Search in a row wise and column wise sorted matrix

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Given an n x n matrix and a number x, find position of x in the matrix if it is present in it. Else print “Not Found”. In the given matrix, every row and column is sorted in increasing order. The designed algorithm should have linear time complexity.

Example :

Input : mat[4][4] = { {10, 20, 30, 40},
                      {15, 25, 35, 45},
                      {27, 29, 37, 48},
                      {32, 33, 39, 50}};
              x = 29
Output : Found at (2, 1)

Input : mat[4][4] = { {10, 20, 30, 40},
                      {15, 25, 35, 45},
                      {27, 29, 37, 48},
                      {32, 33, 39, 50}};
              x = 100
Output : Element not found

A simple solution is to search one by one. Time complexity of this solution is O(n2).

A better solution is to use Divide and Conquer to find the element. Time complexity of this solution is O(n1.58).

Below is an efficient solution that works in O(n) time.
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

Thanks to devendraiiit for suggesting below approach.

Implementation:

// C program to search an element in row-wise
// and column-wise sorted matrix
#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);
  return 0;
}

Output :

n Found at 2, 1

Time Complexity: O(n)

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rakesh

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