Medium · 2D arrays
Matrix Transpose
You already summed a matrix. Transposing it asks a sharper question: when you write a[i][j], are you sure about which of i and j means what?
At a glancePrint column j of the input as row j of the output — the loops swap places, with the outer loop over columns and the inner over rows.
Problem
Input: The first line holds two integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns. The next n lines each hold m integers.
Output: The transpose: m lines, each with n integers. Row j of the output is column j of the input.
Samples:
| Input | Output |
|---|---|
2 3 | 1 4 |
1 1 | 7 |
3 1 | 1 2 3 |
Notice the shape changes: a 2×3 matrix becomes 3×2. So the two loops that print the answer have swapped bounds — that's the whole trick, and the whole trap.
Your solution
Custom input for "Run my input"
Judging runs in your browser against hidden tests. A runaway loop is cut off after 3 seconds — still, make sure every loop has an exit. Your code is saved in this browser as you type.
Stuck? Hints
Hint 1 — You don't need a second array
The obvious approach is to build b[m][n] with b[j][i] = a[i][j] and then print it. That works, but you can print directly: reading the matrix in row order and printing it in column order is the transpose. One array, zero copies.
Hint 2 — Swap the loop bounds, not just the indices
The outer loop runs j from 0 to m - 1, the inner runs i from 0 to n - 1. Writing a[j][i] while keeping the bounds at n and m reads out of bounds the moment the matrix isn't square — and for a square matrix it silently prints the original.
Hint 3 — Say the indices out loud
a[row][column] — the first index is always the row, in C++ and in every language that uses this notation. Before writing any nested loop over a grid, say which loop variable is the row and which is the column. Most 2D-array bugs are not logic errors; they are a swap you didn't notice.
Solution (try it first!)
#include <iostream>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
int a[55][55];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
cin >> a[i][j];
for (int j = 0; j < m; j++) {
for (int i = 0; i < n; i++) {
cout << a[i][j] << " ";
}
cout << endl;
}
return 0;
}Transposing without a second array works because you never needed the transpose to exist — you only needed to read the data in a different order. That distinction, "do I need to rearrange the data, or just traverse it differently?", saves you a copy in a lot of problems. Rotating a matrix 90°, multiplying two matrices, and walking a grid's diagonals are all variations on this. Next: a problem where the data is a number written in a base you choose — Base Conversion.
Deep dive
Why this works
Reading the matrix in row order and printing it in column order *is* the transpose. You never needed the transpose to exist — only to be traversed differently.
The cost
O(n×m) time and zero extra space. Building a second matrix and printing that costs an extra n×m block for nothing.
Where it goes wrong
- Swapping to
a[j][i]while keeping the loop bounds atnandmreads out of bounds whenever the matrix isn't square — and for a square matrix it silently prints the original. - In
a[row][column]the first index is always the row. Say out loud which loop variable is which before writing any nested loop over a grid.
Next step
Now the data is a number written in a base you choose: Base Conversion.
Two things to remember
- Read by rows, print by columns — that's the transpose.
a[row][column]: the first index is always the row.
🐞 This version fails — open it and try
Prints the matrix as-is — the transpose never happened
#include <iostream>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
int a[55][55];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
cin >> a[i][j];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) cout << a[i][j] << " ";
cout << endl;
}
return 0;
}