Given a binary string, count no of substrings with only 1’s. Since the resulting count could be huge, print it modulo .
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
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: where n is the length of string.
- Space Complexity: as we are not using any extra space.
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