Program to count substrings with all 1s in binary string in Python

Suppose we have a binary string s. We have to find the number of substrings that contain only "1"s. If the answer is too large, mod the result by 10^9+7.

So, if the input is like s = "100111", then the output will be 7, because the substrings containing only "1"s are ["1", "1", "1", "1", "11", "11" and "111"].

Algorithm

To solve this, we will follow these steps −

• Initialize a counter a to 0 for tracking consecutive 1s
• Initialize count to 0 for total substrings
• For each character in the string:
  • If character is "0", reset a to 0
  • Otherwise, increment a and add it to count
• Return count

How It Works

When we encounter consecutive 1s, each new 1 can form substrings with all previous 1s in the current sequence. For example, in "111", the first 1 forms 1 substring, the second 1 forms 2 substrings ("1" and "11"), and the third 1 forms 3 substrings ("1", "11", and "111").

Example

Let us see the following implementation to get better understanding −

def solve(s):
    a = 0
    count = 0
    for i in range(len(s)):
        if s[i] == "0":
            a = 0
        else:
            a += 1
            count += a
    return count

s = "100111"
print(solve(s))

The output of the above code is −

7

Step-by-Step Execution

For string "100111", here's how the algorithm works:

With Modulo Operation

For large inputs, we can add modulo operation as mentioned in the problem statement −

def solve_with_mod(s):
    MOD = 10**9 + 7
    a = 0
    count = 0
    for char in s:
        if char == "0":
            a = 0
        else:
            a += 1
            count = (count + a) % MOD
    return count

s = "100111"
print(solve_with_mod(s))

The output remains the same for this small example −

7

Conclusion

This algorithm efficiently counts all-1 substrings by tracking consecutive 1s and adding the count at each position. The time complexity is O(n) and space complexity is O(1), making it optimal for large binary strings.