Overview

Given a route consisting of consecutive points, find the minimum number of points such that the total elevation difference (sum of absolute differences between adjacent points) is at least \(K\).

Analysis

In this problem, the “total elevation difference” for the interval from point \(l\) to point \(r\) is defined as:

\[ \sum_{i=l}^{r-1} |A_{i+1} - A_i| \]

By precomputing the absolute differences in elevation between adjacent points, this can be efficiently calculated as an interval sum.

Issues with the Naive Approach

A straightforward approach would be to try all intervals \([l, r]\) and check whether the total elevation difference is at least \(K\). However, the number of intervals is \(O(N^2)\), and since \(N\) can be up to \(2 \times 10^5\), this is not practical (it would result in TLE).

Solution: Binary Search + Sliding Window

Since we want to find the minimum number of points (i.e., the interval length), we consider the decision problem: “Does there exist an interval with \(x\) points that satisfies the condition?” This can be solved using binary search.

Furthermore, for each number of points, we need to efficiently determine whether a qualifying interval exists. By using a prefix sum of the adjacent differences, we can compute any interval sum in \(O(1)\). Then, by using a sliding window to find the maximum interval sum among all fixed-length intervals and checking whether it is at least \(K\), each decision step can be performed in linear time.

Algorithm

1. Precompute the absolute differences in elevation between adjacent points into an array diff: $\( \text{diff}[i] = |A_{i+1} - A_i| \)$

2. Compute the prefix sum of diff into an array prefix: $\( \text{prefix}[i] = \sum_{j=0}^{i-1} \text{diff}[j] \)$

3. Perform binary search on the number of points \(len\):

   • Determine whether there exists an interval sum of diff with length \(len - 1\) that is at least \(K\).
   • This amounts to checking whether there exist \(i, j\) (where \(j - i = len - 1\)) such that prefix[j] - prefix[i] >= K, using a sliding window.

4. Output the minimum \(len\) that satisfies the condition. If none exists, output -1.

Complexity

• Time complexity: \(O(N \log N)\)
  Each binary search step performs a linear-time check using the sliding window, so the overall complexity is \(O(N \log N)\).
• Space complexity: \(O(N)\)
  The diff and prefix arrays use \(O(N)\) memory.

Implementation Notes

• Use prefix sums to efficiently compute interval sums.

• Set the binary search range from 1 to N, and be careful with the case mid = 1 (since the interval length becomes 0, making it always impossible).

• Use the sliding window to efficiently compute the maximum interval sum and perform the condition check.

  Source Code

import sys
from itertools import accumulate

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    
    N = int(data[0])
    K = int(data[1])
    A = list(map(int, data[2:]))

    # Compute the absolute differences
    diff = [abs(A[i+1] - A[i]) for i in range(N-1)]
    
    # If K is 0, we can always take a single point
    if K == 0:
        print(1)
        return
    
    # Prefix sums of differences
    prefix = [0] + list(accumulate(diff))
    
    # Binary search on the answer (number of points in route)
    # The number of points is len = r - l + 1 => r - l = len - 1
    # So total difference is prefix[r] - prefix[l]
    # We want to find minimal len such that there exists r-l = len-1 with prefix[r] - prefix[l] >= K
    
    def possible(length):
        # Check if there's any subarray of consecutive differences of length (length-1) 
        # whose sum is at least K
        if length == 1:
            return False  # Since K > 0 and sum of 0 terms is 0
        window_len = length - 1
        if window_len > len(prefix) - 1:
            return False
        for i in range(len(prefix) - window_len):
            j = i + window_len
            if prefix[j] - prefix[i] >= K:
                return True
        return False

    # More efficient check using sliding window minimum
    def possible_efficient(length):
        if length == 1:
            return False
        window_len = length - 1
        if window_len > len(prefix) - 1:
            return False
        for i in range(len(prefix) - window_len):
            j = i + window_len
            if prefix[j] - prefix[i] >= K:
                return True
        return False

    # Even more efficient: monotonic queue approach for large inputs -> but since we're doing binary search 
    # over answers and checking feasibility each time, let's optimize the feasibility check.
    
    left = 1
    right = N
    answer = -1
    
    while left <= right:
        mid = (left + right) // 2
        # We want to check if there's a subarray of 'diff' of length (mid - 1) with sum >= K
        window_size = mid - 1
        found = False
        
        if window_size == 0:
            # mid=1 => only one point => sum = 0 < K (since K>0), so not valid
            found = False
        elif window_size > len(diff):
            found = False
        else:
            # Sliding window maximum of size window_size in diff array
            current_sum = sum(diff[:window_size])
            if current_sum >= K:
                found = True
            else:
                for i in range(window_size, len(diff)):
                    current_sum += diff[i] - diff[i - window_size]
                    if current_sum >= K:
                        found = True
                        break
        if found:
            answer = mid
            right = mid - 1
        else:
            left = mid + 1
            
    print(answer)

if __name__ == "__main__":
    main()

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