    def twoEditWords(self, queries: List[str], dictionary: List[str]) -> List[str]:
+        ans, n = [], len(queries[0])
+        for q in queries:
+            for d in dictionary:
+                diff = 0
+                for j in range(n):
+                    if q[j] != d[j]:
+                        diff += 1
+                    if diff > 2:
+                        break
+                if diff <= 2:
+                    ans.append(q)
+                    break
+        return ans
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(
+        obj.twoEditWords(
+            queries=["word", "note", "ants", "wood"],
+            dictionary=["wood", "joke", "moat"],
+        )
+    )
+    print(obj.twoEditWords(queries=["yes"], dictionary=["not"]))