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Problem solution · Python

ABC290 E — Make it Palindrome

ABC290 E — Make it Palindrome: 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
43 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 ABC290 E — Make it Palindrome, 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

  • 43 lines of Python from the credited upstream file abc290_e.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 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 codeABC290 E — Make it Palindrome · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    import sys    from collections import Counter     input = sys.stdin.readline     n = int(input())    a = list(map(int, input().split()))    c = Counter(a)     def nC2(x):        return x * (x - 1) // 2     same = sum([nC2(count) for count in c.values()])    ans = 0     def delete(x):        nonlocal same         same -= nC2(c[x])        c[x] -= 1        same += nC2(c[x])     for left in range(n):        right = n - left - 1         if left >= right:            break         ans += nC2(right - left + 1) - same         delete(a[left])        delete(a[right])     print(ans)  if __name__ == "__main__":    main()

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