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Author: hogan-tech

3479-count-the-number-of-substrings-with-dominant-ones/3479-count-the-number-of-substrings-with-dominant-ones.py

@@ -0,0 +1,29 @@
+# time complexity: O(n ^ (1/n))
+# space complexity: O(n)
+class Solution:
+    def numberOfSubstrings(self, s: str) -> int:
+        n = len(s)
+        prefix = [-1] * (n + 1)
+        for i in range(n):
+            if i == 0 or s[i - 1] == "0":
+                prefix[i + 1] = i
+            else:
+                prefix[i + 1] = prefix[i]
+
+        result = 0
+        for i in range(1, n + 1):
+            zeroes = 1 if s[i - 1] == "0" else 0
+            j = i
+            while j > 0 and zeroes * zeroes <= n:
+                ones = (i - prefix[j]) - zeroes
+                if zeroes * zeroes <= ones:
+                    result += min(j - prefix[j], ones - zeroes * zeroes + 1)
+                j = prefix[j]
+                zeroes += 1
+        return result
+
+
+s = "00011"
+print(Solution().numberOfSubstrings(s))
+s = "101101"
+print(Solution().numberOfSubstrings(s))