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Problem solution · C++

Count the Number of Substrings With Dominant Ones

Count the Number of Substrings With Dominant Ones: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
33 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Count the Number of Substrings With Dominant Ones, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 33 lines of C++ from the credited upstream file 3234.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount the Number of Substrings With Dominant Ones · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int numberOfSubstrings(string s) {    int ans = 0;     // Iterate through all possible number of 0s.    for (int zero = 0; zero + zero * zero <= s.length(); ++zero) {      int lastInvalidPos = -1;      vector<int> count(2);      for (int l = 0, r = 0; r < s.length(); ++r) {        ++count[s[r] - '0'];        // Try to shrink the window to maintain the "minimum" length of the        // valid substring.        for (; l < r; ++l)          if (s[l] == '0' && count[0] > zero) {            --count[0];  // Remove an extra '0'.            lastInvalidPos = l;          } else if (s[l] == '1' && count[1] - 1 >= zero * zero) {            --count[1];  // Remove an extra '1'.          } else {            break;  // Cannot remove more characters.          }        if (count[0] == zero && count[1] >= zero * zero)          // Add valid substrings ending in s[r] to the answer. They are          // s[lastInvalidPos + 1..r], s[lastInvalidPos + 2..r], ..., s[l..r].          ans += l - lastInvalidPos;      }    }     return ans;  }};

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