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Problem solution · Python

ABC375 D — ABA

ABC375 D — ABA: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sliding window or two pointers
Source
KATO-Hiro AtCoder Solutions
Length
41 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 ABC375 D — ABA, 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

  • 41 lines of Python from the credited upstream file abc375_d.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit 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 KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC375 D — ABA · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from string import ascii_uppercase     input = sys.stdin.readline     s = input().rstrip()    n = len(s)    alpha = ascii_uppercase    alpha_count = len(ascii_uppercase)     # 3要素のうち真ん中を固定    # 文字は26種類なので、前計算で左から累積和を取る    acc_from_left = [[0] * n for _ in range(alpha_count)]     for i, a in enumerate(alpha):        for j, si in enumerate(s):            if j >= 1:                acc_from_left[i][j] = acc_from_left[i][j - 1]             if si == a:                acc_from_left[i][j] += 1     ans = 0     for j in range(1, n - 1):        for i in range(alpha_count):            left = acc_from_left[i][j - 1]            right = acc_from_left[i][n - 1] - acc_from_left[i][j]             ans += left * right     print(ans)  if __name__ == "__main__":    main()

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