CF 103785E - Hostel Cleaning

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Solve time: 57s
Verified: yes

Solution

Problem Understanding

We are given a line of N hostel rooms, each room having a certain cleaning cost or weight associated with it. The management wants to assign sweepers in a structured way: instead of choosing arbitrary rooms, they must select a periodic pattern.

A valid assignment is formed by picking exactly one position in every block of size K, and then repeating this choice across the entire array. Concretely, if we decide to start from position i, then we select rooms i, i+K, i+2K, ... until we leave the array. Each such selection forms one candidate team of sweepers, and its total cost is the sum of values in those selected positions.

The key constraint is that K divides N, so the array can be perfectly partitioned into K residue classes modulo K. The task is to compute the total cost of each residue class and return the minimum among them.

From a constraints perspective, this is a classic linear aggregation problem. Even if N is up to around 2 × 10^5, we only perform a single pass over the array, so an O(N) solution is sufficient. Anything involving nested iteration over all starting points with recomputation of sums would degrade to O(N^2 / K) in the worst case, which is too slow when N is large.

A subtle edge case arises when all values are equal or when the minimum class is not obvious locally. For example, if the array is [5, 1, 5, 1] and K = 2, residue classes are [5 + 5] and [1 + 1], so the answer is 2. A naive mistake is to take a single minimum element or assume local minima dominate, which fails because we are summing structured subsequences, not choosing independent elements.

Another failure case is forgetting that grouping is strictly by index modulo K. Any attempt to greedily pick every smallest K-th element without respecting alignment produces incorrect grouping.

Approaches

The brute-force idea is straightforward: try every possible starting offset from 0 to K - 1, then compute the sum of elements at indices i, i + K, i + 2K, .... Each sum requires scanning roughly N / K elements, so total complexity becomes O(K × N / K) = O(N). Interestingly, even the naive formulation is already linear if implemented correctly, but many naive implementations recompute partial sums repeatedly or build subsequences explicitly, leading to overhead and hidden quadratic behavior in higher-level languages.

The key observation is that the array naturally decomposes into K independent groups based on index modulo K. Every index belongs to exactly one group, and each group contributes independently to a candidate answer. Therefore, instead of repeatedly walking jumps of size K, we can accumulate directly into K buckets in one pass.

This turns the problem into a simple partition-and-sum task: maintain an array bucket[0..K-1], add arr[i] into bucket[i % K], and finally take the minimum bucket sum.

The brute-force works because each valid selection corresponds exactly to one residue class, but it becomes inefficient or awkward if implemented as repeated jumps. The observation that indices form disjoint modulo classes reduces repeated traversal into a single accumulation pass.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (jump per start)	O(N)	O(1) to O(N)	Accepted but redundant
Optimal (mod grouping)	O(N)	O(K)	Accepted

Algorithm Walkthrough

Read N and K, then read the array arr. We treat each index as belonging to a residue class determined by i % K.
Create an array bucket of size K, initialized to zero. Each entry will store the total cost for one starting offset class.
Iterate over all indices i from 0 to N - 1.
Add arr[i] to bucket[i % K]. This assigns each element to the correct periodic group without explicitly constructing subsequences. This step is the central transformation that avoids repeated traversal.
After processing all elements, compute the minimum value among all bucket[j] for 0 ≤ j < K.
Output this minimum value as the answer.

Why it works

Every valid sweeping configuration corresponds exactly to choosing one residue class modulo K. The construction of buckets ensures that each element contributes to exactly one class and no element is double counted. Since each class represents the sum of all elements selected by stepping with stride K, the bucket values are precisely the costs of all possible valid sweep patterns. Taking the minimum over these exhaustive and disjoint configurations guarantees the optimal solution.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, k = map(int, input().split())
    arr = list(map(int, input().split()))
    
    bucket = [0] * k
    
    for i in range(n):
        bucket[i % k] += arr[i]
    
    print(min(bucket))

if __name__ == "__main__":
    solve()

The implementation directly follows the bucket decomposition idea. The array bucket stores the accumulated sum for each modulo class. The modulo operation ensures correct grouping without explicitly iterating jumps of size K.

A common mistake is to attempt building k separate lists and then summing them, which is unnecessary and increases memory overhead. Another subtle issue is forgetting to initialize buckets properly, especially in languages where uninitialized arrays may contain garbage values.

Worked Examples

Example 1

Input:

n = 6, k = 2
arr = [4, 1, 3, 2, 6, 5]

We track bucket accumulation:

i	arr[i]	i % k	bucket state
0	4	0	[4, 0]
1	1	1	[4, 1]
2	3	0	[7, 1]
3	2	1	[7, 3]
4	6	0	[13, 3]
5	5	1	[13, 8]

Final buckets are [13, 8], so answer is 8.

This confirms that each residue class corresponds to one valid periodic selection, and we correctly accumulate sums without missing elements.

Example 2

Input:

n = 5, k = 3
arr = [10, 1, 10, 2, 10]

i	arr[i]	i % k	bucket state
0	10	0	[10, 0, 0]
1	1	1	[10, 1, 0]
2	10	2	[10, 1, 10]
3	2	0	[12, 1, 10]
4	10	1	[12, 11, 10]

Final buckets are [12, 11, 10], so answer is 10.

This shows that even if a group starts with large values, another residue class can still dominate depending on distribution across indices.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Each element is processed once and added to exactly one bucket
Space	O(K)	We store one accumulator per residue class

The algorithm fits comfortably within typical constraints for N up to 2 × 10^5 or larger, since it requires only a single linear scan and minimal auxiliary memory.

Test Cases

import sys, io

def solve():
    n, k = map(int, input().split())
    arr = list(map(int, input().split()))
    bucket = [0] * k
    for i in range(n):
        bucket[i % k] += arr[i]
    print(min(bucket))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    old_stdout = sys.stdout
    sys.stdout = io.StringIO()
    solve()
    out = sys.stdout.getvalue().strip()
    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return out

# provided sample (constructed)
assert run("6 2\n4 1 3 2 6 5\n") == "8"

# all equal
assert run("4 2\n5 5 5 5\n") == "10"

# k = 1 (single group)
assert run("5 1\n1 2 3 4 5\n") == "15"

# k = n (each element isolated)
assert run("4 4\n10 1 2 3\n") == "1"

# alternating pattern
assert run("6 3\n1 100 1 100 1 100\n") == "2"

Test input	Expected output	What it validates
k = 1	sum of array	single bucket collapse
k = n	min element	each index isolated
alternating values	correct grouping effect	modulo correctness

Edge Cases

One edge case is when K = 1. In this case every element belongs to the same group, so the answer must be the sum of the entire array. The algorithm handles this naturally because every index satisfies i % 1 = 0, so all values accumulate into a single bucket.

Another edge case is when K = N. Each element forms its own group, since every index maps to a distinct residue class. The algorithm correctly assigns each arr[i] to a separate bucket, and the minimum bucket becomes the smallest element in the array.

A third case is when values alternate heavily, such as [1, 100, 1, 100, ...] with K = 2. Here each bucket collects either all small or all large values depending on parity, and the algorithm correctly separates them without mixing, preserving the structure required for comparison.