Re: [algogeeks] Re: m*n matrix, sorted rows, sorted columns, WAP to find an element efficiently
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 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 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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>
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Re: [algogeeks] Re: m*n matrix, sorted rows, sorted columns, WAP to find an element efficiently
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 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
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
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>
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Re: [algogeeks] Re: m*n matrix, sorted rows, sorted columns, WAP to find an element efficiently
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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[algogeeks] Re: m*n matrix, sorted rows, sorted columns, WAP to find an element efficiently
This search can be done easily in O(n+m) start from top right corner. chek out this link you will understand!! http://www.geeksforgeeks.org/archives/11337 On Sunday, 1 April 2012 21:18:26 UTC+5:30, ashgoel wrote: > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/bK0lXBQtrSEJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
