Check if sums of i-th row and i-th column are same in matrix in Python

In matrix operations, we often need to compare row and column sums. This problem asks us to check if the sum of the i-th row equals the sum of the i-th column for any row-column pair in a given matrix.

Problem Understanding

Given a matrix, we need to verify if there exists at least one index i where the sum of row i equals the sum of column i.

For the example matrix:

Row 0 sum = 2 + 3 + 4 + 5 = 14
Column 0 sum = 2 + 10 + 1 + 1 = 14

Since they match, the function returns True.

Algorithm Steps

• Get matrix dimensions (rows and columns)
• For each row index i:
  • Calculate sum of row i
  • Calculate sum of column i
  • If sums are equal, return True
• If no matching pair found, return False

Implementation

def solve(mat):
    row = len(mat)
    col = len(mat[0])
    
    for i in range(row):
        total_row = 0
        total_col = 0
        
        for j in range(col):
            total_row += mat[i][j]
            total_col += mat[j][i]
        
        if total_row == total_col:
            return True
    
    return False

# Test with the given matrix
matrix = [
    [2, 3, 4, 5],
    [10, 6, 4, 2],
    [1, 4, 6, 7],
    [1, 5, 6, 7]
]

print(solve(matrix))

True

Alternative Implementation Using Built-in Functions

We can simplify the solution using Python's sum() function ?

def solve_optimized(mat):
    rows = len(mat)
    
    for i in range(rows):
        row_sum = sum(mat[i])
        col_sum = sum(mat[j][i] for j in range(rows))
        
        if row_sum == col_sum:
            return True
    
    return False

# Test with another example
matrix2 = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

print(solve_optimized(matrix2))

False

Complete Example with Multiple Test Cases

def check_row_column_sum(matrix):
    """Check if any row sum equals corresponding column sum"""
    rows = len(matrix)
    
    for i in range(rows):
        row_sum = sum(matrix[i])
        col_sum = sum(matrix[j][i] for j in range(rows))
        
        print(f"Row {i} sum: {row_sum}, Column {i} sum: {col_sum}")
        
        if row_sum == col_sum:
            return True
    
    return False

# Test cases
test_matrix = [
    [2, 3, 4, 5],
    [10, 6, 4, 2],
    [1, 4, 6, 7],
    [1, 5, 6, 7]
]

result = check_row_column_sum(test_matrix)
print(f"Result: {result}")

Row 0 sum: 14, Column 0 sum: 14
Result: True

Conclusion

This algorithm efficiently checks if any row sum matches its corresponding column sum by iterating through each row-column pair. The time complexity is O(n²) where n is the matrix dimension, making it suitable for moderate-sized matrices.