Is there O(n) solution available for it?
On Tue, Sep 28, 2010 at 7:19 AM, Nishant Agarwal <
[email protected]> wrote:
> #include<stdio.h>
> #include<stdlib.h>
> int main()
> {
> int a[20],i,n,max,t,j,k;
> printf("Enter the no. of elements\n");
> scanf("%d",&n);
> for(i=0;i<n;i++)
> scanf("%d",&a[i]);
> for(i=0;i<n-1;i++)
> {
> j=n-1;
> max=0;
> k=i;
> while(i<j)
> {
> if(a[j]<a[i]&&a[j]>=max)
> {
> max=a[j];
> k=j;
> j--;
> }
> else
> j--;
> }
> t=a[i];
> a[i]=a[k];
> a[k]=t;
> }
> for(i=0;i<n;i++)
> printf("%d\t",a[i]);
> return 0;
>
> }
>
> On Tue, Sep 28, 2010 at 3:43 AM, Chi <[email protected]> wrote:
>
>> Move-To-The-Front.
>>
>> On Sep 27, 11:58 pm, Anand <[email protected]> wrote:
>> > Given an array of integers, for each index i, you have to swap the
>> value at
>> > i with the first value smaller than A[ i ] that comes after index i
>>
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