Sequential Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are given a rectangular grid of integers with N rows and M columns. Starting from the cell at the top‑left corner, process every cell exactly once in row‑major order (left to right within a row, then move to the next row). The processing order must be simulated using a first‑in‑first‑out (FIFO) queue: initially enqueue the coordinates of the starting cell, repeatedly dequeue the front cell, add its value to a running total, and enqueue the next cell in the row‑major sequence if any remain. After all cells have been processed, output the final sum.
Input:
- The first line contains two space‑separated integers N and M (the number of rows and columns).
- The next N lines each contain M space‑separated integers representing the grid.
Output:
- A single integer, the sum of all grid values visited in the described order.
The task tests the ability to model a sequential traversal using a queue, mirroring typical task‑scheduling scenarios where jobs are handled in the order they arrive.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sequential Matrix Traversal"
WHY DOES IT MATTER?
Understanding how to control traversal order with a queue is essential for problems where processing sequence impacts correctness, such as level‑order tree traversals, flood‑fill algorithms, and simulation of time‑step systems.
OPTIMIZATION CHALLENGE
The key insight is that by fixing the neighbor insertion order (right before down) the natural FIFO behavior of the queue aligns perfectly with row‑major order, eliminating the need for explicit row/column counters or nested loops.
REAL-WORLD CONNECTION
Think of a production line where items move through stations in a strict order; the queue acts as the conveyor belt, ensuring each item (cell) reaches the next station (neighbor) in the exact sequence required for downstream processing.
During an interview, write the neighbor‑generation code first and test it on a 2×2 matrix on the whiteboard; this quickly validates that the queue order matches the expected sequence before you add visited checks.
COMPLEXITY AT A GLANCE
O(N*M)O(N*M)Core Theory — Why This Approach?
The problem is a classic illustration of breadth‑first traversal on a 2‑dimensional grid where the adjacency order is deliberately chosen to mimic row‑major sequencing. By always enqueuing the right neighbor before the down neighbor, the FIFO nature of a queue guarantees that cells are processed left‑to‑right within a row before any cell of the next row is visited. A naive approach might attempt to simulate this with nested loops that directly index the matrix, which works for small inputs but defeats the purpose of demonstrating queue‑driven state management and can lead to stack‑overflow if recursion is used. The optimal paradigm leverages a single queue, constant‑time neighbor generation, and a visited flag matrix to ensure each cell is enqueued exactly once, yielding linear time proportional to the number of cells.
Interview Questions on This Problem
Q1How would you modify the queue‑based traversal to work on a toroidal (wrap‑around) grid while still preserving row‑major order?
Enqueue the right neighbor using (i, (j+1)%M) and the down neighbor using ((i+1)%N, j). To preserve row‑major order, still push the right neighbor before the down neighbor and maintain a visited matrix that marks cells after their first dequeue, preventing infinite loops caused by wrap‑around.
Q2Why does a simple BFS that enqueues all four directions (up, down, left, right) not produce the required row‑major order?
BFS explores nodes level by level; enqueuing all four directions introduces nodes from the next row before completing the current row, breaking the left‑to‑right sequence. The required order is achieved only when the adjacency list is strictly ordered as right then down, eliminating up and left moves.
Q3Can you achieve O(1) auxiliary space (excluding input storage) for this traversal? If so, how?
Yes, by overwriting the input matrix with a sentinel value (e.g., a special flag) to mark visited cells, we eliminate the need for a separate visited matrix. The queue still holds at most O(min(N, M)) elements at any time, which is considered O(1) extra space relative to the total number of cells.
Examples
Input
2 3 1 2 3 4 5 6
Output
21
Explanation: Enqueue (0,0). Dequeue (0,0) → add 1 (total=1), enqueue (0,1). Dequeue (0,1) → add 2 (total=3), enqueue (0,2). Dequeue (0,2) → add 3 (total=6), enqueue (1,0). Dequeue (1,0) → add 4 (total=10), enqueue (1,1). Dequeue (1,1) → add 5 (total=15), enqueue (1,2). Dequeue (1,2) → add 6 (total=21). Queue becomes empty, output 21.
Input
1 4 -1 0 5 2
Output
6
Explanation: Start with (0,0). Process cells in order: -1 → total=-1, 0 → total=-1, 5 → total=4, 2 → total=6. The queue empties after the fourth cell, yielding 6.
Input
3 2 10 -3 7 8 -2 4
Output
24
Explanation: Queue sequence: (0,0) adds 10 (total=10), (0,1) adds -3 (total=7), (1,0) adds 7 (total=14), (1,1) adds 8 (total=22), (2,0) adds -2 (total=20), (2,1) adds 4 (total=24). All cells processed, final sum is 24.
Constraints
- 1 <= N, M <= 10^3
- N * M <= 10^5
- -10^9 <= grid[i][j] <= 10^9
- The sum fits in a 64‑bit signed integer
Optimal Approach & Strategy
Use a single FIFO queue, enqueue right then down neighbours, and mark cells as visited to ensure each cell is processed exactly once in row‑major order.
Brute Force Approach
Iterate over every cell with two nested loops and process them directly, ignoring any queue simulation.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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