we can also apply binary search on rows and columns separately by which we
will decide that in which 1/4th part we need to search.So in this case
complexity will be O(log m + log n).
just check and let me know.......
On Wed, Apr 4, 2012 at 5:27 AM, Ashish Goel <ashg...@gmail.com> wrote:
> why not start from middle(m/2, n/2)
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Wed, Apr 4, 2012 at 12:29 AM, Karthikeyan V.B <kartmu...@gmail.com>wrote:
>
>> Hi,
>>
>> Start from right top corner
>> If matrix element > key move to previous column
>> else if matrix element < key move to next row
>>
>> int search(int** a,int key,int m,int n)
>> {
>> if (key < a[0][0] || key > a[m-1][n-1]) return 0;
>> int min=a[0][n-1],i=0,j=n-1;
>> while(i<=m-1 && j>=0)
>> {
>> if(a[i][j] > key)
>> j--;
>> else if(a[i][j] < key)
>> i++;
>> else
>> return 1;
>> }
>> return 0;
>> }
>>
>> Regards,
>> Karthikeyan.V.B
>>
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Sanjiv Yadav
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