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Count Number of Substrings with Only 1s

ayanb profile image Ayan Banerjee Originally published at notescs.gitlab.io on ・2 min read

Given a binary string, count no of substrings with only 1’s. Since the resulting count could be huge, print it modulo 109+710^9 + 7 .

Example :

Input: “11101101”

Output: 10

Explanation: “1” occurs 6 times, “11” occurs 3 times and “111” occurs 1 time. Thus number of substrings with only 1’s = 6 + 3 + 1 = 10.

Approach: Sliding Window

Intuition

We can see that a ‘0’ divides two blocks of ‘1’s. For a block of ‘1’ with length l has 1+2+3+...+l=l(l+1)/21 + 2 + 3 + ... + l = l * (l + 1) / 2 substrings. Thus, we only need to count the length of each block of 1’s and use the above formula.

For example, the string “11101101” can be divided into blocks of 1’s having 3 characters, 2 characters and 1 character which will give 3 * (3 + 1) / 2 = 6, 2 * (2 + 1) / 2 = 3 and 1 * (1 + 1) / 2 = 1 substring respectively.

Implementation

C++ code:

#include <iostream>
#include <string>
using namespace std;

int subCount(string s) {
    const int MOD = 1e9 + 7;
    int i = 0, l = s.length();
    int ans = 0;
    while (i < l) {
        int count = 0; // count gives length of block of 1's
        if (s[i] == '1') {
            while(i < l and s[i] == '1') {
                ++count;
                ++i;
            }
            long substringCount = 1L * count * (count + 1) / 2 % MOD;
            ans = (ans + substringCount) % MOD;
        } else {
            ++i;
        }
    }
    return ans;
}

int main() {
    cout << "Number of substring with only 1's for 11101101 is " << subCount("11101101") << endl;
    return 0;
}

Python code:

def sub_count(s):
    MOD = 10**9 + 7
    i = 0
    l = len(s)
    ans = 0
    while i < l:
        if s[i] == '1':
            count = 0 # count gives length of block of 1's
            while i < l and s[i] == '1':
                count += 1
                i += 1
            substring_count = count * (count + 1) // 2 % MOD
            ans = (ans + substring_count) % MOD
        else:
            i += 1
    return ans

if __name__ == ' __main__':
    print("Number of substring with only 1's for 11101101 is", sub_count('11101101'))

Complexity Analysis

  • Time Complexity: O(n)O(n) where n is the length of string.
  • Space Complexity: O(1)O(1) as we are not using any extra space.

Discussion

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