Understanding the Efficiency of Finding the Diagonal Difference in HackerRank

Question

Understanding the Efficiency of Finding the Diagonal Difference in HackerRank

Looking for help to determine the time complexity of this code snippet!

Context: I came across this algorithm problem on HackerRank where the solution provided in the editorial section claims a time complexity of O(n^2), but I personally think it's actually O(n).

I argue that the time complexity is O(n) because we only iterate through the 2D array once with a single loop, avoiding nested loops and reducing the number of iterations.

Question:

Is my assumption correct or am I overlooking something?

Solution:

//arr represents a 2D array with equal rows and columns
function diagonalDifference(arr) {
  let diff = 0;
  const length = arr.length - 1;
  for (let i = 0; i < arr.length; i++) {
    diff += arr[i][i] - arr[i][length - i];
  }
  return Math.abs(diff);
}

Challenge:

Given a square matrix (or 2D Array), find the absolute difference between the sums of its diagonals.

For example, consider the square matrix arr:

1 2 3 4 5 6 9 8 9

The sum of left-to-right diagonal = 1 + 5 + 9 = 15. The sum of right-to-left diagonal = 3 + 5 + 9 = 17. The absolute difference between them is |15 - 17| = 2.

javascript arrays algorithm multidimensional-array

Answer 1

Answer №1

Do I have the correct understanding, or is there something I am overlooking?

I agree with you, but it seems like their classification of O(n^2) might be based on interpreting n as the length of the matrix's side. In this scenario, the total number of elements in the matrix equals exactly n^2, leading to any solution (since all n^2 inputs must be processed) being considered as Ω(n^2) (where Ω essentially denotes "at least").

Your classification of the solution as O(n) is accurate if we define n as the size of the entire input matrix.

Answer 2

Do I have the correct understanding, or is there something I am overlooking?

I agree with you, but it seems like their classification of O(n^2) might be based on interpreting n as the length of the matrix's side. In this scenario, the total number of elements in the matrix equals exactly n^2, leading to any solution (since all n^2 inputs must be processed) being considered as Ω(n^2) (where Ω essentially denotes "at least").

Your classification of the solution as O(n) is accurate if we define n as the size of the entire input matrix.

Answer 3

Answer №2

The efficiency of this code is linearO(n) where n represents the size of the matrix. As correctly mentioned, the loop iterates only once and its length equals the number of rows/columns, which is equal to the size of the matrix. There are no nested loops present. Therefore, the time complexity remains constant at O(n) for all scenarios including best, worst, and average cases.

Answer 4

The efficiency of this code is linearO(n) where n represents the size of the matrix. As correctly mentioned, the loop iterates only once and its length equals the number of rows/columns, which is equal to the size of the matrix. There are no nested loops present. Therefore, the time complexity remains constant at O(n) for all scenarios including best, worst, and average cases.

Answer 5

Answer №3

function findDiagonalDifference(matrix) { const dimensions = matrix.length;

let primaryDiagSum = 0;
let secondaryDiagSum = 0;

for (let j = 0; j < dimensions; j++) {
    primaryDiagSum += matrix[j][j]; // Main diagonal
    secondaryDiagSum += matrix[j][dimensions - j - 1]; // Other diagonal
}

return Math.abs(primaryDiagSum - secondaryDiagSum);

}

Answer 6

function findDiagonalDifference(matrix) { const dimensions = matrix.length;

let primaryDiagSum = 0;
let secondaryDiagSum = 0;

for (let j = 0; j < dimensions; j++) {
    primaryDiagSum += matrix[j][j]; // Main diagonal
    secondaryDiagSum += matrix[j][dimensions - j - 1]; // Other diagonal
}

return Math.abs(primaryDiagSum - secondaryDiagSum);

}

Answer 7

Answer №4

Here is my proposed solution:

function calculateDiagonalDifference(matrix) {
let primaryDiagSum = 0;
let secondaryDiagSum = 0;
for (let i = 0; i < matrix.length; i++) {
    primaryDiagSum += matrix[i][i];
    secondaryDiagSum += matrix[i][matrix[i].length - (i + 1)];
}
let difference = Math.abs(primaryDiagSum - secondaryDiagSum);
return difference;

}

Answer 8

Here is my proposed solution:

function calculateDiagonalDifference(matrix) {
let primaryDiagSum = 0;
let secondaryDiagSum = 0;
for (let i = 0; i < matrix.length; i++) {
    primaryDiagSum += matrix[i][i];
    secondaryDiagSum += matrix[i][matrix[i].length - (i + 1)];
}
let difference = Math.abs(primaryDiagSum - secondaryDiagSum);
return difference;

}

Answer 9

Answer №5

Give this method a shot:

`

function calculateDiagonalDifference(matrix) {
    let firstDiagonalSum = matrix[0][0] + matrix[1][1] + matrix[2][2];
    let secondDiagonalSum = matrix[0][2] + matrix[1][1] + matrix[2][0];
    let difference = (secondDiagonalSum) - (firstDiagonalSum);
    
    return difference;
}`

Answer 10

Give this method a shot:

`

function calculateDiagonalDifference(matrix) {
    let firstDiagonalSum = matrix[0][0] + matrix[1][1] + matrix[2][2];
    let secondDiagonalSum = matrix[0][2] + matrix[1][1] + matrix[2][0];
    let difference = (secondDiagonalSum) - (firstDiagonalSum);
    
    return difference;
}`

Answer 11

Answer №6

Python Code Example:

def calculateDiagonalDifference(matrix):
    
    diagonal1 = 0
    diagonal2 = 0
    size = len(matrix)

    for i in range(0, size):
        for j in range(0, size):
            
            if (i == j):
                diagonal1 += matrix[i][j]
    
            if (i == size - j - 1):
                diagonal2 += matrix[i][j]
        
    return abs(diagonal1 - diagonal2)

Answer 12

Python Code Example:

def calculateDiagonalDifference(matrix):
    
    diagonal1 = 0
    diagonal2 = 0
    size = len(matrix)

    for i in range(0, size):
        for j in range(0, size):
            
            if (i == j):
                diagonal1 += matrix[i][j]
    
            if (i == size - j - 1):
                diagonal2 += matrix[i][j]
        
    return abs(diagonal1 - diagonal2)

Answer 13

Answer №7

Here's another approach you can take

def calculateDiagonalDifference(matrix):
    left_diag_sum = 0
    right_diag_sum = 0
    
    for row in range(len(matrix)):
        for col in range(len(matrix)):
            if row == col:
                left_diag_sum += matrix[row][col]
            if row == abs(col - (len(matrix) - 1)):
                right_diag_sum += matrix[row][col]
    
    return abs(left_diag_sum - right_diag_sum)

Answer 14

Here's another approach you can take

def calculateDiagonalDifference(matrix):
    left_diag_sum = 0
    right_diag_sum = 0
    
    for row in range(len(matrix)):
        for col in range(len(matrix)):
            if row == col:
                left_diag_sum += matrix[row][col]
            if row == abs(col - (len(matrix) - 1)):
                right_diag_sum += matrix[row][col]
    
    return abs(left_diag_sum - right_diag_sum)

Understanding the Efficiency of Finding the Diagonal Difference in HackerRank

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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