Given a binary string s (a string consisting only of ‘0’ and ‘1’s).

Return the number of substrings with all characters 1’s.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.


Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.


Example 4:

Input: s = "000"
Output: 0


Constraints:

• s[i] == '0' or s[i] == '1'
• 1 <= s.length <= 10^5

Solution: DP/ Prefix Sum

dp[i] := # of all 1 subarrays end with s[i].
dp[i] = dp[i-1] if s[i] == ‘1‘ else 0
ans = sum(dp)
s=1101
dp[0] = 1 // 1
dp[1] = 2 // 11, *1
dp[2] = 0 // None
dp[3] = 1 // ***1
ans = 1 + 2 + 1 = 5

Time complexity: O(n)
Space complexity: O(n)

C++

dp[i] only depends on dp[i-1], we can reduce the space complexity to O(1)

Python3

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