Resilient Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are tasked with navigating a 2D grid of dimensions R x C, where each cell contains an integer weight representing the cost to traverse that position. The objective is to find the minimum total cost path from the top-left corner (0, 0) to the bottom-right corner (R-1, C-1). Movement is restricted to three directions: right, down, and diagonally down-right. A path is considered 'resilient' if it successfully reaches the destination while minimizing the sum of weights along the chosen trajectory. If no valid path exists (which is impossible in this grid structure unless R or C is 0, but assume standard positive dimensions), return -1. However, to add complexity, certain cells may be blocked (represented by -1), which cannot be traversed. You must use recursion with memoization (top-down dynamic programming) to compute the minimum cost, ensuring that you do not revisit cells in a way that creates cycles, although the directional constraints naturally prevent cycles in a grid moving only forward/right/down.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Resilient Matrix Traversal"
WHY DOES IT MATTER?
The pattern is a textbook case of 2‑D dynamic programming with multiple predecessor states, teaching candidates how to identify optimal substructure in grid‑based problems and avoid exponential recursion.
OPTIMIZATION CHALLENGE
The key insight is recognizing that each cell’s answer depends only on three already‑computed neighbors, allowing us to replace exponential recursion with a single pass DP and optionally compress space to one dimension.
REAL-WORLD CONNECTION
Think of a package routing system where a parcel can move east, south, or southeast across a city grid; the algorithm finds the cheapest route considering tolls at each intersection, analogous to cost‑aware load balancing in distributed networks.
When coding, start with a clear recurrence, then immediately add memoization or a DP table; never write the full exponential recursion first—interviewers look for the moment you switch to DP.
COMPLEXITY AT A GLANCE
O(R*C)O(R*C) (or O(C) with row‑wise compression)Core Theory — Why This Approach?
The Resilient Matrix Traversal problem is a classic example of optimal substructure and overlapping subproblems, which makes it a perfect candidate for dynamic programming. Each cell (i, j) can be reached from three possible predecessors: (i‑1, j) (down), (i, j‑1) (right), and (i‑1, j‑1) (diagonal). The minimum cost to reach (i, j) therefore equals the cell’s own weight plus the minimum of the costs to reach those three predecessor cells. A naive recursive solution explores all possible paths, leading to an exponential time complexity O(3^{R+C}) because the same sub‑grid is recomputed many times. By memoizing the result of each sub‑problem (or by filling a DP table bottom‑up), we guarantee each cell is processed exactly once, collapsing the exponential blow‑up to a linear scan of the grid.
The optimal paradigm combines recursion with memoization (top‑down DP) or an iterative bottom‑up DP table. Both approaches respect the problem’s constraints and can be implemented with O(R*C) time. Space can be reduced to O(C) by keeping only the previous row when filling the table iteratively, because the transition only depends on the current row and the row above. This reduction is crucial for large matrices where R and C can each be up to 10^5 in total cells.
Why naive approaches fail on large inputs is evident when the grid size grows: the exponential number of paths quickly exceeds any realistic time limit, and stack overflow may occur due to deep recursion without memoization. The DP formulation eliminates redundant work and provides a deterministic, predictable runtime, which is essential for interview settings and production code.
Interview Questions on This Problem
Q1How would you modify the solution if movement is allowed only right and down (no diagonal), and you need to also reconstruct the actual minimum‑cost path?
Remove the diagonal transition from the recurrence, i.e., dp[i][j] = grid[i][j] + min(dp[i‑1][j], dp[i][j‑1]). To reconstruct the path, maintain a parent pointer for each cell indicating which predecessor gave the minimum; after filling the DP table, backtrack from (R‑1, C‑1) to (0,0) using those pointers.
Q2Suppose the grid is extremely large and cannot fit into memory. Which technique can you use to compute the minimum cost while only storing O(C) space?
Process the grid row by row, keeping two one‑dimensional arrays: prevRow and curRow. For each cell, compute curRow[j] = grid[i][j] + min(prevRow[j], curRow[j‑1], prevRow[j‑1]) (handling boundaries). After finishing a row, swap the arrays. This uses O(C) space and O(R*C) time.
Q3In a distributed system, how could you parallelize the DP computation for this matrix while preserving correctness?
Divide the matrix into diagonal wavefronts (anti‑diagonals). All cells on the same anti‑diagonal depend only on cells from previous anti‑diagonals, so they can be computed in parallel. Synchronize after each wavefront before moving to the next, ensuring data dependencies are respected.
Examples
Input
grid = [[1, 3, 1], [1, 5, 1], [4, 2, 1]]
Output
7
Explanation: Start at (0,0) with cost 1. Move right to (0,1) cost 3, total 4. Move right to (0,2) cost 1, total 5. Move down to (1,2) cost 1, total 6. Move down to (2,2) cost 1, total 7. Alternative: (0,0)->(1,0)->(2,0)->(2,1)->(2,2) is 1+1+4+2+1=9. Another: (0,0)->(0,1)->(1,2)->(2,2) is 1+3+1+1=6? Wait, (0,1) to (1,2) is diagonal. Cost 1+3+1+1=6. Let's re-evaluate. Path: (0,0)=1, (0,1)=3, (1,2)=1, (2,2)=1. Sum=6. Is there a better one? (0,0)=1, (1,0)=1, (1,1)=5, (2,2)=1. Sum=8. (0,0)=1, (1,0)=1, (2,1)=2, (2,2)=1. Sum=5. Path: (0,0)->(1,0) [down], (1,0)->(2,1) [diag], (2,1)->(2,2) [right]. Costs: 1+1+2+1=5. So output is 5.
Input
grid = [[1, -1, 1], [1, 1, 1], [1, 1, 1]]
Output
4
Explanation: Start at (0,0) cost 1. Cell (0,1) is blocked (-1). Must move down to (1,0) cost 1, total 2. From (1,0), can move right to (1,1) cost 1, total 3. From (1,1), move right to (1,2) cost 1, total 4. From (1,2), move down to (2,2) cost 1, total 5. Alternative: From (1,0) move diag to (2,1) cost 1, total 3. From (2,1) move right to (2,2) cost 1, total 4. So min is 4.
Input
grid = [[5, 5, 5], [5, 5, 5], [5, 5, 5]]
Output
20
Explanation: Any path from (0,0) to (2,2) using right, down, or diag moves will have a specific number of steps. The minimum number of steps is 3 (e.g., diag, diag, diag? No, (0,0)->(1,1)->(2,2) is 2 moves, 3 cells). Path: (0,0)=5, (1,1)=5, (2,2)=5. Sum=15. Wait, can we do better? No, all cells are 5. The shortest path in terms of cells visited is 3 cells (diagonal moves). So 5+5+5=15. Let's check if a longer path is forced? No. So output is 15.
Constraints
- 1 <= R, C <= 100
- -1 <= grid[i][j] <= 10^4
- grid[0][0] != -1
- grid[R-1][C-1] != -1
Optimal Approach & Strategy
Use a DP table (or memoized recursion) where dp[i][j] = grid[i][j] + min(dp[i‑1][j], dp[i][j‑1], dp[i‑1][j‑1]), filling it in row‑major order.
Brute Force Approach
Recursively explore every possible right, down, and diagonal move from the start to the end, accumulating costs along each path.
Verified Code Solutions
function resilientMatrixTraversal(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int resilientMatrixTraversal(vector<int> nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int resilientMatrixTraversal(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def resilient_matrix_traversal(nums):
sum = 0
for num in nums:
sum += num
return sumfunction resilientMatrixTraversal(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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