Shifted Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are given a rectangular matrix of integers with dimensions m rows and n columns. Your task is to compute the total sum of all elements in the matrix. The input begins with two integers m and n, followed by m lines each containing n integers that represent the matrix rows. Output a single integer – the sum of all matrix entries. The solution should be implemented using a two‑pointer approach that iterates over the matrix in a single pass, ensuring O(m·n) time and O(1) additional space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Matrix Traversal"
WHY DOES IT MATTER?
The two‑pointer (or dual‑index) pattern enables linear traversal of multi‑dimensional data without extra memory, a skill that directly translates to cache‑optimal code and is frequently tested in interviews to assess a candidate's ability to write space‑efficient solutions.
OPTIMIZATION CHALLENGE
The key insight is recognizing that every element must be visited exactly once, so any additional loops or data structures are unnecessary. By advancing row and column pointers in lockstep, we eliminate nested overhead and achieve O(1) auxiliary space.
REAL-WORLD CONNECTION
Think of a warehouse robot that moves along rows and columns simultaneously, picking items as it passes. It never revisits a location, mirroring how the two pointers sweep the matrix once, which is analogous to streaming data pipelines that process logs row by row.
During the interview, write the outer loop for rows and the inner loop for columns, but explicitly name the indices (e.g., r and c) as pointers. This signals to the interviewer that you are consciously applying the two‑pointer mindset and helps avoid off‑by‑one errors.
COMPLEXITY AT A GLANCE
O(m*n)O(1)Core Theory — Why This Approach?
The problem reduces to aggregating a scalar value across a two‑dimensional grid. A naive solution might iterate over each row and, inside that loop, over each column, accumulating the sum – this is already optimal in terms of asymptotic time because every element must be visited at least once. However, many candidates mistakenly introduce extra data structures (e.g., auxiliary arrays) or perform redundant passes, inflating both time and space. The two‑pointer paradigm treats the row index and column index as independent pointers that advance synchronously through the matrix, allowing a single pass without auxiliary storage. By maintaining two pointers – one for the current row and one for the current column – we can compute the sum in O(m·n) time while keeping the auxiliary space constant, which is the optimal paradigm for this class of aggregation problems.
When the matrix size grows to millions of elements, the constant‑factor overhead of extra copies or nested function calls becomes noticeable. The two‑pointer approach eliminates function‑call recursion and leverages cache‑friendly linear traversal, which is crucial for large inputs. Moreover, this pattern generalizes to many other grid‑based aggregations (e.g., counting, finding min/max) where a single linear scan suffices, reinforcing its importance in interview settings.
Interview Questions on This Problem
Q1How would you compute the sum of a massive m×n matrix stored in a streaming fashion where you cannot hold the entire matrix in memory?
Maintain a running total variable and read each row (or chunk) sequentially from the stream, adding each element to the total. Since each element is processed exactly once and no extra storage is needed, the solution runs in O(totalElements) time and O(1) extra space.
Q2Explain how you could adapt the two‑pointer traversal to compute the sum of only the border elements of the matrix.
Use two pointers for rows (top and bottom) and two for columns (left and right). Iterate over the top row and bottom row fully, then iterate over the leftmost and rightmost columns excluding the corners already counted. This visits each border element once, achieving O(m+n) time and O(1) space.
Q3In a distributed system where each node holds a sub‑matrix, how would you aggregate the global sum efficiently?
Each node computes the local sum using the two‑pointer scan, then a reduction (e.g., MPI_Reduce or a map‑reduce shuffle) aggregates these local sums into the global total. This approach keeps per‑node work linear in its sub‑matrix size and uses only O(1) extra space per node.
Examples
Input
2 3 1 2 3 4 5 6
Output
undefined
Explanation: Initialize two pointers: one at the first element (row 0, col 0) and one at the last element (row 1, col 2). Add the values 1 and 6 to the running total (7). Move the first pointer to the next element (row 0, col 1) and the second pointer to the previous element (row 1, col 1). Add 2 and 5 (total 14). Continue: add 3 and 4 (total 21). All elements have been processed, so the final sum is 21.
Input
3 2 -1 4 -2 -3 5 0
Output
undefined
Explanation: Start with pointers at (-1) and (0). Sum = -1 + 0 = -1. Move inward: add 4 and 5 → sum = 8. Next: add -2 and -3 → sum = 3. All elements accounted for; final sum is 3.
Input
1 1 -100
Output
undefined
Explanation: Only one element exists. The two pointers both point to -100. Sum = -100. No further elements to process.
Input
4 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Output
undefined
Explanation: Traverse the matrix from both ends: (1+16)=17, (2+15)=17 → total 34. Next pair (3+14)=17, (4+13)=17 → total 68. Continue with (5+12)=17, (6+11)=17 → total 102. Finally (7+10)=17, (8+9)=17 → total 136. All 16 elements summed to 136.
Constraints
- 1 <= m, n <= 1000
- -10^9 <= matrix[i][j] <= 10^9
- The resulting sum fits within a 64‑bit signed integer
Optimal Approach & Strategy
Use two pointers (row and column indices) to traverse the matrix in a single linear scan, accumulating the sum while keeping only a constant‑size accumulator variable.
Brute Force Approach
Iterate over each row, and inside that loop iterate over each column, adding each element to a sum variable – essentially a double nested loop without any optimization.
Verified Code Solutions
function main() {
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
const input = lines.join('\n').split(/\s+/).filter(Boolean);
let idx = 0;
const m = parseInt(input[idx++]);
const n = parseInt(input[idx++]);
let totalSum = 0;
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
totalSum += parseInt(input[idx++]);
}
}
console.log(totalSum);
});
}
main();#include <iostream>
#include <vector>
using namespace std;
int main() {
int m, n;
cin >> m >> n;
long long totalSum = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int val;
cin >> val;
totalSum += val;
}
}
cout << totalSum << endl;
return 0;
}import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
long totalSum = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
totalSum += sc.nextInt();
}
}
System.out.println(totalSum);
}
}import sys
def main():
data = sys.stdin.read().split()
idx = 0
m = int(data[idx]); idx += 1
n = int(data[idx]); idx += 1
total_sum = 0
for i in range(m):
for j in range(n):
total_sum += int(data[idx])
idx += 1
print(total_sum)
if __name__ == "__main__":
main()function main() {
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
const input = lines.join('\n').split(/\s+/).filter(Boolean);
let idx = 0;
const m = parseInt(input[idx++]);
const n = parseInt(input[idx++]);
let totalSum = 0;
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
totalSum += parseInt(input[idx++]);
}
}
console.log(totalSum);
});
}
main();Asked in Top Tech Interviews
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