Overview

This problem asks us to minimize the number of points whose elevations need to be rewritten (= maximize the number of points that retain their original elevation) when traveling from point \(1\) to point \(N\), while ensuring that the elevation decrease between any two adjacent points is at most \(K-1\).

Analysis

Reformulating the Condition

The condition “feeling fear when the elevation drops by \(K\) or more” means that to avoid feeling fear, the following inequality must hold for all \(i\): $\(H_i - H_{i+1} \leq K - 1\)\( Setting \)D = K - 1\(, the condition can be written as \)H_{i+1} \geq H_i - D\(. Extending this relationship from point \)i\( to point \)j\( (\)i < j\(), we get: \)\(H_j \geq H_i - (j - i)D\)$

Rearranging this expression, we can transform it into a form combining index \(i\) and elevation \(H_i\): $\(H_j + j \cdot D \geq H_i + i \cdot D\)$

Sequence Transformation

Defining a new sequence \(A\) as \(A_i = H_i + i \cdot D\), the condition becomes “for any chosen point indices \(i, j\) (\(i < j\)), \(A_i \leq A_j\) must hold”, meaning the sequence \(A\) needs to be non-decreasing.

Fixed Endpoints

The elevations at points \(1\) and \(N\) cannot be changed. Therefore, we first need to verify the following conditions: 1. \(A_1 \leq A_N\) must hold. If this is not satisfied, no matter how we rewrite the intermediate points, the condition cannot be met, so the answer is -1. 2. To keep an intermediate point \(i\) (\(1 < i < N\)) unchanged, it must satisfy \(A_1 \leq A_i \leq A_N\).

After listing the \(A_i\) values of intermediate points that satisfy this condition, the problem reduces to finding the maximum number of elements we can select from them such that they form a non-decreasing subsequence.

Algorithm

1. Set \(D = K - 1\).
2. For each point \(i\), compute \(A_i = H_i + (i-1) \cdot D\) (or \(A_i = H_i + i \cdot D\) if 0-indexed).
3. Compare \(A_0\) and \(A_{N-1}\); if \(A_0 > A_{N-1}\), output -1.
4. For \(1 \leq i \leq N-2\), extract in order those \(A_i\) satisfying \(A_0 \leq A_i \leq A_{N-1}\) to form a new list target.
5. Compute the Longest Increasing Subsequence (LIS), more precisely the Longest Non-Decreasing Subsequence length for target.
   • This can be computed efficiently using binary search (bisect_right).
6. The answer is (number of rewritable points \(N-2\)) - (length of the longest non-decreasing subsequence).

Complexity

• Time Complexity: \(O(N \log N)\)
  • Constructing sequence \(A\) takes \(O(N)\).
  • Computing the longest non-decreasing subsequence takes \(O(N \log N)\).
• Space Complexity: \(O(N)\)
  • Needed to store sequence \(A\) and the arrays used for LIS computation.

Implementation Notes

• Computing the Longest Non-Decreasing Subsequence: In the standard LIS (strictly increasing), bisect_left is used, but since equal consecutive values are allowed in this problem, we use bisect_right.

• Non-negative integer constraint on elevations: The problem states that elevations can be “changed to non-negative integer values,” but since there is no upper bound on the modified elevations, we can always adjust values to be sufficiently large to satisfy \(A_i \leq A_{i+1}\). Therefore, the non-negativity constraint does not need to be directly considered in the LIS computation.

• Index handling: In the implementation, calculations are done 0-indexed, so point \(i\) (\(1 \leq i \leq N\)) is treated as \(0 \leq i \leq N-1\), and computing \(A_i = H_i + i \cdot D\) makes things straightforward.

  Source Code

import sys
from bisect import bisect_right

def solve():
    # Fast input reading
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    # Check if we have at least N and K
    if len(input_data) < 2:
        return
        
    N = int(input_data[0])
    K = int(input_data[1])
    
    # Altitudes H_1, ..., H_N
    H = list(map(int, input_data[2:]))
    
    # Descent limit: H_i - H_{i+1} <= K - 1
    # Let D = K - 1. The condition is H_i - H_{i+1} <= D.
    # This is equivalent to H_{i+1} >= H_i - D.
    # By induction, for any i < j, H_j >= H_i - (j - i) * D.
    # This can be rewritten as H_j + j * D >= H_i + i * D.
    # Let A[i] = H[i] + i * D (using 0-based indexing for i).
    # The condition becomes A[j] >= A[i] for any kept indices i < j.
    
    D = K - 1
    
    # Points 1 and N are fixed (indices 0 and N-1)
    # They must satisfy the descent condition: H_N >= H_1 - (N - 1) * D
    A_start = H[0] + 0 * D
    A_end = H[N - 1] + (N - 1) * D
    
    if A_start > A_end:
        print("-1")
        return
        
    # We want to keep the maximum number of original H[i] for 1 <= i <= N-2
    # such that A_start <= A[i] <= A_end and the kept A[i] are non-decreasing.
    target = []
    for i in range(1, N - 1):
        val = H[i] + i * D
        # A point i can be kept only if it's reachable from H_1 and can reach H_N
        if A_start <= val <= A_end:
            target.append(val)
    
    # Longest Non-Decreasing Subsequence (LNDS) of target
    # tails[i] will store the smallest tail of all non-decreasing subsequences of length i+1
    tails = []
    for x in target:
        # Use bisect_right to find the position to maintain non-decreasing order
        idx = bisect_right(tails, x)
        if idx < len(tails):
            tails[idx] = x
        else:
            tails.append(x)
    
    max_kept_middle = len(tails)
    # The answer is the number of middle points minus the maximum number we could keep
    print((N - 2) - max_kept_middle)

if __name__ == "__main__":
    solve()

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