CF 1678B1 - Tokitsukaze and Good 01-String (easy version)

Rating: 800
Tags: implementation
Solve time: 1m 30s
Verified: yes

Solution

Problem Understanding

We are given a binary string and allowed to flip any individual character at unit cost. After all changes, the string is interpreted by compressing it into maximal runs of identical characters. A run is just a contiguous block of equal bits, and we care about the lengths of these runs.

A string is considered valid when every run length is even. This forces a very rigid structure: runs cannot have odd size, and since runs alternate between 0 and 1, the entire string must be partitioned into alternating blocks, each block having even length.

The task is to compute the minimum number of flips needed to transform the input string into any string satisfying this constraint.

The constraint that the total length over all test cases is at most 2×10^5 means we need a linear or near-linear solution per test case. Any approach that tries to recompute possibilities for each position independently or explores transformations explicitly will be too slow because even O(n^2) across tests is immediately impossible.

A few edge situations are worth keeping in mind.

A string already consisting of runs of even length is already valid, so the answer is zero. For example, "1100" is valid because it becomes two runs of length 2 and 2.

A single long run is only valid if its length is even. "1111" is valid, but "111" is not, and fixing it requires changing at least one character to break or adjust parity.

Another subtle case is when runs alternate but only one boundary is “wrong”. For example, "1110011000" is invalid even though many runs are even except the first and last, showing that local correctness does not imply global validity.

Approaches

A brute-force approach would try to consider every possible final binary string of length n, compute its run decomposition, and check whether all run lengths are even. For each candidate string we would compute its difference from the input to count flips. The number of binary strings is 2^n, and even evaluating each string takes O(n), making this entirely infeasible.

The key observation is that the condition “all runs have even length” is equivalent to requiring that every pair of adjacent positions inside a run must match, and that run boundaries can only occur at even indices if we think in a structured way. Instead of thinking in terms of arbitrary runs, we can enforce a pairing structure: positions (1,2), (3,4), (5,6), and so on must behave consistently.

More concretely, we can process the string in blocks of two characters. If we want every run to have even length, then inside each valid final configuration, positions i and i+1 must belong to the same run, meaning they must be equal. This reduces the problem to deciding, for each pair, whether it should become "00" or "11". Each pair is independent in cost once we fix this interpretation, because any valid final string is fully determined by the choice of target value for each pair.

Thus the problem becomes: for each pair (s[2k], s[2k+1]), we pay 0 if they already match our chosen target, otherwise we pay 1 per mismatched character relative to the chosen pair value. Since each pair independently chooses whether it becomes all zeros or all ones, we take the cheaper option per pair.

This transforms a global run-parity constraint into a local per-pair optimization.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n · n)	O(n)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

We process the string from left to right in steps of two positions.

Split the string into consecutive pairs (s[i], s[i+1]). This is valid because n is guaranteed even.
For each pair, compute how many flips are needed to make it "00" and how many flips are needed to make it "11".

The first cost is the number of ones in the pair, the second is the number of zeros in the pair. 3. Add the smaller of these two costs to the answer. 4. Repeat for all pairs and output the accumulated total.

Each step is justified by the fact that in any valid final configuration, both characters in a pair must be identical. If they were not, the run would split inside the pair, immediately creating a run of length 1, which is invalid because all runs must be even.

Why it works

The crucial property is that in any valid final string, indices (2k−1, 2k) must belong to the same run. If they were in different runs, then one of the runs would start or end at an odd-length boundary inside the pair, forcing an odd run somewhere in the decomposition.

Once we accept that every valid solution is fully determined by choosing a value per pair, the cost decomposes into independent subproblems. Each pair contributes independently to the total number of flips, and no cross-pair dependency can improve the result because changing one pair does not affect the run validity of another pair.

This independence guarantees that locally optimal choices per pair combine into a globally optimal solution.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input().strip())
    s = input().strip()

    ans = 0

    for i in range(0, n, 2):
        a = s[i]
        b = s[i + 1]

        ones = (a == '1') + (b == '1')
        zeros = 2 - ones

        ans += min(ones, zeros)

    print(ans)

The solution relies on iterating over disjoint pairs, so there is no need for additional data structures or state tracking. The only subtlety is ensuring correct grouping of indices, since a shift by one position would completely invalidate the pairing logic.

Worked Examples

We trace two inputs: the first sample and a small custom case.

Example 1

Input string: "1110011000"

We process pairs:

Pair index	Pair	Cost to 00	Cost to 11	Chosen cost
1	11	0	0	0
2	10	1	1	1
3	01	1	1	1
4	10	1	1	1
5	00	0	2	0

Total = 3

This shows how each pair independently contributes, even though the original run structure spans multiple pairs.

Example 2

Input string: "100110"

Pairs:

Pair index	Pair	Cost to 00	Cost to 11	Chosen cost
1	10	1	1	1
2	01	1	1	1
3	10	1	1	1

Total = 3

This confirms that even alternating patterns reduce to uniform per-pair decisions.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each test case scans the string once in steps of two
Space	O(1)	Only a running counter is maintained

The linear scan matches the constraint that the total input size is at most 2×10^5, ensuring fast execution even in the worst case.

Test Cases

import sys, io

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

    t = int(input())
    out = []
    for _ in range(t):
        n = int(input())
        s = input().strip()

        ans = 0
        for i in range(0, n, 2):
            a = s[i]
            b = s[i+1]
            ones = (a == '1') + (b == '1')
            ans += min(ones, 2 - ones)

        out.append(str(ans))

    return "\n".join(out)

# provided samples
assert run("""5
10
1110011000
8
11001111
2
00
2
11
6
100110
""") == """3
0
0
0
3"""

# all equal
assert run("""1
6
000000
""") == "0"

# alternating
assert run("""1
6
101010
""") == "3"

# minimum size
assert run("""1
2
01
""") == "1"

# already valid alternating pairs
assert run("""1
4
1100
""") == "0"

Test input	Expected output	What it validates
000000	0	already valid configuration
101010	3	worst-case alternating structure
01	1	minimal non-trivial correction
1100	0	valid two-run structure

Edge Cases

For a string already composed of uniform pairs like "1100", the algorithm processes (11) and (00), both contributing zero cost. This matches the fact that the string already decomposes into even-length runs.

For a fully alternating string like "101010", every pair is mixed, so each contributes exactly one flip. The algorithm produces three, which corresponds to converting each pair independently into either all zeros or all ones.

For a single pair like "01", the only valid fix is flipping one bit, and the algorithm correctly chooses min(1,1) = 1, reflecting that either direction costs the same.

These cases confirm that the pairing assumption does not miss interactions across boundaries, since no correction in one pair influences the feasibility of another.