CF 1197C - Array Splitting

Rating: 1400
Tags: greedy, sortings
Solve time: 1m 27s
Verified: yes

Solution

Problem Understanding

We are given a non-decreasing array and asked to split it into exactly $k$ contiguous parts. Each part contributes a cost equal to the difference between its largest and smallest element, and since the array is sorted, this is simply the difference between its last and first element in that segment.

The task is to choose $k-1$ cut positions so that the total sum of these segment ranges is minimized.

The constraints go up to $n = 3 \cdot 10^5$, which immediately rules out any approach that tries all ways of placing cuts or uses quadratic dynamic programming. Anything beyond $O(n \log n)$ risks timing out in practice, and an $O(n^2)$ DP over segment boundaries is far too slow.

A subtle aspect of the problem is that although we are “splitting into segments”, the cost depends only on segment endpoints, not on internal structure. This means the entire optimization is driven by where we place the cuts.

One failure case for naive intuition is assuming we should cut where adjacent differences are largest globally. For example, taking the largest $k-1$ gaps does not always work if we do not account for how these gaps interact with segment formation, since each cut removes internal contributions to a segment’s range rather than independently adding cost.

Another pitfall is thinking greedy segmentation from left to right using local decisions is optimal; local minimal range segments can block better global partitions.

Approaches

A brute-force solution would try all ways to place $k-1$ cuts among $n-1$ possible gaps. Each configuration defines a partition and we compute its cost in $O(n)$, leading to $O(\binom{n}{k} \cdot n)$, which is completely infeasible even for small $n$.

A more structured DP approach defines $dp[i][j]$ as the minimum cost to split the prefix $1 \dots i$ into $j$ segments. Transitioning requires trying all previous cut points, giving $O(n^2 k)$ time in straightforward form, which still exceeds limits.

The key observation is that since the array is sorted, merging two adjacent segments always changes cost in a predictable way. If we take the entire array as one segment, the cost is $a_n - a_1$. Every time we split between $i$ and $i+1$, we effectively “remove” the contribution of connecting $a_i$ to $a_{i+1}$, because those two elements no longer belong to the same segment.

If the array were one segment, the cost would include the full span. Each cut at position $i$ removes the gap contribution $(a_{i+1} - a_i)$. Thus, instead of thinking about building segments, we can think about starting from the full span and subtracting savings obtained by cuts.

To minimize cost, we want to maximize the total removed contribution, which means selecting the $k-1$ largest adjacent differences.

This transforms the problem into a simple sorting selection problem over $n-1$ values.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	O(n)	Too slow
DP over partitions	O(n²k)	O(nk)	Too slow
Greedy gap selection	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Compute all adjacent differences $d_i = a_{i+1} - a_i$ for $1 \le i < n$.

These represent potential savings if we place a cut between $i$ and $i+1$. 2. Compute the total cost as $a_n - a_1$.

This corresponds to the cost if we do not split at all. 3. Sort the list of differences in descending order.

This allows us to pick the most beneficial cut positions first. 4. Take the sum of the largest $k-1$ differences.

Each selected difference corresponds to a cut that reduces total cost. 5. Subtract this sum from the initial total cost and output the result.

Why it works

The algorithm relies on the invariant that any segmentation cost can be expressed as the full range minus the sum of differences removed by cuts. Since each cut eliminates exactly one adjacency contribution from the global span, and these contributions are independent, choosing cuts does not interfere across positions. Therefore, maximizing removed contributions is equivalent to selecting the largest adjacent gaps, guaranteeing optimality.

Python Solution

import sys
input = sys.stdin.readline

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

    if k == 1:
        print(a[-1] - a[0])
        return

    diffs = []
    for i in range(n - 1):
        diffs.append(a[i + 1] - a[i])

    diffs.sort(reverse=True)

    saved = sum(diffs[:k - 1])
    print((a[-1] - a[0]) - saved)

if __name__ == "__main__":
    solve()

The solution begins by handling the trivial case where no cuts are made. The main computation builds the list of adjacent differences, sorts them to prioritize large gaps, and then selects the best $k-1$ of them. The final subtraction converts the “maximum savings” interpretation back into the original cost formulation.

A common mistake is forgetting that only $k-1$ gaps can be chosen, or incorrectly summing all differences instead of selecting the largest ones.

Worked Examples

Example 1

Input:

6 3
4 8 15 16 23 42

Adjacent differences are:

i	a[i]	a[i+1]	diff
1	4	8	4
2	8	15	7
3	15	16	1
4	16	23	7
5	23	42	19

We sort diffs: $[19, 7, 7, 4, 1]$.

We take $k-1 = 2$ largest: $19 + 7 = 26$.

Initial cost is $42 - 4 = 38$.

Final answer is $38 - 26 = 12$.

This confirms that the optimal cuts correspond exactly to the largest gaps.

Example 2

Input:

5 2
1 2 3 10 11

i	diff
1	1
2	1
3	7
4	1

Sorted diffs: $[7, 1, 1, 1]$.

Pick $k-1 = 1$: take 7.

Initial cost: $11 - 1 = 10$.

Final cost: $10 - 7 = 3$.

We effectively cut between 3 and 10, producing segments $[1,2,3]$ and $[10,11]$, whose costs are $2$ and $1$, totaling $3$.

This shows how a single dominant gap dictates the optimal partition.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n \log n)$	Sorting $n-1$ differences dominates
Space	$O(n)$	Storage for differences array

The algorithm comfortably fits within limits for $n \le 3 \cdot 10^5$, as sorting and a single pass over the array are efficient in Python.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from __main__ import solve
    import contextlib, io as sio
    out = sio.StringIO()
    with contextlib.redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided sample
assert run("6 3\n4 8 15 16 23 42\n") == "12"

# k = 1
assert run("4 1\n1 5 9 10\n") == "9"

# all equal
assert run("5 3\n7 7 7 7 7\n") == "0"

# maximum cuts
assert run("5 5\n1 2 3 4 5\n") == "0"

# strong single gap
assert run("5 2\n1 2 3 100 101\n") == "2"

# alternating gaps
assert run("6 3\n1 2 10 11 20 21\n") == "8"

Test input	Expected output	What it validates
k = 1 case	full range	no splitting logic
all equal	0	zero-cost stability
max cuts	0	extreme segmentation
strong gap	small answer	dominant cut correctness
alternating gaps	balanced selection	greedy choice correctness

Edge Cases

A key edge case is when $k = 1$. The algorithm must avoid selecting any differences and directly return $a_n - a_1$. In this case, the diff array is irrelevant, and any accidental subtraction would incorrectly reduce the result.

Another edge case is when all elements are equal. Every adjacent difference is zero, so sorting and selecting any subset must still produce zero savings. The final result remains zero regardless of $k$, confirming that the algorithm does not overcount cuts.

When $k = n$, every element becomes its own segment. The cost must be zero because each segment has equal min and max. The algorithm handles this correctly because it sums the largest $n-1$ gaps, which equals the full range $a_n - a_1$, resulting in a final answer of zero.