Ummm..
Algorithm:
Start from the right of the array,
make the last element of B to 0,
initialize a variables counter to 0 and max to A[end];
LOOP:
and now move from right to left,
if the next element of the left element > max
increment counter and assign it to that B[ n - element ] index.
max = that element.
if the next element is smaller, assign 0 and repeat LOOP
if the next element's index is 0, check and exit
\/
A : ( 4, 3, 1, 2 )
B : ( 0, 0, 0, 0 )
counter = 0;
max = 2;
\/
A : ( 4, 3, 1, 2 )
B : ( 0, 0, 0, 0 )
counter = 1;
max = 2;
\/
A : ( 4, 3, 1, 2 )
B : ( 0, 0, 0, 0 )
counter = 2;
max = 2;
3 > max i.e. 2
b[n - 3] = counter => b[1] = 2
max = 3;
\/
A : ( 4, 3, 1, 2 )
B : ( 0, 0, 0, 0 )
counter = 3;
max = 3;
4 > max i.e. 3
b[n - 4] = counter => b[0] = 3
Edge Cases: It shall remain all 0's for all same numbers as well as
ascending numbers.
On Tue, Oct 4, 2011 at 7:13 AM, DIVIJ WADHAWAN <divij...@gmail.com> wrote:
> int B[sizeof(A)];
> int k=0 , j ;
> for(j=i;j<n;j++)
> {
> if (A[j] < A[i])
> B[k++]=A[j];
>
> }
>
>
> On Tue, Oct 4, 2011 at 5:30 AM, Abraham <freedomsou...@gmail.com> wrote:
>
>> Hi Vikram!
>> Obviously The naivest solution is O(n2). Could you give a hint for
>> this problem?
>>
>>
>> On Oct 3, 4:39 pm, Vikram Singh <singhvikram...@gmail.com> wrote:
>> > Given an array say A=(4,3,1,2). An array B is formed out of this in
>> > such a way that B[i] = no. of elements in A, occuring on rhs of A[i],
>> > which are less then A[i].
>> > eg.for the A given, B is (3,2,0,0).
>> > Here A of length n only contains elements from 1 to n that too
>> > distinct..
>> > Now the problem is:
>> > 1). You are given with any such A. Find out corresponding B?
>> >
>> > 2). You are given with any such B. Find out corresponding A?
>> >
>> > Please provide solution in O(n),if possible??
>>
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