Table Recovery

Grid Reconstruction via Frequency Analysis

Algorithm: Construction / Frequency Analysis

1. Problem Understanding

We are given an $N \times N$ grid.

Key Clue: This grid is generated from a sequence of length $2N-1$. Each row represents a part of the sequence, and there is significant overlap between rows.

Goal:
Reconstruct the relative order of the original sequence by analyzing the frequency of numbers in the grid, and output the lexicographically smallest solution.

2. Key Observation

Think: If we cut a long sequence into $N$ segments of length $N$, how many times does each number appear?

Suppose sequence is: $A, B, C, D, E$ ($N=3$)

  • Row 1: [A, B, C]
  • Row 2: [B, C, D]
  • Row 3: [C, D, E]

Frequency Count:

  • A: Appears 1 time (only in Row 1)
  • B: Appears 2 times (Rows 1, 2)
  • C: Appears 3 times (Rows 1, 2, 3) -> Center Peak
  • D: Appears 2 times
  • E: Appears 1 time
Conclusion: Frequency = Distance from Edge
The frequency depends entirely on the number's position in the original sequence. Closer to the center = higher frequency; closer to the edges = lower frequency.

3. Algorithm Visualization

Freq: 1 Edge Freq: 2 Freq: N=3 Center Freq: 2 Freq: 1 Edge Original Sequence Position

Numbers appearing once are at the ends. The number appearing N times is in the exact center.

4. Example Walkthrough

Input Matrix:
3 4 2
5 2 3
6 3 5
Count Frequencies (cnt):
4: 1 time $\to$ Edge (Pos 1 or 5)
6: 1 time $\to$ Edge (Pos 5 or 1)
2: 2 times $\to$ Adjacent (Pos 2 or 4)
5: 2 times $\to$ Adjacent (Pos 4 or 2)
3: 3 times $\to$ Center (Pos 3)
Construct Sequence:
Possible sequences:
$4 \to 2 \to 3 \to 5 \to 6$
Or reversed: $6 \to 5 \to 3 \to 2 \to 4$

5. Lexicographical Strategy

We derive two possible original sequences (forward or reverse).

The problem requires the reconstructed grid, so we must choose the lexicographically smallest one.

1. Find start (Freq=1) and end (Freq=1) elements.

2. Map all numbers to their positions in the sequence ($1 \dots 2N-1$).

3. Check the first element of the grid. If its corresponding position is large (e.g., $> N+1$), we might be looking at it "backwards", so flip the mapping.

6. Code Implementation: Input


#include <bits/stdc++.h>
using namespace std;

int n;
int a[1001][1001];
int cnt[2001]; // Frequency count
int perm[2001]; // Map: number -> final position value

int main() {
    scanf("%d",&n);
    // Read input and count frequencies
    for (int i=0; i<n; i++) 
        for (int j=0; j<n; j++) {
            scanf("%d",&a[i][j]);
            cnt[a[i][j]]++;
        }

6. Code Implementation: Endpoints


    // Handle special case
    if (n == 1) { printf("2\n"); return 0; }

    // Find the two numbers that appear only once (Start and End)
    int first = 0, last = 0;
    for (int i=2; i<=2*n; i++) {
        if (cnt[i] == 1) {
            if (first == 0) first = i;
            else last = i;
        }
    }
    
    // Find lines containing these numbers to establish reference
    int firstline, lastline;
    for (int i=0; i<n; i++) {
        for (int j=0; j<n; j++) {
            if (a[i][j] == first) firstline = i;
            if (a[i][j] == last) lastline = i;
        }
    }

6. Code Implementation: Mapping

Use firstline and lastline to fill the mapping table.


    // Core Logic: Fill the mapping table
    // firstline contains 1st half, lastline contains 2nd half
    for (int j=0; j<n; j++) {
        // For lines near the start, pos = count + 1
        perm[a[firstline][j]] = cnt[a[firstline][j]] + 1;
        // For lines near the end, pos = TotalLen - count + 1
        perm[a[lastline][j]] = 2*n - cnt[a[lastline][j]] + 1;
    }

    // Lexicographical check: if first element maps to a large value
    // it means the order is reversed
    if (perm[a[0][0]] > n+1 || 
       (perm[a[0][0]] == n+1 && perm[a[0][1]] > n+1)) {
        for (int i = 2; i <= 2*n; i++)
            perm[i] = 2*n + 2 - perm[i]; // Reverse mapping values
    }

6. Code Implementation: Output


    // Output the transformed grid
    for (int i=0; i<n; i++) {
        for (int j=0; j<n; j++) {
            printf("%d%s", perm[a[i][j]], j==n-1?"":" ");
        }
        printf("\n");
    }

    return 0;
}

7. Summary

  • Observation: Utilize the overlap property of sliding windows.
  • Frequency is Coordinate: Occurrence count corresponds directly to the number's position in the sequence.
  • Complexity: $O(N^2)$ for I/O, logic is only $O(N)$.
  • Technique: Construct one valid sequence first, then flip if necessary based on lexicographical order.