CF 2220D1 - Unique Values (Easy version)

In the original interactive problem, there is a hidden array of length $2n+1$. Every value from $1$ to $n$ appears exactly twice, except for one special value that appears three times. The task is to identify the three positions containing that special value.

codeforcescompetitive-programmingbinary-searchinteractive

CF 2220D1 - Unique Values (Easy version)

Rating: 1900
Tags: binary search, interactive
Solve time: 1m 50s
Verified: no

Solution

Problem Understanding

In the original interactive problem, there is a hidden array of length $2n+1$. Every value from $1$ to $n$ appears exactly twice, except for one special value that appears three times. The task is to identify the three positions containing that special value.

For the non-interactive version used in hacks, the entire array is given as input. We no longer need to reconstruct anything through queries. We simply receive the array and must output the three indices where the value with frequency three occurs.

The length of the array is $2n+1$, and the sum of all $n$ values over the test file is at most $2 \cdot 10^4$. That means the total number of array elements is only about $4 \cdot 10^4$. Any linear or near-linear solution is easily fast enough.

The main pitfall is remembering that we need positions, not the value itself. For example, in

1 2 1 3 2 1 3

the repeated-three-times value is 1, but the required answer is the positions

1 3 6

A solution that prints the value instead of its indices is incorrect.

Another easy mistake is using zero-based indices internally and forgetting to convert back to one-based indexing when printing. For

1 1 2 2 1

the correct answer is

1 2 5

not

0 1 4

Approaches

The most direct brute-force idea is to check every distinct value and count how many times it appears. Since the values are restricted to the range $1 \ldots n$, we could scan the array repeatedly and find the value whose frequency equals three. After identifying that value, we scan again and collect its positions.

If there are $n$ possible values and the array length is $2n+1$, this approach performs $O(n(2n+1)) = O(n^2)$ work.

The structure of the input gives a much simpler option. Every value appears exactly twice except one value, which appears exactly three times. A single frequency table immediately reveals the special value.

While reading the array, we count occurrences of every number. After the frequency table is built, exactly one value has frequency three. A second pass through the array collects its three positions.

This reduces the running time to linear complexity.

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

Algorithm Walkthrough

  1. Read $n$.
  2. Read the array of length $2n+1$.
  3. Build a frequency table for all values in the array.
  4. Find the unique value whose frequency equals three.
  5. Scan the array again and record every position where that value occurs.
  6. Output the three recorded positions using one-based indexing.

Why it works

The problem guarantees that exactly one value appears three times and every other value appears exactly twice. The frequency table uniquely identifies that value. Once the value is known, every position containing it belongs to the answer. Since it occurs exactly three times, the collected positions are precisely the required three indices.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())

    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))

        freq = {}
        for x in a:
            freq[x] = freq.get(x, 0) + 1

        target = None
        for x, cnt in freq.items():
            if cnt == 3:
                target = x
                break

        ans = []
        for i, x in enumerate(a, start=1):
            if x == target:
                ans.append(i)

        print(*ans)

solve()

The first loop constructs the frequency table. Because the statement guarantees the existence of exactly one value appearing three times, the search for cnt == 3 always succeeds.

After finding the special value, the second pass collects its positions. The call to enumerate(..., start=1) is important because the problem uses one-based indexing. Forgetting start=1 is the most common off-by-one error here.

The algorithm never performs nested scans over the array. Every element is processed a constant number of times.

Worked Examples

Example 1

Input:

n = 2
a = [1, 1, 1, 2, 2]
Step Action Result
1 Count frequencies freq[1] = 3, freq[2] = 2
2 Find frequency 3 target = 1
3 Collect positions [1, 2, 3]

Output:

1 2 3

This example shows the simplest possible configuration. The special value is found immediately from the frequency table.

Example 2

Input:

n = 3
a = [2, 1, 3, 2, 1, 3, 2]
Step Action Result
1 Count frequencies freq[1] = 2, freq[2] = 3, freq[3] = 2
2 Find frequency 3 target = 2
3 Collect positions [1, 4, 7]

Output:

1 4 7

This example demonstrates that the three occurrences can be spread anywhere in the array. The frequency table still identifies them immediately.

Complexity Analysis

Measure Complexity Explanation
Time $O(n)$ Two linear passes over an array of length $2n+1$
Space $O(n)$ Frequency table stores at most $n$ distinct values

The total input size is small, with $\sum n \le 2 \cdot 10^4$. A linear solution runs comfortably within the limits.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

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

    def solve():
        input = sys.stdin.readline
        t = int(input())
        out = []

        for _ in range(t):
            n = int(input())
            a = list(map(int, input().split()))

            freq = {}
            for x in a:
                freq[x] = freq.get(x, 0) + 1

            target = next(x for x, c in freq.items() if c == 3)

            pos = []
            for i, x in enumerate(a, start=1):
                if x == target:
                    pos.append(str(i))

            out.append(" ".join(pos))

        return "\n".join(out)

    return solve()

# minimum size
assert run("1\n2\n1 1 1 2 2\n") == "1 2 3"

# triple value at scattered positions
assert run("1\n3\n2 1 3 2 1 3 2\n") == "1 4 7"

# triple value at end
assert run("1\n3\n1 2 1 2 3 3 3\n") == "5 6 7"

# multiple test cases
assert run(
    "2\n"
    "2\n1 2 1 2 1\n"
    "3\n2 2 3 1 3 1 2\n"
) == "1 3 5\n1 2 7"
Test input Expected output What it validates
1 1 1 2 2 1 2 3 Minimum valid size
2 1 3 2 1 3 2 1 4 7 Non-adjacent occurrences
1 2 1 2 3 3 3 5 6 7 Triple value at the end
Multiple test cases Two answers Proper reset between cases

Edge Cases

Consider the smallest valid instance:

n = 2
a = [1, 1, 1, 2, 2]

The frequency table becomes {1: 3, 2: 2}. The algorithm selects 1 as the special value and outputs positions 1 2 3. Nothing in the logic depends on larger arrays.

Consider a case where the three occurrences are far apart:

n = 3
a = [2, 1, 3, 2, 1, 3, 2]

The positions are 1, 4, and 7. Since the algorithm identifies the value first and only then collects positions, adjacency is irrelevant.

Consider a case where the triple value is the largest value:

n = 4
a = [1, 2, 1, 2, 3, 4, 3, 4, 4]

The frequency table is

1 -> 2
2 -> 2
3 -> 2
4 -> 3

The algorithm correctly selects 4 and outputs 6 8 9. The actual numeric value does not matter, only its frequency.