leetcode

commit 758bb05572b1e5948c3a9b9ce4413ffede87362b
parent ef565e321f29ebb749d8fdad74a0f7d7105ddee0
Author: Jack Sangdahl <jack@pngu.org>
Date:   Wed, 18 Feb 2026 20:41:44 -0700

696. count binary substrings

Diffstat:
A
count_binary_substr.c
 | 
33
+++++++++++++++++++++++++++++++++
A
count_binary_substr.py
 | 
26
++++++++++++++++++++++++++

2 files changed, 59 insertions(+), 0 deletions(-)
diff --git a/count_binary_substr.c b/count_binary_substr.c
@@ -0,0 +1,33 @@
+// 696. Count Binary Substrings
+#include "leetcode.h"
+
+/*
+ * given a binary string 's', return the number of non-empty substrings that
+ * have the same number of 0's and 1's, and all the 0's and all the 1's in these
+ * substrings are group consecutively. substrings that occur multiple times are
+ * counted the number of times they occur.
+ */
+
+int countBinarySubstrings(char *s) {
+  int curr = 1, pre = 0, ans = 0, n = strlen(s);
+  for (int i = 1; i < n; i++) {
+    if (s[i] == s[i - 1])
+      curr++;
+    else {
+      ans += fmin(curr, pre);
+      pre = curr;
+      curr = 1;
+    }
+  }
+  return ans + fmin(curr, pre);
+}
+
+int main() {
+  char *s1 = "00110011", *s2 = "10101";
+  int r1 = countBinarySubstrings(s1);
+  int r2 = countBinarySubstrings(s2);
+  printf("%d\n", r1); // expect: 6
+  assert(r1 == 6);
+  printf("%d\n", r2); // expect: 4
+  assert(r2 == 4);
+}
diff --git a/count_binary_substr.py b/count_binary_substr.py
@@ -0,0 +1,26 @@
+# 696. Count Binary Substrings
+
+"""
+given a binary string 's', return the number of non-empty substrings that
+have the same number of 0's and 1's, and all the 0's and all the 1's in these
+substrings are group consecutively. substrings that occur multiple times are
+counted the number of times they occur.
+"""
+
+
+class Solution:
+    def countBinarySubstrings(self, s: str) -> int:
+        curr, prev, ans = 1, 0, 0
+        for i in range(1, len(s)):
+            if s[i] == s[i - 1]:
+                curr += 1
+            else:
+                ans += min(curr, prev)
+                prev, curr = curr, 1
+        return ans + min(curr, prev)
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.countBinarySubstrings(s="00110011"))
+    print(obj.countBinarySubstrings(s="10101"))