Time: 23 ms (29.21%), Space: 18.3 MB (40.82%) - LeetHub

Author: hogan-tech

1636-number-of-substrings-with-only-1s/1636-number-of-substrings-with-only-1s.py

@@ -0,0 +1,23 @@
+# time complexity: O(n)
+# space complexity: O(1)
+class Solution:
+    def numSub(self, s: str) -> int:
+        MOD = 10**9 + 7
+        left = 0
+        result = 0
+        for right in range(len(s)):
+            if s[right] == '1':
+                result += (right - left + 1)
+            else:
+                left = right + 1
+        return result % MOD
+
+
+s = "0110111"
+print(Solution().numSub(s))
+s = "101"
+print(Solution().numSub(s))
+s = "000"
+print(Solution().numSub(s))
+s = "111111"
+print(Solution().numSub(s))