Prime Complexity @primeComplexity

Joined September 2023

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Prime Complexity @primeComplexity

12 months ago

@samtoosoon Find the longest non-decreasing suffix (fix). Keep a running non-decreasing prefix (running). For each running, if last(running) ≤ first(fix), removal = fixStart−runningEnd−1. While fix < running, move fix ahead. Take min removal. Edge cases: remove full prefix/suffix.

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Prime Complexity @primeComplexity

12 months ago

A: Count 1s in first zero positions → min ops. B: Use y=2x or y=8x from the equation derivation.

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Prime Complexity @primeComplexity

12 months ago

Universal construction: 10^d ≡ 1 mod 3 → 3x | x*(10^d-1) → choose s = 3x → y = 2x 10^d ≡ 1 mod 9 → 9x | x*(10^d-1) → choose s = 9x → y = 8x Both y=2x and y=8x always work.

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Prime Complexity @primeComplexity

12 months ago

CF Round 1049 (Div. 2) Quick Thread 🧵(A and B) will up solve later ones

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Prime Complexity @primeComplexity

a year ago

Else all paths are monotone. Split in Si<Ti and Si>Ti, their intervals don’t overlap. Right-goers safe in descending Si since each departs before blocking anyone behind. Left-goers safe in ascending Si since each clears before the next arrives. Combine both gives no fight order.

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Prime Complexity @primeComplexity

a year ago

Idea: Key: Sort by starts. If targets not sorted too, then ∃i<j with Si<Sj but Ti>Tj. In that case whichever moves first, the other’s target lies inside its path. This forces a collision, so no permutation exists and the answer is No.

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Prime Complexity @primeComplexity

a year ago

Quick thread on AtCoder Regular Contest 205 (Div 2) Solved A & B. Will up solve later ones.

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Prime Complexity @primeComplexity

a year ago

E. Min-Max Deque (Greedy) Only border values of B matter. Once an element leaves border, never comes back. At step i, Ai must be border. So: Alice replaces min(x,y) with Ai (keeps both large). Bob replaces max(x,y) with Ai (pushes small). Finally ends = (An, x). Ans = min(An, x).

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Prime Complexity @primeComplexity

a year ago

Binary Search on answer : If Alice can secure value x, then she can also secure any ≤ x. If she fails at x, she fails at all > x. Since monotonic so, binary search the largest feasible x. That’s exactly f(A), Alice’s guaranteed final score.

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Prime Complexity @primeComplexity

a year ago

Quick thread 🧵 on Starters 202 (Solved A–E) Will up solve Min-Max Deque Hard later.

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Prime Complexity @primeComplexity

a year ago

A: If X == 4 then H else 24 * (X − 4) + H B: If (50 - X) lies in [2Y, 2Y+10] then Yes else No C: Brute over small N and try all (2i + 3j = N), take min cost. D: If (A[i] > A[i + 1]) then [i, i+1] works else -1 E: Minimax Binary search + greedy if Alice can force score ≥ x