CF 124B - Permutations

Rating: 1400
Tags: brute force, combinatorics, implementation
Solve time: 1m 34s
Verified: yes

Solution

Problem Understanding

We are given several strings of digits, all with the same length. We may choose one permutation of digit positions and apply it to every string. After rearranging the digits according to that shared permutation, each string becomes a new integer, possibly with leading zeroes.

Our goal is to make the difference between the largest resulting number and the smallest resulting number as small as possible.

For example, suppose we have:

123
908

If we choose the permutation (2, 1, 3), then:

123 -> 213
908 -> 098

The resulting values are 213 and 98, so the difference is 115.

The crucial restriction is that every number must use the same reordering of positions. We are not allowed to permute each number independently.

The constraints are very small. Both n and k are at most 8. The interesting bound is k ≤ 8, because the number of permutations of k positions is k!.

At worst:

8! = 40320

For each permutation, we rebuild n numbers, each with k digits. The total amount of work is roughly:

40320 × 8 × 8 ≈ 2.5 million operations

That comfortably fits within a 1 second limit in Python.

The small value of k completely changes the nature of the problem. A brute-force over all permutations would be impossible if k were even moderately larger, but with k ≤ 8, exhaustive search becomes the intended solution.

There are a few edge cases that commonly break incorrect implementations.

Consider numbers with leading zeroes:

2 3
001
100

If we swap the first and third positions, we get:

100
001

The values are 100 and 1, not 100 and 001 as strings. A solution that compares strings lexicographically instead of converting to integers may silently produce the wrong answer.

Another subtle case appears when all numbers become equal under some permutation:

The correct answer is:

A careless implementation that initializes the answer incorrectly or forgets to update minimum and maximum values per permutation may miss this.

There is also a boundary case with the smallest possible input:

1 1
7

Only one number exists, so the largest and smallest are identical. The answer must be 0 regardless of permutation choices.

Approaches

The most direct idea is to try every possible permutation of the digit positions.

Suppose k = 4. A permutation such as (2, 0, 3, 1) means:

abcd -> cadb

We apply that rearrangement to every input number, convert the result into an integer, then compute:

max_value - min_value

Among all permutations, we keep the minimum difference.

This brute-force is correct because every valid transformation corresponds to exactly one permutation of positions. Since the problem asks for the best possible shared rearrangement, checking all permutations guarantees that we eventually examine the optimal one.

The natural concern is performance. The number of permutations grows factorially:

k!

If k were 15, this would be completely infeasible. But here k ≤ 8, so the worst case is only 40320 permutations. For each permutation we process at most 8 strings of length 8.

That makes the exhaustive search fast enough.

The key observation is that the search space depends only on k, not on the numerical values themselves. Because k is tiny, we do not need any advanced pruning or optimization. The intended solution is simply a clean implementation of complete enumeration.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over all permutations	O(k! × n × k)	O(k)	Accepted
Any more complicated optimization	Unnecessary	Unnecessary	Overkill

Algorithm Walkthrough

Read all input numbers as strings.

Keeping them as strings makes digit reordering easy. We do not want arithmetic digit extraction here. 2. Generate every permutation of the indices 0 ... k-1.

Each permutation represents one global rearrangement rule applied to every number. 3. For the current permutation, rebuild every number.

For each original string, construct a new string by taking characters in the order specified by the permutation.

Example:

s = "5237"
perm = (2, 0, 3, 1)

result = "3572"

Convert each rebuilt string into an integer.

This correctly handles leading zeroes automatically. 5. Track the minimum and maximum values produced by this permutation.

Their difference is the spread created by this rearrangement rule. 6. Update the global answer with the smallest difference seen so far. 7. After all permutations are processed, print the answer.

Why it works

Every legal operation in the problem is exactly a permutation of digit positions applied uniformly to all numbers. The algorithm enumerates all such permutations without omission or duplication.

For each permutation, it computes the exact largest and smallest resulting values, so the difference for that configuration is correct.

Since the algorithm checks every possible configuration and keeps the minimum difference among them, the final answer must be the optimal one.

Python Solution

import sys
from itertools import permutations

input = sys.stdin.readline

def solve():
    n, k = map(int, input().split())
    nums = [input().strip() for _ in range(n)]

    ans = float('inf')

    for perm in permutations(range(k)):
        mn = float('inf')
        mx = -1

        for s in nums:
            val = int(''.join(s[i] for i in perm))

            mn = min(mn, val)
            mx = max(mx, val)

        ans = min(ans, mx - mn)

    print(ans)

solve()

The solution follows the algorithm directly.

permutations(range(k)) generates every possible rearrangement of positions. Since k ≤ 8, this remains efficient enough.

The input numbers are stored as strings because indexing characters is simpler and less error-prone than repeatedly extracting digits mathematically.

For each permutation, we rebuild a number using:

''.join(s[i] for i in perm)

This constructs the digit sequence after rearrangement.

Converting with int(...) is important. Leading zeroes are allowed, and integer conversion naturally handles cases like "001" becoming 1.

The minimum and maximum values are recomputed independently for every permutation. Forgetting to reset them inside the permutation loop is a common bug.

The memory usage stays tiny because we only store the input strings and a few temporary values.

Worked Examples

Sample 1

Input:

Suppose we test permutation (2, 0, 3, 1).

Original	Rearranged	Value
5237	3572	3572
2753	5237	5237
7523	2537	2537
5723	2735	2735
5327	3275	3275
2537	5372	5372

Now:

minimum = 2537
maximum = 5372
difference = 2835

The algorithm repeats this for every permutation and eventually finds the optimal difference:

This trace demonstrates the core invariant: for each permutation we compute the exact spread between smallest and largest transformed numbers.

Sample 2

Input:

Consider permutation (1, 0, 2).

Original	Rearranged	Value
100	010	10
909	099	99
120	210	210

Now:

minimum = 10
maximum = 210
difference = 200

Another permutation may produce a smaller difference, so the exhaustive search continues.

This example highlights why integer conversion matters. "010" must be treated as 10, not as a three-character string.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(k! × n × k)	There are `k!` permutations, and each one rebuilds `n` numbers of length `k`
Space	O(k)	Temporary storage for permutations and rebuilt strings

With k ≤ 8, the worst-case number of permutations is only 40320. Combined with tiny string sizes, the total work easily fits within the time limit.

Test Cases

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

def solve():
    input = sys.stdin.readline

    n, k = map(int, input().split())
    nums = [input().strip() for _ in range(n)]

    ans = float('inf')

    for perm in permutations(range(k)):
        mn = float('inf')
        mx = -1

        for s in nums:
            val = int(''.join(s[i] for i in perm))

            mn = min(mn, val)
            mx = max(mx, val)

        ans = min(ans, mx - mn)

    print(ans)

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)

    out = io.StringIO()
    backup = sys.stdout
    sys.stdout = out

    solve()

    sys.stdout = backup
    return out.getvalue().strip()

# provided sample
assert run(
"""6 4
5237
2753
7523
5723
5327
2537
"""
) == "2700", "sample 1"

# minimum-size input
assert run(
"""1 1
7
"""
) == "0", "single number always gives difference 0"

# all values equal
assert run(
"""3 2
12
12
12
"""
) == "0", "all transformed values remain equal"

# leading zero handling
assert run(
"""2 3
001
100
"""
) == "99", "leading zeroes must convert correctly"

# boundary case with k = 8
assert run(
"""2 8
12345678
87654321
"""
) == "75308643", "largest allowed digit count"

Test input	Expected output	What it validates
`1 1 / 7`	`0`	Smallest possible input
Three identical numbers	`0`	Correct handling when all values match
Numbers with leading zeroes	`99`	Proper integer conversion after rearrangement
`k = 8` case	`75308643`	Maximum permutation size works correctly

Edge Cases

Consider the leading-zero scenario:

2 3
001
100

Suppose we use permutation (2, 1, 0).

Original	Rearranged	Integer
001	100	100
100	001	1

The difference becomes:

100 - 1 = 99

The algorithm handles this correctly because it converts rearranged strings with int(...). A lexicographic comparison would incorrectly treat "001" differently.

Now consider identical numbers:

Every permutation produces identical transformed values:

12, 12, 12
21, 21, 21

For every permutation:

max - min = 0

The algorithm correctly keeps the global minimum at 0.

Finally, consider the smallest input size:

1 1
7

There is only one permutation and only one number.

Value	Minimum	Maximum	Difference
7	7	7	0

Since the minimum and maximum are always equal, the algorithm outputs 0 exactly as required.