Given an n x n matrix, where every row and column is sorted in increasing order. Given a number x, how to decide whether this x is in the matrix. The designed algorithm should have linear time complexity.

Thanks to devendraiiit for suggesting below approach.

1) Start with top right element
2) Loop: compare this element e with x
….i) if they are equal then return its position
…ii) e < x then move it to down (if out of bound of matrix then break return false)
..iii) e > x then move it to left (if out of bound of matrix then break return false)
3) repeat the i), ii) and iii) till you find element or returned false

Implementation:

#include<stdio.h>

/* Searches the element x in mat[][]. If the element is found, 
    then prints its position and returns true, otherwise prints 
    "not found" and returns false */
int search(int mat[4][4], int n, int x)
{
   int i = 0, j = n-1;  //set indexes for top right element
   while ( i < n && j >= 0 )
   {
      if ( mat[i][j] == x )
      {
         printf("\n Found at %d, %d", i, j);
         return 1;
      }
      if ( mat[i][j] > x )
        j--;
      else //  if mat[i][j] < x
        i++;
   }

   printf("\n Element not found");
   return 0;  // if ( i==n || j== -1 )
}

// driver program to test above function
int main()
{
  int mat[4][4] = { {10, 20, 30, 40},
                    {15, 25, 35, 45},
                    {27, 29, 37, 48},
                    {32, 33, 39, 50},
                  };
  search(mat, 4, 29);
  getchar();
  return 0;
}

Time Complexity: O(n)

The above approach will also work for m x n matrix (not only for n x n). Complexity would be O(m + n).

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

         

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