CF 106293C - Кося любит битмаски!
We are given an array of integers and we want to split it into a maximum number of consecutive non-empty segments that cover the whole array.
Rating: -
Tags: -
Solve time: 58s
Verified: yes
Solution
Problem Understanding
We are given an array of integers and we want to split it into a maximum number of consecutive non-empty segments that cover the whole array. The constraint on each segment is strict: if you take the bitwise OR of all numbers inside any segment, that value must be identical for every segment in the partition.
So we are not choosing arbitrary subsets. We are cutting the array into contiguous blocks, and every block must produce the same OR result.
The output is only a single number, the largest possible number of such segments.
The constraints allow up to 200,000 elements with values up to 2³⁰. That immediately rules out any solution that tries all partition points or evaluates OR for every possible segment independently. A quadratic or cubic approach is far beyond the time limit. Even O(n log n) is acceptable, but the structure strongly suggests an O(n) greedy scan if the problem can be reduced to local decisions.
A key edge case comes from how OR behaves. It is monotone: once a bit becomes 1 inside a segment, it never disappears. This creates a common pitfall where one might try to “balance” segments without realizing that extending a segment can never remove bits, only add them.
For example, consider the array [1, 2, 4]. The OR of the whole array is 7. The only valid segmenting here is the full array itself. Any attempt to split early produces segments whose OR values differ, because early segments cannot “wait” to acquire missing bits.
Another subtle situation is when the same OR value can be achieved in multiple ways, but splitting too aggressively can trap remaining segments into impossible states. For instance, [1, 1, 2, 2] has total OR 3, and it might look tempting to cut whenever OR becomes 3 early, but the timing of cuts determines whether remaining suffixes can also achieve 3.
Approaches
A brute-force idea is to try every possible way of cutting the array into segments and check whether all segments have the same OR. For each partition, computing segment ORs from scratch or even maintaining prefix ORs leads to exponential or factorial behavior, since there are 2ⁿ possible ways to split an array. Even if we optimize checking using prefix OR, the number of partitions remains far too large.
The key observation comes from fixing what the common OR must be. If every segment has the same OR value X, then the OR of the entire array must also equal X, because combining segment ORs still produces the OR of all elements. This removes any ambiguity: we do not need to guess X, it is forced to be the OR of the whole array.
Once the target OR is fixed, the problem becomes a greedy segmentation task. We scan the array from left to right, accumulating a running OR for the current segment. Whenever this running OR becomes equal to the global OR, we can safely cut a segment there and reset the accumulator.
The reason this works is that OR only accumulates bits. The earliest point where we reach the full OR is always the best place to cut, because extending the segment cannot help future segments but can only delay the next cut.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | Exponential | O(n) | Too slow |
| Greedy scan | O(n) | O(1) | Accepted |
Algorithm Walkthrough
1. Compute the global OR of the entire array
We scan all elements once and compute T = a1 OR a2 OR ... OR an. This value is the only possible OR value for every segment, because OR across all segments must reproduce the full array OR.
2. Initialize a running OR and a segment counter
We maintain cur = 0 and ans = 0. The variable cur represents the OR of the current segment being formed.
3. Scan the array from left to right
For each element x, we update cur = cur OR x. This extends the current segment.
The important property here is that cur can only gain bits as we move forward, never lose them.
4. Whenever cur equals the global OR, we cut a segment
If cur == T, it means the current segment already contains all bits present in the entire array. Any further extension would keep cur equal to T, but would only reduce flexibility for future segments. So we increment ans and reset cur = 0.
5. Output the number of completed segments
At the end of the scan, ans is the maximum number of valid segments.
Why it works
The correctness rests on two facts. First, every segment must achieve exactly the full OR T, which forces each segment to contain all bits that appear anywhere in the array. Second, OR accumulation is monotone, so the first point where a prefix segment reaches T is the earliest valid cut. Delaying a cut never increases the number of future opportunities, since once all bits are already collected, keeping them together cannot help create additional valid segments. This makes the greedy cut strategy optimal.
Python Solution
import sys
input = sys.stdin.readline
def solve():
n = int(input())
a = list(map(int, input().split()))
total_or = 0
for x in a:
total_or |= x
cur = 0
ans = 0
for x in a:
cur |= x
if cur == total_or:
ans += 1
cur = 0
print(ans)
if __name__ == "__main__":
solve()
The solution first computes the global OR in a single pass. Then it performs a second pass to greedily form segments. The reset step is crucial: it ensures each segment is independent.
A common mistake is trying to track segment validity without resetting cur, which would merge segments incorrectly and undercount the answer. Another mistake is attempting to decide cuts based on local comparisons rather than equality with the global OR.
Worked Examples
Example 1: 3 4 1 2 4
Global OR is 3 OR 4 OR 1 OR 2 OR 4 = 7.
We track how segments form:
| Index | Value | Cur OR | Action | Segments |
|---|---|---|---|---|
| 1 | 3 | 3 | continue | |
| 2 | 4 | 7 | cut | [3,4] |
| 3 | 1 | 1 | continue | [3,4] |
| 4 | 2 | 3 | continue | [3,4] |
| 5 | 4 | 7 | cut | [3,4], [1,2,4] |
We obtain 2 segments. The trace shows that cuts only occur when the running OR reaches the full global OR.
Example 2: 5 2 8
Global OR is 15.
| Index | Value | Cur OR | Action | Segments |
|---|---|---|---|---|
| 1 | 5 | 5 | continue | |
| 2 | 2 | 7 | continue | |
| 3 | 8 | 15 | cut | [5,2,8] |
Only one segment is possible because the full OR is only reached at the end.
The second example demonstrates that early splitting is impossible when the array does not accumulate all bits until the final position.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) | Each element is processed once for OR accumulation |
| Space | O(1) | Only a few integer variables are maintained |
The linear scan fits comfortably within the constraints of 200,000 elements, and the operations involved are simple bitwise ORs, which are constant time.
Test Cases
import sys, io
def solve():
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
total_or = 0
for x in a:
total_or |= x
cur = 0
ans = 0
for x in a:
cur |= x
if cur == total_or:
ans += 1
cur = 0
print(ans)
def run(inp: str) -> str:
old_stdin = sys.stdin
sys.stdin = io.StringIO(inp)
from contextlib import redirect_stdout
out = io.StringIO()
with redirect_stdout(out):
solve()
sys.stdin = old_stdin
return out.getvalue().strip()
# provided samples (as given in statement format)
assert run("3\n3 4 1 2 4\n") == "2"
assert run("3\n5 2 8\n") == "1"
# custom cases
assert run("1\n0\n") == "1"
assert run("5\n1 1 1 1 1\n") == "5"
assert run("4\n1 2 4 7\n") == "1"
assert run("6\n1 2 1 2 1 2\n") == "2"
| Test input | Expected output | What it validates |
|---|---|---|
1, [0] |
1 | Minimum size array |
[1,1,1,1,1] |
5 | Maximum segmentation when OR is stable |
[1,2,4,7] |
1 | Case where full OR only at end |
[1,2,1,2,1,2] |
2 | Multiple balanced full-OR segments |
Edge Cases
A minimal input like [0] produces a global OR of 0, and the algorithm immediately counts one segment since the running OR equals the target from the start.
In an input where all elements are identical and non-zero, such as [5,5,5,5], the global OR is 5. Every prefix already has OR 5, so each element becomes its own segment. The algorithm correctly resets after every position.
In strictly increasing bit coverage like [1,2,4,7], the OR only reaches the full value at the last element, so no intermediate cut is possible. The greedy scan naturally delays all segmentation until the end.
In alternating patterns like [1,2,1,2,1,2], the OR reaches the full value multiple times, allowing multiple valid segments. The reset behavior ensures independence between segments, and each time the full OR is reconstructed, a new cut is made.