excellent solution sunny
On Tue, Jul 5, 2011 at 11:44 PM, amit kumar <amitthecoo...@gmail.com> wrote:
> #include<stdio.h>
> #define n 11
> int main()
> {
> int ar[n]={1,0,1,1,1,0,0,0,0,0,1};
> int a[n][n];
> int l,i,k,maxlength=0,start,end;
> for(i=0;i<n;i++)
> {
> ar[i]==0?(a[i][i]=-1):(a[i][i]=1);
> }
> for(l=2;l<=n;l++)
> {
> for(i=0;i<=n-l;i++)
> {
> for(k=i;k<i+l-1;k++)
> {
> a[i][i+l-1]=a[i][k]+a[k+1][i+l-1];
> if(a[i][i+l-1]==0 && l>maxlength)
> {
> maxlength=l;
> start=i;
> end=i+l-1;
> }
> }
> }
> }
> printf("maxm length=%d\n",maxlength);
> for(i=start;i<=end;i++)
> printf("%d ",ar[i]);
>
> }
>
> On Tue, Jul 5, 2011 at 11:49 AM, Kunal Patil <kp101...@gmail.com> wrote:
>
>> @Sunny : Excellent !!! Keep posting such nice solutions !! :)
>>
>>
>> On Sat, Jul 2, 2011 at 1:47 AM, Anika Jain <anika.jai...@gmail.com>wrote:
>>
>>> ohh ok i got it.. thanx :)
>>>
>>>
>>> On Sat, Jul 2, 2011 at 12:48 AM, sunny agrawal
>>> <sunny816.i...@gmail.com>wrote:
>>>
>>>> it is (2,4] not [2,4].
>>>> open interval, close interval ......... consider (i,j] as [i+1,j]
>>>>
>>>> because a[i] = sum of values from 0 to i
>>>> and a[j] = sum of values from 0 to j
>>>> if a[i] = a[j] then it means sum does not changes between a[i] to a[j]
>>>>
>>>> so from i+1 to j there are equal no of 1's and -1's
>>>> or in other words equal no of 0's and 1's
>>>>
>>>>
>>>> On Sat, Jul 2, 2011 at 12:42 AM, Anika Jain <anika.jai...@gmail.com>wrote:
>>>>
>>>>> @sunny: in a[2,4] has 2 1s and one 0 then how is it a solution? i mean
>>>>> i didnt get wen a[i]==a[j] then a[i,j] is a solution case..
>>>>>
>>>>> On Fri, Jul 1, 2011 at 4:13 PM, sunny agrawal <sunny816.i...@gmail.com
>>>>> > wrote:
>>>>>
>>>>>> Take an array of size of the length of the string.
>>>>>> fill the array positions with one where string contains 1, and -1
>>>>>> where it is 0
>>>>>>
>>>>>> Now for each i (1,n-1) perform the following operation
>>>>>> a[i] = a[i-1] + a[i]
>>>>>> now a[i] will contains sum of values from a[0] to a[i] in original
>>>>>> array.
>>>>>>
>>>>>> Now only thing remains to find is two indexes in this array such that
>>>>>> a[i] = a[j]
>>>>>> where ever this condition is met ( i,j ] is a solution
>>>>>>
>>>>>> or if for some i a[i] = 0 in this case [0,i] will be a solution
>>>>>>
>>>>>> this can be done using hashing
>>>>>> hash the values in the array of size 2*n+1. as range of values is -n
>>>>>> to n and keep track of max interval.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Fri, Jul 1, 2011 at 3:22 PM, Anantha Krishnan <
>>>>>> ananthakrishnan....@gmail.com> wrote:
>>>>>>
>>>>>>> Given a string containing 0's and 1's. One would need to find the
>>>>>>> longest sub-string such that the count of 0's is equal to the count of
>>>>>>> 1's
>>>>>>> in the largest sub string.
>>>>>>>
>>>>>>> Regards
>>>>>>> Anantha Krishnan
>>>>>>>
>>>>>>> --
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>>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Sunny Aggrawal
>>>>>> B-Tech IV year,CSI
>>>>>> Indian Institute Of Technology,Roorkee
>>>>>>
>>>>>>
>>>>>> --
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>>>>>>
>>>>>
>>>>> --
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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Sunny Aggrawal
>>>> B-Tech IV year,CSI
>>>> Indian Institute Of Technology,Roorkee
>>>>
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>>>>
>>>
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>>
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>
>
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