Overview

This problem asks you to find the minimum value \(X\) of a simultaneously applied impact needed to topple all \(N\) pillars. You need to consider the chain rule where the collapse of a left-neighboring pillar deals additional damage to the right-neighboring pillar.

Analysis

Since the condition is to topple all pillars, let’s organize the conditions required for each pillar to fall.

1. First pillar (\(A_1\)) There is no left-neighboring pillar, so there is no additional damage from chaining. Therefore, for it to fall on its own, we need \(X \ge A_1\).
2. Second and subsequent pillars (\(A_i\), \(i \ge 2\)) Since our goal is to “topple all pillars,” when considering the second pillar, we can assume that the first pillar has already collapsed. When the first pillar collapses, the second pillar receives chain damage of \(1\). In other words, the total damage becomes \(X + 1\). This needs to be at least the durability \(A_i\), so the condition is \(X + 1 \ge A_i\), i.e., \(X \ge A_i - 1\).
3. About cumulative chaining The problem statement says “the durability of the immediately right-neighboring pillar decreases by 1 more.” This is a bonus from the left neighbor collapsing, and even if pillars further to the left have also collapsed, the additional damage is always \(1\). Therefore, for any \(i \ge 2\), the required impact takes the simple form \(X \ge A_i - 1\).

From the above, the minimum \(X\) to topple all pillars is the smallest integer satisfying all of the following conditions: - \(X \ge A_1\) - \(X \ge A_2 - 1\) - \(X \ge A_3 - 1\) - \(\vdots\) - \(X \ge A_N - 1\)

Expressed as a formula, this is \(X = \max(A_1, A_2-1, A_3-1, \dots, A_N-1)\).

Algorithm

1. Initialize the candidate answer variable ans with \(A_1\).
2. Iterate through the pillars from \(i = 2\) to \(N\), updating ans to the larger of ans and \(A_i - 1\).
3. Output the final ans.

Complexity

• Time complexity: \(O(N)\) The processing completes in a single loop over \(N\) pillars, making it very fast.
• Space complexity: \(O(N)\) \(O(N)\) is required when storing the input as a list.

Implementation Notes

• Since \(N\) can be as large as \(5 \times 10^5\), in Python it is advisable to read input quickly using sys.stdin.read().split() or similar methods.

• Note that the condition for the first pillar is simply \(A_1\) as-is, while for the second pillar onward it becomes \(A_i - 1\).

  Source Code

import sys

def solve():
    # Read all input data at once for speed
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    # N: number of pillars
    n = int(input_data[0])
    # a: list of durabilities A_1, A_2, ..., A_N
    a = list(map(int, input_data[1:]))
    
    # The condition for all pillars to collapse:
    # 1. Pillar 1 must collapse from the initial shock X:
    #    X >= A_1
    # 2. For each pillar i (where i > 1), it collapses if its durability 
    #    minus (initial shock X + 1 if the previous pillar collapsed) is <= 0.
    #    Since we require all pillars to collapse, Pillar i-1 will have collapsed,
    #    giving Pillar i an additional 1 damage.
    #    So: X + 1 >= A_i  =>  X >= A_i - 1
    
    # Start with the requirement for the first pillar
    ans = a[0]
    
    # Check the requirement for all subsequent pillars
    for i in range(1, n):
        # The required X for pillar i+1 is A_{i+1} - 1
        if a[i] - 1 > ans:
            ans = a[i] - 1
            
    # Output the minimum force X
    print(ans)

if __name__ == '__main__':
    solve()

This editorial was generated by gemini-3-flash-thinking.