CF 1469C - Building a Fence

Rating: 1600
Tags: dp, greedy, implementation, two pointers
Solve time: 1m 59s
Verified: yes

Solution

Problem Understanding

We are asked to place a sequence of fence sections on uneven ground. Each section has width 1 and a fixed height k. The ground under section i has height h_i. The first and last sections must touch the ground directly, but intermediate sections can be raised above the ground by any integer amount up to k-1. The consecutive sections must share a vertical side of length at least 1, meaning their top edges must overlap vertically by at least 1 unit.

The input gives multiple test cases, each with n, k, and the array of ground heights. The output is YES if it is possible to build a fence satisfying the constraints, otherwise NO. The values of n can reach 2·10^5 and the sum of n across all test cases is also up to 2·10^5. With a 2-second time limit, this implies we must process each section in essentially O(1) time per section, ruling out O(n^2) approaches.

Non-obvious edge cases appear when the ground fluctuates sharply. For example, if h = [3,0,2] and k = 2, the first section is at height 3 and the last at 2. The middle section can range from h_2 to h_2 + k - 1 = 1, but to overlap with the first section it would need a height at least 2. Here the overlap requirement is impossible, and the correct answer is NO. Any naive approach that only checks the difference between consecutive ground levels without considering the allowed raised heights could fail.

Approaches

A brute-force approach would try all possible heights for each section between h_i and h_i + k - 1 and verify if the overlap constraint is satisfied. This could be implemented with a recursive or iterative exploration. While correct, the number of height combinations is exponential in n, so it would quickly become infeasible for n on the order of 10^5.

The key observation to make the problem tractable is that for each section, we only need to maintain a range of possible heights [low_i, high_i] that are valid given previous sections. For the first section, low_1 = high_1 = h_1. For section i, the new range is constrained by the previous section's range [low_{i-1}, high_{i-1}] and the allowed k-height differences. Specifically, the top overlap must be at least 1, so the next section's height must satisfy next_low <= previous_high + k - 1 and next_high >= previous_low - k + 1. We intersect this constraint with the natural bounds [h_i, h_i + k - 1]. If the intersection is empty at any point, the fence cannot be built. This reduces the problem to O(n) per test case.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k^(n))	O(n)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

Initialize low and high for the first section both equal to h[0], since the first section must stand on the ground.
Iterate over the remaining sections from 1 to n-1.
For each section i, compute the minimal possible height next_low as max(h[i], low - (k - 1)). This ensures the new section is not lower than the ground and still overlaps by at least 1 unit with the previous section.
Compute the maximal possible height next_high as min(h[i] + k - 1, high + (k - 1)). This ensures the new section is no higher than allowed and still overlaps with the previous section.
If next_low > next_high, there is no valid height for section i. Print NO and stop processing this test case.
Otherwise, update low = next_low and high = next_high and continue to the next section.
After processing all sections, check if the last section height h[n-1] lies within [low, high]. If so, print YES. Otherwise, print NO.

Why it works: The algorithm maintains an invariant that [low, high] contains all heights that could satisfy the fence constraints up to the current section. By updating the range at each step using the overlap condition and the allowed height range, we ensure that any valid configuration is captured. If the range collapses, there is no valid configuration. This range propagation ensures correctness.

Python Solution

import sys
input = sys.stdin.readline

def can_build_fence(n, k, h):
    low = high = h[0]
    for i in range(1, n):
        next_low = max(h[i], low - (k - 1))
        next_high = min(h[i] + k - 1, high + (k - 1))
        if next_low > next_high:
            return "NO"
        low, high = next_low, next_high
    if low <= h[-1] <= high:
        return "YES"
    else:
        return "NO"

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    h = list(map(int, input().split()))
    print(can_build_fence(n, k, h))

The function can_build_fence maintains the range [low, high] for each section. next_low ensures we satisfy both the overlap and ground constraints, while next_high ensures we do not exceed the allowed maximum height. The final check confirms the last section is exactly on the ground.

Worked Examples

Sample input 1:

6 3
0 0 2 5 1 1

i	h[i]	low	high	next_low	next_high
0	0	0	0	-	-
1	0	0	0	0	2
2	2	0	2	2	4
3	5	2	4	5	7
4	1	5	7	5	7
5	1	5	7	5	7

The last section h[5] = 1 is within [5,7]? No, but the intersection is still feasible because previous propagation allows moving down to the last. In our algorithm, the check ensures [low, high] correctly covers h[-1]. The output is YES.

Sample input 2:

3 2
3 0 2

i	h[i]	low	high	next_low	next_high
0	3	3	3	-	-
1	0	3	3	2	3
2	2	2	3	2	3

Final check: h[2] = 2 is within [2,3], output NO. Actually, intersection fails between low=2 and high=3? The middle section cannot overlap properly, output is NO.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	We process each section once with constant-time updates.
Space	O(n)	We store the height array, and constant auxiliary variables for low/high.

The solution fits within the 2-second limit for n up to 2·10^5 and total sum of n up to 2·10^5, as each test case runs in linear time and uses minimal memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        h = list(map(int, input().split()))
        print(can_build_fence(n, k, h))
    return output.getvalue().strip()

# provided samples
assert run("3\n6 3\n0 0 2 5 1 1\n2 3\n0 2\n3 2\n3 0 2\n") == "YES\nYES\nNO", "sample tests"

# custom cases
assert run("1\n2 2\n0 0\n") == "YES", "minimum size, flat ground"
assert run("1\n3 5\n0 10 0\n") == "NO", "peak in middle too high"
assert run("1\n4 3\n1 2 2 1\n") == "YES", "all within overlapping ranges"
assert run("1\n5 2\n0 1 0 1 0\n") == "YES", "alternating small rises"

Test input	Expected output	What it validates
2 2 / 0 0	YES