In this Forming a Magic Square HackerRank solution, We define a magic square to be an matrix of distinct positive integers from to where the sum of any row, column, or diagonal of length is always equal to the same number: the magic constant.
You will be given a matrix of integers in the inclusive range . We can convert any digit to any other digit in the range at cost of . Given , convert it into a magic square at minimal cost. Print this cost on a new line.
Note: The resulting magic square must contain distinct integers in the inclusive range .
Example
$s = [[5, 3, 4], [1, 5, 8], [6, 4, 2]]
The matrix looks like this:
5 3 4
1 5 8
6 4 2
We can convert it to the following magic square:
8 3 4
1 5 9
6 7 2
This took three replacements at a cost of .
Function Description
Complete the formingMagicSquare function in the editor below.
formingMagicSquare has the following parameter(s):
- int s[3][3]: a array of integers
Returns
- int: the minimal total cost of converting the input square to a magic square
Input Format
Each of the lines contains three space-separated integers of row .
Constraints
Sample Input
4 9 2
3 5 7
8 1 5
Sample Output
1
Explanation
Matrix initially looks like this:
4 9 2
3 5 7
8 1 5
Observe that it’s not yet magic, because not all rows, columns, and center diagonals sum to the same number.
If we change the bottom right value, , from to at a cost of , becomes a magic square at the minimum possible cost. Thus, we print the cost, , on a new line.
Forming a Magic Square HackerRank Solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in C++
int formingMagicSquare(vector<vector<int>> s) {
vector<vector<int>> Square{
{8,3,4,1,5,9,6,7,2},
{4,3,8,9,5,1,2,7,6},
{8,1,6,3,5,7,4,9,2},
{6,1,8,7,5,3,2,9,4},
{2,9,4,7,5,3,6,1,8},
{4,9,2,3,5,7,8,1,6},
{2,7,6,9,5,1,4,3,8},
{6,7,2,1,5,9,8,3,4}
};
// int len = Square[0].size();
// cout<<"The size is : "<<len<<endl;
int min = INT_MAX;
// cout<<"The value of INTMAX is : "<<min<<endl;
for (int i = 0; i < Square.size(); i++) {
int total = 0;
for (int j = 0; j < Square[i].size(); j++) {
// cout<<"The value of s["<<j/3<<"]["<<j%3<<"] is "<<s[j/3][j%3]<<endl;
total = total + abs(Square[i][j] - s[j/3][j%3]);
}
// cout<<"The value of total is : "<<total<<endl;
if (total < min) {
min = total;
}
}
return min;
}Problem Solution in Python
def formingMagicSquare(s):
magic_squares = [
[[8, 1, 6], [3, 5, 7], [4, 9, 2]],
[[6, 1, 8], [7, 5, 3], [2, 9, 4]],
[[4, 9, 2], [3, 5, 7], [8, 1, 6]],
[[2, 9, 4], [7, 5, 3], [6, 1, 8]],
[[8, 3, 4], [1, 5, 9], [6, 7, 2]],
[[4, 3, 8], [9, 5, 1], [2, 7, 6]],
[[6, 7, 2], [1, 5, 9], [8, 3, 4]],
[[2, 7, 6], [9, 5, 1], [4, 3, 8]]
]
min_moves = float('inf')
for magic_square in magic_squares:
moves = 0
for i in range(3):
for j in range(3):
moves += abs(s[i][j] - magic_square[i][j])
if moves < min_moves:
min_moves = moves
return min_movesProblem Solution in C#
public static int FormingMagicSquare(List<List<int>> s)
{
List<List<int>> magic = new List<List<int>> {
new List<int> {8, 1, 6, 3, 5, 7, 4, 9, 2},
new List<int> {6, 1, 8, 7, 5, 3, 2, 9, 4},
new List<int> {4, 9, 2, 3, 5, 7, 8, 1, 6},
new List<int> {2, 9, 4, 7, 5, 3, 6, 1, 8},
new List<int> {8, 3, 4, 1, 5, 9, 6, 7, 2},
new List<int> {4, 3, 8, 9, 5, 1, 2, 7, 6},
new List<int> {2, 7, 6, 9, 5, 1, 4, 3, 8},
new List<int> {6, 7, 2, 1, 5, 9, 8, 3, 4}};
List<int> flatList = new List<int>();
foreach (var oneList in s)
foreach (int num in oneList)
flatList.Add(num);
int response = int.MaxValue;
foreach (var term in magic)
{
int temp = getCost(term, flatList);
if (response > temp)
response = temp;
}
return response;
}
private static int getCost(List<int> magic, List<int> data)
{
int sum = 0;
for (int i = 0; i < data.Count; i++)
if (data[i] != magic[i])
sum += Math.Abs(data[i] - magic[i]);
return sum;
}Problem Solution in JavaScript
function formingMagicSquare(s) {
const possibleSquares = [
[[8, 1, 6], [3,5,7], [4,9,2]],
[[6, 1, 8], [7,5,3], [2,9,4]],
[[8, 3, 4], [1,5,9], [6,7,2]],
[[6, 7, 2], [1,5,9], [8,3,4]],
[[4, 3, 8], [9,5,1], [2,7,6]],
[[2, 7, 6], [9,5,1], [4,3,8]],
[[4, 9, 2], [3,5,7], [8,1,6]],
[[2, 9, 4], [7,5,3], [6,1,8]],
];
let minCost = Number.MAX_VALUE;
possibleSquares.forEach(squareMatrix => {
let cost = 0;
for(let i = 0; i < 3; i++){
for(let j = 0; j < 3; j++){
cost += Math.abs(squareMatrix[i][j] - s[i][j])
}
}
if(cost < minCost){
minCost = cost;
}
})
return minCost;
}