2352. Equal Row and Column Pairs

Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.

A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).

Understand the problem — what are we counting?

You're given a square grid of numbers. Count how many rows look exactly like some column.

Take this 3×3 example:

(row 0)    3    2    1
(row 1)    1    7    6
(row 2)    2    7    7

Row: (3, 2, 1), (1, 7, 6), (2, 7, 7)
Col: (3, 1, 2), (2, 7, 7), (1, 6, 7)

Bottom row (2, 7, 7) = middle column (2, 7, 7) → 1 pair

Approach — keep a ledger

Comparing every row against every column is slow and messy. Better: two phases.

1. Phase 1 "ledger": scan every row and record "this shape appeared how many times?"
2. Phase 2 "check": for each column, ask the ledger "does this shape exist?" If yes, add that count to the score

Walkthrough — ledger and columns

After phase 1 the ledger is {(3, 2, 1): 1, (1, 7, 6): 1, (2, 7, 7): 1}. Then for j in range(n) build each column:

Loop ends with pairs = 1 — exact match.

Try it yourself first

Ledger and walkthrough table are above — write it yourself, then open the fold below when stuck or ready to compare.

from collections import Counter

def equal_pairs(grid):
    # --- Phase 1: ledger the rows ---
    n = len(grid)
    row_count = Counter(tuple(row) for row in grid)
    # {(3, 2, 1): 1, (1, 7, 6): 1, (2, 7, 7): 1}

    pairs = 0

    # --- Phase 2: build each column and query the ledger ---
    for j in range(n):
        # lock column j, let row i walk top → bottom
        col = tuple(grid[i][j] for i in range(n))
        pairs += row_count[col]

    return pairs

print(equal_pairs([[3, 2, 1], [1, 7, 6], [2, 7, 7]]))  # 1
print(equal_pairs([[3, 1, 2, 2], [1, 4, 4, 5],
                   [2, 4, 2, 2], [2, 4, 2, 2]]))        # 3

1
3

col = tuple(grid[i][j] for i in range(n))  # j = 0 → (3, 1, 2)