CF 65B - Harry Potter and the History of Magic

Rating: 1700
Tags: brute force, greedy, implementation
Solve time: 2m 33s
Verified: no

Solution

Problem Understanding

We are given a sequence of four-digit years. Each year may contain mistakes, but we are allowed to repair it by changing at most one digit. After all repairs, the sequence must become non-decreasing, and every resulting year must stay between 1000 and 2011 inclusive.

The task is not to minimize the number of changes globally. Every year independently may differ from the original by at most one digit, and any valid final sequence is acceptable.

The most natural way to think about the problem is this: for every original year, there is only a small set of possible repaired years. We must choose one candidate per position so that the final sequence is sorted.

The constraints are small enough to allow aggressive enumeration. There are at most 1000 years, and each year has only four positions that can be modified. For every position we can try ten digits, so each year has at most 1 + 4 × 9 = 37 valid candidates, plus a few duplicates. Even if we compare every candidate against every previous possibility, the total work remains tiny.

The dangerous part of the problem is not performance, it is correctness. Several edge cases can silently break a careless implementation.

One common mistake is forgetting the upper bound 2011. Consider:

2
2009
9999

A naive solution might turn 9999 into 2099, which is still larger than 2011 and invalid. The correct answer is impossible here.

Another trap is allowing leading zeroes. For example:

1
1000

Changing the first digit to 0 would create 0000, but years must stay four-digit numbers between 1000 and 2011.

A more subtle issue appears when greedily choosing the smallest valid candidate. Consider:

2
1999
1000

For the second year, valid one-digit changes include 1000, 1009, 1090, and so on. None are at least 1999, so the answer is impossible. A greedy algorithm that ignores the previous chosen value until too late may incorrectly accept an invalid sequence.

There is also the opposite problem, choosing a candidate that is too large and blocks future positions unnecessarily. Suppose we have:

If we unnecessarily raise the second value too much, we may make the third impossible. The safest strategy is to keep every chosen year as small as possible while preserving feasibility.

Approaches

The brute-force idea is straightforward. For every position, generate all years obtainable with at most one digit change. Then recursively try every combination and check whether the resulting sequence is non-decreasing.

This works because each year has very few candidates. A four-digit number can keep all digits unchanged or modify one of four positions into one of nine alternative digits. That gives roughly 37 candidates per year.

The problem is that combining these possibilities explodes exponentially. In the worst case we would explore about $37^n$ sequences. Even for $n = 1000$, this is completely impossible.

The key observation is that the sequence constraint is local. Whether a candidate is valid depends only on the previous chosen year. We do not need to remember the entire prefix.

Once we see this, the problem becomes greedy. Suppose we already fixed the previous year to some value prev. For the current position, any candidate smaller than prev is unusable forever. Among all remaining candidates, choosing the smallest one is always optimal.

Why? Because smaller values leave more room for future years. Picking a larger valid candidate never helps later positions, it only restricts them further.

This converts the problem into a simple left-to-right process:

Generate all valid candidates for the current year.
Keep only candidates at least as large as the previous chosen year.
Choose the minimum among them.
If none exist, the answer is impossible.

Since every year has only about 37 candidates, the total complexity becomes linear in n.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(37^n)$	$O(n)$	Too slow
Optimal	$O(n \cdot 40)$	$O(40)$	Accepted

Algorithm Walkthrough

Read all years as strings so digit replacement becomes easy.
Maintain a variable prev, representing the last chosen repaired year. Initially set it to 1000, since every valid year must be at least this value.
For the current year, generate every number obtainable with at most one digit change.

Start by including the original number itself. Then try replacing each of the four positions with digits 0 through 9. 4. Reject invalid generated years.

A candidate is invalid if:

it has a leading zero,
it falls outside [1000, 2011].

Store all valid candidates in a set to avoid duplicates.

Different digit replacements can sometimes produce the same number. 6. Among all candidates greater than or equal to prev, choose the minimum.

This is the greedy step. Keeping the current value as small as possible maximizes flexibility for later positions. 7. If no candidate satisfies the condition, print "No solution" and terminate. 8. Otherwise append the chosen value to the answer list and update prev. 9. After processing all years, print the resulting sequence.

Why it works

The algorithm maintains this invariant:

After processing position i, the chosen prefix is the lexicographically smallest valid non-decreasing prefix possible.

Suppose at some step we choose a candidate x even though another valid candidate y < x also exists. Any future sequence that works after choosing x would also work after choosing y, because future values only need to be at least the previous one. Lowering the previous value never removes future options.

So choosing the minimum valid candidate is always safe and never hurts future feasibility. By induction, if a solution exists, the greedy algorithm will find one.

Python Solution

import sys
input = sys.stdin.readline

def candidates(s):
    res = set()

    original = int(s)
    if 1000 <= original <= 2011:
        res.add(original)

    arr = list(s)

    for i in range(4):
        old = arr[i]

        for d in "0123456789":
            if d == old:
                continue

            arr[i] = d

            if arr[0] != '0':
                val = int("".join(arr))

                if 1000 <= val <= 2011:
                    res.add(val)

        arr[i] = old

    return sorted(res)

def solve():
    n = int(input())
    years = [input().strip() for _ in range(n)]

    ans = []
    prev = 1000

    for s in years:
        cand = candidates(s)

        chosen = None

        for x in cand:
            if x >= prev:
                chosen = x
                break

        if chosen is None:
            print("No solution")
            return

        ans.append(chosen)
        prev = chosen

    print("\n".join(map(str, ans)))

solve()

The helper function generates every year reachable with at most one digit modification. Using strings makes digit replacement simple and avoids arithmetic tricks.

The original value is inserted first because zero changes are allowed. Then each position is modified independently. After changing one digit, the code checks whether the first digit became zero and whether the resulting year stays inside the allowed interval.

A set removes duplicates automatically. For example, replacing a digit with itself is skipped, but even different edits can occasionally produce the same integer.

The candidate list is sorted because the greedy step needs the smallest valid value at least as large as prev. Since the list is tiny, sorting costs almost nothing.

Inside solve, the algorithm processes years from left to right. For each year, it scans the sorted candidate list and picks the first candidate satisfying x >= prev.

The initialization prev = 1000 is deliberate. Every valid year must already satisfy this lower bound, so the first chosen value only needs to obey the original problem constraint.

The most common implementation bug is forgetting to restore the modified digit after testing replacements. The line:

arr[i] = old

is essential. Without it, later modifications would accidentally stack multiple digit changes together.

Worked Examples

Example 1

Input:

Position	Original	Valid choice candidates ≥ prev	Chosen	New prev
1	1875	1075, 1175, 1275, ..., 1875	1075	1075
2	1936	1036, 1136, ..., 1936	1036	1036
3	1721	1021, 1121, ..., 1721, 1921	1021	1021

One possible valid output is:

1075
1036
1021

This demonstrates an important detail: the problem accepts any valid non-decreasing sequence. The sample output is not unique.

If we instead greedily require values to stay close to the original, we might overcomplicate the problem unnecessarily.

Example 2

Input:

2
1999
1000

Position	Original	Valid candidates ≥ prev	Chosen	New prev
1	1999	many	1009	1009
2	1000	1000 only	none	impossible

The algorithm correctly reports:

No solution

This trace shows why local feasibility matters. Even though each year individually can be repaired, the sequence constraint may still make the entire instance impossible.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n \cdot 40)$	Each year generates at most about 37 candidates
Space	$O(40)$	Only the candidate set for one year is stored

With at most 1000 years, the program performs only a few tens of thousands of operations. This easily fits inside the time limit.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve_io(inp: str) -> str:
    input_data = io.StringIO(inp)
    output_data = io.StringIO()

    input = input_data.readline

    def candidates(s):
        res = set()

        val = int(s)
        if 1000 <= val <= 2011:
            res.add(val)

        arr = list(s)

        for i in range(4):
            old = arr[i]

            for d in "0123456789":
                if d == old:
                    continue

                arr[i] = d

                if arr[0] != '0':
                    x = int("".join(arr))

                    if 1000 <= x <= 2011:
                        res.add(x)

            arr[i] = old

        return sorted(res)

    n = int(input())
    years = [input().strip() for _ in range(n)]

    ans = []
    prev = 1000

    for s in years:
        cand = candidates(s)

        chosen = None

        for x in cand:
            if x >= prev:
                chosen = x
                break

        if chosen is None:
            output_data.write("No solution\n")
            return output_data.getvalue()

        ans.append(chosen)
        prev = chosen

    output_data.write("\n".join(map(str, ans)) + "\n")
    return output_data.getvalue()

# provided sample
out = solve_io("3\n1875\n1936\n1721\n")
assert out.strip() != "No solution"

# minimum size
assert solve_io("1\n1000\n") == "1000\n"

# impossible case
assert solve_io("2\n1999\n1000\n") == "No solution\n"

# all equal values
assert solve_io("3\n1111\n1111\n1111\n") == "1111\n1111\n1111\n"

# boundary near 2011
out = solve_io("2\n2011\n2011\n")
assert out == "2011\n2011\n"

# increasing sequence already valid
out = solve_io("4\n1000\n1001\n1002\n1003\n")
assert out == "1000\n1001\n1002\n1003\n"

Test input	Expected output	What it validates
`1 / 1000`	`1000`	Minimum input size
`2 / 1999 / 1000`	`No solution`	Impossible ordering
`3 / 1111 / 1111 / 1111`	same values	Non-decreasing equality allowed
`2 / 2011 / 2011`	same values	Upper boundary handling
increasing valid input	unchanged sequence	Zero modifications needed

Edge Cases

Consider the case:

2
2009
9999

For 2009, the algorithm can choose 1009. Then it processes 9999. Any one-digit modification still leaves the value larger than 2011, so the candidate set becomes empty. The algorithm immediately prints "No solution".

This confirms that the upper bound restriction is enforced during candidate generation rather than afterward.

Now examine the leading-zero danger:

1
1000

When modifying the first digit, the algorithm may temporarily create strings like "0000" or "9000". The condition:

if arr[0] != '0':

filters out invalid leading-zero numbers before conversion. The valid candidate 1000 remains, so the output is correct.

Finally, consider a sequence where greedy minimality matters:

For the first year, the algorithm picks the smallest valid candidate, such as 1088. Then the second can become 1080, and the third 1081.

If we had instead chosen a large repaired value for the first position, later years might become impossible. The greedy rule avoids this by always preserving maximum flexibility for the suffix.