leetcode

commit 8c511efa9fb6e02856033dc22bf938a1a5830025
parent 6da1a3a48cb3ac0bf2e4cacbf1ecaef106dee100
Author: Jack Sangdahl <jack@pngu.org>
Date:   Mon, 22 Jun 2026 08:46:28 -0600

1189. maximum number of balloons

Diffstat:
A
max_num_balloons.c
 | 
38
++++++++++++++++++++++++++++++++++++++
A
max_num_balloons.py
 | 
22
++++++++++++++++++++++

2 files changed, 60 insertions(+), 0 deletions(-)
diff --git a/max_num_balloons.c b/max_num_balloons.c
@@ -0,0 +1,38 @@
+// 1189. Maximum Number of Balloons
+#include "leetcode.h"
+
+/*
+ * given a string 'text', you want to use the characters of 'text' to form as
+ * many instances of the word "balloon" as possible. you can use each character
+ * in 'text' at most once. return the maximum number of instances that can be
+ * formed.
+ */
+
+int maxNumberOfBalloons(char *text) {
+  int freq[5] = {0};
+  char *s = "balon";
+  for (int i = 0; text[i]; i++)
+    for (int j = 0; j < 5; j++)
+      freq[j] += (text[i] == s[j]);
+  int min = freq[0];
+  min = fmin(min, freq[1]);
+  min = fmin(min, freq[2] >> 1);
+  min = fmin(min, freq[3] >> 1);
+  min = fmin(min, freq[4]);
+  return min;
+}
+
+int main() {
+  char *t1 = "nlaebolko";
+  char *t2 = "loonbalxballpoon";
+  char *t3 = "leetcode";
+  int r1 = maxNumberOfBalloons(t1);
+  int r2 = maxNumberOfBalloons(t2);
+  int r3 = maxNumberOfBalloons(t3);
+  printf("%d\n", r1);
+  assert(r1 == 1);
+  printf("%d\n", r2);
+  assert(r2 == 2);
+  printf("%d\n", r3);
+  assert(r3 == 0);
+}
diff --git a/max_num_balloons.py b/max_num_balloons.py
@@ -0,0 +1,22 @@
+# 1189. Maximum Number of Balloons
+from collections import Counter
+
+"""
+given a string 'text', you want to use the characters of 'text' to form as
+many instances of the word "balloon" as possible. you can use each character
+in 'text' at most once. return the maximum number of instances that can be
+formed.
+"""
+
+
+class Solution:
+    def maxNumberOfBalloons(self, text: str) -> int:
+        freq = Counter(text)
+        return min(freq["b"], freq["a"], freq["l"] >> 1, freq["o"] >> 1, freq["n"])
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.maxNumberOfBalloons(text="nlaebolko"))
+    print(obj.maxNumberOfBalloons(text="loonbalxballpoon"))
+    print(obj.maxNumberOfBalloons(text="leetcode"))