# Count of cells in a matrix which give a Fibonacci number when the count of adjacent cells is added

Given an M x N matrix mat[][]. The task is to count the number of good cells in the matrix. A cell will be good if the sum of the cell value and the number of the adjacent cells is a Fibonacci number.

Examples:

Input: mat[][] = {
{1, 2},
{3, 4}}
Output: 2
Only the cells mat[0][0] and mat[1][0] are good.
i.e. (1 + 2) = 3 and (3 + 2) = 5 are both Fibonacci numbers.

Input: mat[][] = {
{1, 0, 5, 3},
{2, 17, 5, 6},
{5, 8, 15, 11}};
Output: 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Iterate the entire matrix and for each cell find the count of adjacent cells. There can be 3 types of cells, one with 2 adjacent cells, one with 3 adjacent cells and the rest with 4 adjacent cells. Sum this count with the value at the current cell and check whether the result is a Fibonacci number. If yes then increment the count.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define M 3 ` `#define N 4 ` ` `  `// Function that returns true if ` `// x is a perfect square ` `bool` `isPerfectSquare(``long` `double` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x ` `    ``long` `double` `sr = ``sqrt``(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - ``floor``(sr)) == 0); ` `} ` ` `  `// Function that returns true ` `// if n is a Fibonacci number ` `bool` `isFibonacci(``int` `n) ` `{ ` `    ``return` `isPerfectSquare(5 * n * n + 4) ` `           ``|| isPerfectSquare(5 * n * n - 4); ` `} ` ` `  `// Function to return the count of good cells ` `int` `goodCells(``int` `mat[M][N]) ` `{ ` ` `  `    ``// To store the required count ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < M; i++) { ` `        ``for` `(``int` `j = 0; j < N; j++) { ` ` `  `            ``int` `sum = mat[i][j]; ` ` `  `            ``// Corner cells of the matrix ` `            ``// have only 2 adjacent cells ` `            ``if` `((i == 0 && j == 0) ` `                ``|| (i == M - 1 && j == 0) ` `                ``|| (i == 0 && j == N - 1) ` `                ``|| (i == M - 1 && j == N - 1)) { ` `                ``sum += 2; ` `            ``} ` ` `  `            ``// All the boundary elements ` `            ``// except the corner elements ` `            ``// have only 3 adjacent cells ` `            ``else` `if` `(i == 0 || j == 0 ` `                     ``|| i == M - 1 || j == N - 1) { ` `                ``sum += 3; ` `            ``} ` ` `  `            ``// Rest of the elements have 4 adjacent cells ` `            ``else` `{ ` `                ``sum += 4; ` `            ``} ` ` `  `            ``// If the sum is a Fibonacci number ` `            ``if` `(isFibonacci(sum)) ` `                ``count++; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `mat[M][N] = { { 1, 0, 5, 3 }, ` `                      ``{ 2, 17, 5, 6 }, ` `                      ``{ 5, 8, 15, 11 } }; ` `    ``cout << goodCells(mat); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `int` `M = ``3``; ` `static` `int` `N = ``4``; ` ` `  `// Function that returns true if ` `// x is a perfect square ` `static` `boolean` `isPerfectSquare(``long` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x ` `    ``double` `sr = Math.sqrt(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - Math.floor(sr)) == ``0``); ` `} ` ` `  `// Function that returns true ` `// if n is a Fibonacci number ` `static` `boolean` `isFibonacci(``int` `n) ` `{ ` `    ``return` `isPerfectSquare(``5` `* n * n + ``4``) ` `        ``|| isPerfectSquare(``5` `* n * n - ``4``); ` `} ` ` `  `// Function to return the count of good cells ` `static` `int` `goodCells(``int` `mat[][]) ` `{ ` ` `  `    ``// To store the required count ` `    ``int` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < M; i++)  ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < N; j++)  ` `        ``{ ` ` `  `            ``int` `sum = mat[i][j]; ` ` `  `            ``// Corner cells of the matrix ` `            ``// have only 2 adjacent cells ` `            ``if` `((i == ``0` `&& j == ``0``) ` `                ``|| (i == M - ``1` `&& j == ``0``) ` `                ``|| (i == ``0` `&& j == N - ``1``) ` `                ``|| (i == M - ``1` `&& j == N - ``1``)) ` `            ``{ ` `                ``sum += ``2``; ` `            ``} ` ` `  `            ``// All the boundary elements ` `            ``// except the corner elements ` `            ``// have only 3 adjacent cells ` `            ``else` `if` `(i == ``0` `|| j == ``0` `                    ``|| i == M - ``1` `|| j == N - ``1``)  ` `            ``{ ` `                ``sum += ``3``; ` `            ``} ` ` `  `            ``// Rest of the elements have 4 adjacent cells ` `            ``else`  `            ``{ ` `                ``sum += ``4``; ` `            ``} ` ` `  `            ``// If the sum is a Fibonacci number ` `            ``if` `(isFibonacci(sum)) ` `                ``count++; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `mat[][] = { { ``1``, ``0``, ``5``, ``3` `}, ` `                    ``{ ``2``, ``17``, ``5``, ``6` `}, ` `                    ``{ ``5``, ``8``, ``15``, ``11` `} }; ` `        ``System.out.println( goodCells(mat)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

 `# Python implementation of the approach ` `from` `math ``import` `ceil,sqrt,floor ` `M ``=` `3` `N ``=` `4` ` `  `# Function that returns true if ` `# x is a perfect square ` `def` `isPerfectSquare(x): ` ` `  `    ``# Find floating povalue of ` `    ``# square root of x ` `    ``sr ``=` `(sqrt(x)) ` ` `  `    ``# If square root is an integer ` `    ``return` `((sr ``-` `floor(sr)) ``=``=` `0``) ` ` `  `# Function that returns true ` `# if n is a Fibonacci number ` `def` `isFibonacci(n): ` `    ``return` `isPerfectSquare(``5` `*` `n ``*` `n ``+` `4``) ``or` `isPerfectSquare(``5` `*` `n ``*` `n ``-` `4``) ` ` `  `# Function to return the count of good cells ` `def` `goodCells(mat): ` ` `  `    ``# To store the required count ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(M): ` `        ``for` `j ``in` `range``(N): ` ` `  `            ``sum` `=` `mat[i][j] ` ` `  `            ``# Corner cells of the matrix ` `            ``# have only 2 adjacent cells ` `            ``if` `((i ``=``=` `0` `and` `j ``=``=` `0``) ` `                ``or` `(i ``=``=` `M ``-` `1` `and` `j ``=``=` `0``) ` `                ``or` `(i ``=``=` `0` `and` `j ``=``=` `N ``-` `1``) ` `                ``or` `(i ``=``=` `M ``-` `1` `and` `j ``=``=` `N ``-` `1``)): ` `                ``sum` `+``=` `2` ` `  `            ``# All the boundary elements ` `            ``# except the corner elements ` `            ``# have only 3 adjacent cells ` `            ``elif` `(i ``=``=` `0` `or` `j ``=``=` `0` `or` `i ``=``=` `M ``-` `1` `or` `j ``=``=` `N ``-` `1``): ` `                ``sum` `+``=` `3` ` `  `            ``# Rest of the elements have 4 adjacent cells ` `            ``else``: ` `                ``sum` `+``=` `4` ` `  `            ``# If the sum is a Fibonacci number ` `            ``if` `(isFibonacci(``sum``)): ` `                ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver code ` ` `  `mat ``=` `[ [ ``1``, ``0``, ``5``, ``3` `], ` `    ``[ ``2``, ``17``, ``5``, ``6` `], ` `    ``[ ``5``, ``8``, ``15``, ``11` `] ] ` `print``(goodCells(mat)) ` ` `  `# This code is contributed by mohit kumar 29 `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `int` `M = 3; ` `static` `int` `N = 4; ` ` `  `// Function that returns true if ` `// x is a perfect square ` `static` `bool` `isPerfectSquare(``long` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x ` `    ``double` `sr = Math.Sqrt(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - Math.Floor(sr)) == 0); ` `} ` ` `  `// Function that returns true ` `// if n is a Fibonacci number ` `static` `bool` `isFibonacci(``int` `n) ` `{ ` `    ``return` `isPerfectSquare(5 * n * n + 4) ` `        ``|| isPerfectSquare(5 * n * n - 4); ` `} ` ` `  `// Function to return the count of good cells ` `static` `int` `goodCells(``int` `[,]mat) ` `{ ` ` `  `    ``// To store the required count ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < M; i++)  ` `    ``{ ` `        ``for` `(``int` `j = 0; j < N; j++)  ` `        ``{ ` ` `  `            ``int` `sum = mat[i,j]; ` ` `  `            ``// Corner cells of the matrix ` `            ``// have only 2 adjacent cells ` `            ``if` `((i == 0 && j == 0) ` `                ``|| (i == M - 1 && j == 0) ` `                ``|| (i == 0 && j == N - 1) ` `                ``|| (i == M - 1 && j == N - 1)) ` `            ``{ ` `                ``sum += 2; ` `            ``} ` ` `  `            ``// All the boundary elements ` `            ``// except the corner elements ` `            ``// have only 3 adjacent cells ` `            ``else` `if` `(i == 0 || j == 0 ` `                    ``|| i == M - 1 || j == N - 1)  ` `            ``{ ` `                ``sum += 3; ` `            ``} ` ` `  `            ``// Rest of the elements have 4 adjacent cells ` `            ``else` `            ``{ ` `                ``sum += 4; ` `            ``} ` ` `  `            ``// If the sum is a Fibonacci number ` `            ``if` `(isFibonacci(sum)) ` `                ``count++; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[,]mat = { { 1, 0, 5, 3 }, ` `                ``{ 2, 17, 5, 6 }, ` `                ``{ 5, 8, 15, 11 } }; ` `    ``Console.WriteLine( goodCells(mat)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:
```7
```

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Improved By : vt_m, mohit kumar 29

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