# Search ingredient in Matrix – GeeksforGeeks

0
7 Given an N×N matrix mat[][] and an integer Ok, the duty is to go looking Ok in mat[][] and if discovered return its indices.

Examples:

Enter: mat[][] = {{1, 2, 3}, {7, 6, 8}, {9, 2, 5}}, Ok = 6
Output: {1, 1}
Clarification: mat = 6

Enter: mat[][] = {{3, 4, 5, 0}, {2, 9, 8, 7}}, Ok = 10
Output: Not Discovered

Method:

Traverse the matrix utilizing nested loops and if the goal is discovered then, return its indices

Comply with the steps to resolve this downside:

• Take the matrix (2-D array) and goal as enter.
• Apply two nested for-loop with i and j variables respectively.
• If mat[i][j] = Ok return the indices pair {i, j}
• If the goal just isn’t discovered print “Not Discovered”.

Beneath is the implementation of this strategy

## C++

 ` `  `#embrace ` `utilizing` `namespace` `std;` ` `  `pair<``int``, ``int``> FindPosition(vector > M,` `                            ``int` `Ok)` `{` `    ``for` `(``int` `i = 0; i < M.measurement(); i++) {` `        ``for` `(``int` `j = 0; j < M[i].measurement(); j++) {` `            ``if` `(M[i][j] == Ok)` `                ``return` `{ i, j };` `        ``}` `    ``}` `    ``return` `{ -1, -1 };` `}` ` `  `int` `primary()` `{` `    ` `    ``vector > M{ { 1, 2, 3 },` `                            ``{ 7, 6, 8 },` `                            ``{ 9, 2, 5 } };` `    ``int` `Ok = 6;` ` `  `    ` `    ``pair<``int``, ``int``> p = FindPosition(M, Ok);` `    ``if` `(p.first == -1)` `        ``cout << ``"Not Discovered"` `<< endl;` `    ``else` `        ``cout << p.first << ``" "` `<< p.second;` `    ``return` `0;` `}`

Time Complexity: O(N2)
Auxiliary House: O(1)