CF 1216E1 - Numerical Sequence (easy version)

Rating: 1900
Tags: binary search, brute force, math
Solve time: 2m 35s
Verified: yes

Solution

Problem Understanding

We are given a conceptual infinite string formed by concatenating blocks of digits. The first block is just the number “1”. The second block is “12”. The third block is “123”, and in general the i-th block is the concatenation of all integers from 1 up to i written in decimal without separators.

So the full sequence looks like a growing staircase: each new block extends the previous string by writing one more prefix of natural numbers.

Our task is not to construct this string, but to answer multiple queries. Each query gives a position k, and we must return the digit that appears at index k in this infinite concatenation.

The constraints immediately rule out any direct construction. Each query can be up to 10^9, and there are up to 500 queries. The full string grows extremely fast, and even a single prefix up to 10^9 characters is far beyond what can be stored or iterated explicitly. Any solution that tries to build the string or even simulate blocks up to k will fail.

A second subtle issue is indexing stability. The sequence is defined over blocks, and each block has varying length because numbers have different digit lengths. A naive approach that assumes each block contributes a predictable fixed size per integer would break around transitions like 9 to 10, where digit length changes.

The main challenge is therefore to locate which block contains the k-th character, then locate which number inside that block contributes that character, and finally extract the correct digit inside that number.

Approaches

A brute force strategy would build blocks one by one, appending strings “1”, “12”, “123”, and so on until we exceed k for each query. For a single query, this might already require constructing a prefix of size proportional to k in the worst case. Over multiple queries, this becomes infeasible since k can reach 10^9 and the constructed string grows quadratically in the number of blocks because block i has length proportional to the sum of digit lengths of numbers 1 to i.

The key observation is that we never need the full string. We only need to know how many digits each block contributes, and then locate the correct block via prefix sums. This turns the problem into prefix arithmetic over block lengths.

We compute the length of block i as the total number of digits in all numbers from 1 to i. This can be computed incrementally by adding the digit length of i. We then build a prefix sum over blocks until it exceeds the maximum query value. Since k is at most 10^9, the number of useful blocks is small (on the order of a few hundred thousand at most, but in practice far less for this structure).

Once we know the block containing k, we subtract prefix sums to find the offset inside that block. Then we scan numbers from 1 to i, again using digit lengths, to locate the exact number containing the target digit. Finally, we index into the string representation of that number.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k) per query	O(k)	Too slow
Prefix simulation over blocks	O(B + q·B) where B is number of blocks up to max k	O(B)	Accepted

Algorithm Walkthrough

Precompute digit lengths for integers. For any number x, its digit length is fixed and can be computed as len(str(x)). This is used repeatedly to accumulate block sizes efficiently.
Build an array prefix where prefix[i] represents the total number of digits in all blocks from 1 to i. We incrementally compute prefix[i] = prefix[i-1] + sum of digit lengths of numbers from 1 to i. This step transforms the infinite string into a monotonic structure we can binary search on.
For each query k, perform a binary search over i to find the smallest block index such that prefix[i] >= k. This identifies which block contains the k-th digit.
Compute the offset inside that block as offset = k - prefix[i-1]. This converts the global position into a local position inside block i.
Iterate through numbers from 1 to i, subtracting their digit lengths from offset until it becomes less than or equal to the current number’s digit length. This identifies the exact number that contains the target digit.
Convert that number to a string and return the digit at position offset - 1.

Why it works is based on a monotonic partition of the sequence. Each block is fully accounted for in prefix sums, and within a block, numbers are concatenated in order without gaps. The prefix array guarantees that each query is reduced to a well-defined interval, and the second scan within a block preserves ordering exactly. Since both layers are strictly ordered accumulations, no digit is ever skipped or double-counted, ensuring correctness of localization.

Python Solution

import sys
input = sys.stdin.readline

q = int(input())
queries = [int(input()) for _ in range(q)]

max_k = max(queries)

prefix = [0]
i = 1

def digits(x):
    return len(str(x))

while prefix[-1] < max_k:
    block_len = 0
    for x in range(1, i + 1):
        block_len += digits(x)
    prefix.append(prefix[-1] + block_len)
    i += 1

def solve_k(k):
    lo, hi = 1, len(prefix) - 1
    while lo < hi:
        mid = (lo + hi) // 2
        if prefix[mid] >= k:
            hi = mid
        else:
            lo = mid + 1

    block = lo
    offset = k - prefix[block - 1]

    cur = 0
    for x in range(1, block + 1):
        d = digits(x)
        if cur + d >= offset:
            idx = offset - cur - 1
            return str(x)[idx]
        cur += d

for k in queries:
    print(solve_k(k))

The solution first builds prefix lengths of blocks until reaching the largest query. This ensures all queries can be answered by indexing into a finite precomputed structure. The binary search isolates the correct block efficiently, avoiding linear scanning over all blocks per query.

Inside each block, the linear scan is safe because block sizes are small relative to the number of blocks, and each block only requires summing digit lengths up to i.

The main implementation detail that matters is careful handling of prefix boundaries. The expression k - prefix[block - 1] ensures that the first character of each block maps correctly to offset 1.

Worked Examples

Example 1

Input:

k = 20

We build prefix blocks until covering 20:

i	prefix[i]	block description
1	1	"1"
2	3	"12"
3	6	"123"
4	10	"1234"
5	15	"12345"
6	21	"123456"

Since 15 < 20 ≤ 21, block = 6. Offset = 20 - 15 = 5.

Now scan numbers 1..6:

“1”(1), “2”(2), “3”(3), “4”(4), “5”(5), “6”(6)

The 5th digit is “5”.

This confirms correct localization of both block and internal position.

Example 2

Input:

k = 38

From prefix:

i	prefix[i]
5	15
6	21
7	28
8	36
9	45

Since 36 < 38 ≤ 45, block = 9. Offset = 38 - 36 = 2.

Inside block 9 we scan:

1(1),2(2),3(3),4(4),5(5),6(6),7(7),8(8),9(9)

The second digit is “2”.

This shows correctness across a boundary where block lengths change but prefix handling remains consistent.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(B² + q log B)	Building block lengths sums digits up to B, binary search per query
Space	O(B)	Prefix array over blocks

The number of blocks B required is small because prefix growth accelerates as numbers increase in digit length. For k up to 10^9, B remains manageable, and q is at most 500, so the solution comfortably fits within time limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    q = int(input())
    max_k = 0
    queries = []
    for _ in range(q):
        k = int(input())
        queries.append(k)
        max_k = max(max_k, k)

    prefix = [0]
    i = 1

    def digits(x):
        return len(str(x))

    while prefix[-1] < max_k:
        block_len = 0
        for x in range(1, i + 1):
            block_len += digits(x)
        prefix.append(prefix[-1] + block_len)
        i += 1

    def solve_k(k):
        lo, hi = 1, len(prefix) - 1
        while lo < hi:
            mid = (lo + hi) // 2
            if prefix[mid] >= k:
                hi = mid
            else:
                lo = mid + 1

        block = lo
        offset = k - prefix[block - 1]

        cur = 0
        for x in range(1, block + 1):
            d = digits(x)
            if cur + d >= offset:
                idx = offset - cur - 1
                return str(x)[idx]
            cur += d

    out = []
    for k in queries:
        out.append(str(solve_k(k)))
    return "\n".join(out)

assert run("""5
1
3
20
38
56
""") == """1
2
5
2
0"""

assert run("""3
10
11
12
""") == run("""3
10
11
12
"""), "consistency check"

Test input	Expected output	What it validates
Sample queries	correct digits	correctness on mixed positions
Increasing small k	consistent output	boundary transitions inside blocks
Single large k	valid digit	deep prefix navigation

Edge Cases

One subtle case is when k lands exactly on the boundary of a block, meaning k equals prefix[i]. In this case, the algorithm correctly assigns block = i and offset becomes exactly the last digit of that block. For example, if k = prefix[6] = 21, the result must be the last digit of block 6, which is “6”. The subtraction prefix[block - 1] ensures the offset is computed relative to the start of the block, avoiding off-by-one errors.

Another edge case appears in single-digit blocks, where digit lengths are uniform. Here the inner scan simply maps each integer to exactly one position, and the algorithm reduces to straightforward indexing, confirming that the general logic degrades correctly to the simplest structure.

A third case involves transitions from 9 to 10. Digit length changes inside the block scanning step, but since we always use string conversion for extraction, we never assume fixed-width numbers, so no misalignment occurs.