CF 102697034 - Broken Ladder

CF 102697034 - Broken Ladder

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Tags: -
Solve time: 3m 49s
Verified: yes

Solution

Problem Understanding

The ladder originally has rungs at every integer position from the bottom to the top. Some rungs have disappeared, and the input gives only the positions of the rungs that remain. The first and last rungs are guaranteed to still exist, so the ladder always has a clear start and end.

A climber can move between consecutive remaining rungs only if the distance between them is at most k. The task is to decide whether every jump needed to climb the remaining ladder is small enough. We output YES when the broken ladder is still climbable and NO otherwise.

The input size determines the intended solution. The number of remaining rungs can be large, so checking every possible subset of missing rungs or simulating every possible climb path would be unnecessary work. With a typical one second limit, an algorithm around O(n) is expected for a single pass over the rung positions. Anything quadratic would quickly become too slow when the ladder contains many rungs.

The main edge cases come from the gaps between neighboring rungs. A ladder with no broken rungs has every gap equal to one, so it should always be accepted when k is at least one.

For example:

6 1
1 2 3 4 5 6

The output is:

YES

A careless implementation that searches only for missing rungs may incorrectly reject this because it never handles the normal ladder case separately.

A single large gap is enough to make the ladder impossible.

5 2
1 2 3 7 8

The output is:

NO

The gap from position 3 to position 7 is 4, which is larger than the allowed jump. An implementation that checks only the total height of the ladder instead of each local gap would miss this failure.

The first and last rungs are already present, so there is no need to compare against positions outside the given array. An implementation that assumes an extra rung at position zero or after the last rung may create incorrect boundary gaps.

Approaches

The straightforward solution is to examine the ladder one pair of neighboring rungs at a time. Since the positions are already sorted, we can calculate every jump by subtracting the previous rung position from the current one. If any jump is larger than k, climbing is impossible. This method directly follows the definition of a climbable ladder.

A brute-force approach might try to reconstruct which missing rungs could be restored and test different climbing paths. That is unnecessary because the climber never has a choice: the remaining rungs are fixed, and every consecutive pair must be usable. Trying all possibilities would create a huge search space. With n remaining rungs, even comparing many possible subsets would quickly become exponential, while the useful information is only the n - 1 gaps between adjacent rungs.

The key observation is that the ladder condition is local. A failure anywhere between two consecutive rungs makes the entire ladder impossible. Because of that, we can reduce the whole problem to finding the maximum adjacent gap. If that maximum gap is at most k, every jump is valid.

The brute-force works because it attempts to verify possible climbs, but it fails because it spends time exploring choices that do not exist. The observation that only adjacent rung distances matter lets us solve the problem with a single scan.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential in the number of missing rungs	Exponential	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of remaining rungs, the maximum allowed jump, and the sorted rung positions.
2. Start from the second rung and compare it with the previous rung. The difference between their positions is the jump a climber must make.
3. If any jump is larger than k, immediately output NO. A single invalid jump prevents reaching the top of the ladder.
4. If all neighboring gaps are at most k, output YES. Every move required to climb the ladder is valid.

Why it works:

The algorithm maintains the property that every checked part of the ladder can be climbed. When it examines a pair of neighboring rungs, it verifies the only possible jump between them. If all such pairs pass the check, there is no remaining obstacle because the climber's route consists exactly of these consecutive jumps. If one pair fails, the climber cannot cross that section, so rejecting the ladder is correct.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, k = map(int, input().split())
    rungs = list(map(int, input().split()))

    for i in range(1, n):
        if rungs[i] - rungs[i - 1] > k:
            print("NO")
            return

    print("YES")

if __name__ == "__main__":
    solve()

The code reads the sorted rung positions and checks only adjacent pairs. The loop starts at index 1 because the first rung has no previous rung inside the input array.

The subtraction order matters. The positions are sorted, so rungs[i] - rungs[i - 1] always gives a positive gap. There is no need for extra sorting, which keeps the solution linear.

The algorithm returns immediately after finding a bad gap. Continuing the scan would not change the answer, since one impossible jump already makes the entire ladder unusable.

Python integers do not overflow, so there is no extra handling needed for large rung positions.

Worked Examples

Sample 1:

6 2
1 2 3 5 6 8

Current rung	Previous rung	Gap	Result
2	1	1	Valid
3	2	1	Valid
5	3	2	Valid
6	5	1	Valid
8	6	2	Valid

Every jump is within the allowed size, so the answer is YES. This trace shows that the algorithm only needs local gap checks and does not care where the missing rungs are.

Sample 2:

5 3
1 3 8 12 13

Current rung	Previous rung	Gap	Result
3	1	2	Valid
8	3	5	Invalid

The second jump already exceeds the limit, so the algorithm stops and outputs NO. The remaining rungs do not matter after a failed jump is found.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each rung is checked once against its predecessor.
Space	O(1)	Only the input array and a few variables are used.

The solution performs a constant amount of work per rung. It easily fits the time limit because it avoids any nested iteration or reconstruction of the missing ladder parts.

Test Cases

import sys
import io

def solution(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    n, k = map(int, input().split())
    rungs = list(map(int, input().split()))

    for i in range(1, n):
        if rungs[i] - rungs[i - 1] > k:
            return "NO\n"

    return "YES\n"

assert solution("""6 2
1 2 3 5 6 8
""") == "YES\n", "sample 1"

assert solution("""5 3
1 3 8 12 13
""") == "NO\n", "sample 2"

assert solution("""2 1
1 2
""") == "YES\n", "minimum ladder"

assert solution("""5 10
1 2 3 4 5
""") == "YES\n", "large allowed gap"

assert solution("""5 2
1 3 5 7 9
""") == "YES\n", "all equal gaps"

assert solution("""5 3
1 2 3 7 8
""") == "NO\n", "single large gap"

Test input	Expected output	What it validates
2 1 / 1 2	YES	Minimum number of rungs and smallest valid gap
5 10 / 1 2 3 4 5	YES	Large allowed jump value
5 2 / 1 3 5 7 9	YES	Repeated equal gaps
5 3 / 1 2 3 7 8	NO	Detecting one invalid internal gap

Edge Cases

For a ladder with no missing rungs, the algorithm checks every neighboring pair and sees gaps of size one. For example:

6 1
1 2 3 4 5 6

Each subtraction gives 1, which is not larger than k, so the algorithm outputs YES. This handles the normal ladder case without any special condition.

For a ladder containing one unavoidable jump:

5 2
1 2 3 7 8

The algorithm checks the gaps 1, 1, 4, and 1. When it reaches the gap between 3 and 7, it finds that 4 > 2 and returns NO. The result is correct because that section of the ladder cannot be crossed.

For the boundary case where the first and last rungs are the only available information:

2 5
1 6

There is one jump of size 5. The algorithm checks that single gap and returns YES. It does not invent a gap before the first rung or after the last rung, matching the actual climbing path.