Summary

This problem asks us to choose exactly one tile from each row of an R x C grid. Consecutive choices must touch by an edge or a corner, so from column j in one row, the next row can use column j - 1, j, or j + 1.

Each tile has a cost generated by a repeating row-major sequence:

1, 2, 3, ..., M, 1, 2, 3, ..., M, ...

The goal is to find the minimum possible total cost from the first row to the last row.

The key idea is dynamic programming. We keep the best cost to reach each column in the current row. Since each row only depends on the previous row, we can use rolling arrays and reduce memory to O(C).

Understanding the Problem

We have an R x C grid. We must start in row 0 and end in row R - 1, choosing exactly one tile in every row.

If we choose column j in the previous row, then in the next row we may choose:

• j - 1
• j
• j + 1

as long as that column exists.

The tile costs are not given directly. They are generated in row-major order:

1, 2, 3, ..., M, 1, 2, 3, ..., M, ...

For a 0-based cell (r, c), its row-major index is:

index = r * C + c

So the cost is:

cost(r, c) = (index % M) + 1

We need to output the minimum total cost of any legal path.

Core DSA Concepts Involved

• Dynamic Programming

Use a state to represent the best cost to reach a position.

• State Transition

Each current cell can only be reached from at most three neighboring cells in the previous row.

• Rolling Array

The current row depends only on the previous row, so we do not need to store the full R x C DP table.

• Grid Path Optimization

This is a classic grid DP pattern with local transitions between rows.

Dynamic Programming Idea

Define:

dp[r][j] = minimum total cost to reach row r, column j

The base row is:

dp[0][j] = cost(0, j)

For later rows:

dp[r][j] = cost(r, j) + min(
    dp[r - 1][j - 1],
    dp[r - 1][j],
    dp[r - 1][j + 1]
)

Only valid columns are included.

The final answer is:

answer = min(dp[R - 1][j])

Why Rolling Arrays Work

A full 2D DP table would use:

O(R * C)

space.

But when computing row r, we only need row r - 1.

So we keep two arrays:

• prev[j]: best cost to reach column j in the previous row
• cur[j]: best cost to reach column j in the current row

After finishing one row, we swap them:

prev, cur = cur, prev

The memory becomes:

O(C)

Computing Tile Costs

A common source of mistakes is the cost pattern.

The costs continue across the whole grid in row-major order. They do not restart at 1 for each row.

For example, if C = 4 and M = 5, the grid costs begin:

row 0: 1 2 3 4
row 1: 5 1 2 3
row 2: 4 5 1 2
...

So the first cost in each new row advances by C positions in the repeating sequence.

The code uses:

row_start = ((row_start - 1 + C) % M) + 1

to compute the first tile cost of the next row.

Then each column increases the cost by 1, wrapping back to 1 after M.

Python Solution

# CCC 2025 J5 - Connecting Territories
# Dynamic Programming with rolling arrays
# Time:  O(R * C)
# Space: O(C)

R = int(input())
C = int(input())
M = int(input())

# prev[j] = best cost to reach column j in the previous row
prev = [0] * C
cur = [0] * C

# Row 0 costs: 1, 2, 3, ..., M, 1, 2, ...
v = 1
for j in range(C):
    prev[j] = v
    v += 1
    if v == M + 1:
        v = 1

# row_start is the first tile value in the current row.
# Row 0 starts at 1. Each next row advances by C positions.
row_start = 1

for r in range(1, R):
    row_start = ((row_start - 1 + C) % M) + 1
    v = row_start

    if C == 1:
        # Only one column: must go straight down
        cur[0] = prev[0] + v
    else:
        # First column: can only come from prev[0] or prev[1]
        best = min(prev[0], prev[1])
        cur[0] = best + v

        v += 1
        if v == M + 1:
            v = 1

        # Middle columns
        for j in range(1, C - 1):
            best = min(prev[j - 1], prev[j], prev[j + 1])
            cur[j] = best + v

            v += 1
            if v == M + 1:
                v = 1

        # Last column: can only come from prev[C - 2] or prev[C - 1]
        best = min(prev[C - 2], prev[C - 1])
        cur[C - 1] = best + v

    prev, cur = cur, prev

print(min(prev))

C++ Solution

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// CCC 2025 J5 - Connecting Territories
// Dynamic Programming with rolling arrays
// Time:  O(R * C)
// Space: O(C)

int main() {
    int R, C, M;
    cin >> R >> C >> M;

    vector<long long> prev(C, 0), cur(C, 0);

    // Row 0 costs: 1, 2, 3, ..., M, 1, 2, ...
    int v = 1;
    for (int j = 0; j < C; j++) {
        prev[j] = v;
        v++;
        if (v == M + 1) v = 1;
    }

    // row_start is the first tile value in the current row.
    // Row 0 starts at 1. Each next row advances by C positions.
    int row_start = 1;

    for (int r = 1; r < R; r++) {
        row_start = ((row_start - 1 + C) % M) + 1;
        v = row_start;

        if (C == 1) {
            // Only one column: must go straight down
            cur[0] = prev[0] + v;
        } else {
            // First column: can only come from prev[0] or prev[1]
            long long best = min(prev[0], prev[1]);
            cur[0] = best + v;

            v++;
            if (v == M + 1) v = 1;

            // Middle columns
            for (int j = 1; j < C - 1; j++) {
                best = min(prev[j - 1], min(prev[j], prev[j + 1]));
                cur[j] = best + v;

                v++;
                if (v == M + 1) v = 1;
            }

            // Last column: can only come from prev[C - 2] or prev[C - 1]
            best = min(prev[C - 2], prev[C - 1]);
            cur[C - 1] = best + v;
        }

        prev.swap(cur);
    }

    cout << *min_element(prev.begin(), prev.end()) << "\n";
    return 0;
}

Complexity Analysis

There are R rows and C columns.

Each row computes C states, so the time complexity is:

O(R * C)

We only store two arrays of length C, so the space complexity is:

O(C)

Common Mistakes

• Restarting the tile costs from 1 on every row

The cost sequence continues across the whole grid in row-major order.

• Forgetting the boundary columns

Column 0 cannot use j - 1; column C - 1 cannot use j + 1.

• Forgetting the C == 1 case

With only one column, the path must go straight down.

• Using a full 2D DP table unnecessarily

A 2D DP table works conceptually, but rolling arrays are more memory efficient.

• Using int for the total cost in C++

The total cost may be large. long long is safer.

Learning Outcomes

After this problem, students should be able to:

• Recognize a grid path dynamic programming problem
• Use dp[row][col] to represent an optimal value at a position
• Transition from local neighboring states
• Optimize DP memory with rolling arrays
• Handle boundary columns and special cases
• Understand row-major order and repeating tile costs

Review Questions

1. Why does dp[r][j] depend only on the previous row?

1. Why do we need a special case when C == 1?

1. Why should tile costs not restart at 1 for each row?

1. Why can rolling arrays reduce memory from O(R * C) to O(C)?

1. If the path could only move straight down and not diagonally, how would the transition formula change?