1. use radix sort and sort the array.
2. take two pointers(say i and j). Point the first to head and the second to
the tail of the array. then check for a[i] + a[j]. If this sum is greater
the decrement the pointer j else if the sum is less than k increment the i
pointer.
If you get the sum equal to k then stop else move untill j > i.
I think this solution will have O(n) time complexity and O(n space
complexity).
On Fri, May 20, 2011 at 7:22 PM, Aakash Johari <aakashj....@gmail.com>wrote:
> One possible solution is using maps. But i think that won't be in O(n)
>
>
> On Fri, May 20, 2011 at 6:49 AM, Gunjan Sharma
> <gunjan.khan...@gmail.com>wrote:
>
>> First of all there is an infinite loop in this code....
>> Secondly it works only for sorted array.
>>
>>
>> On Fri, May 20, 2011 at 7:16 PM, hari <rajakin...@gmail.com> wrote:
>>
>>> In while loop have i,j which points first and last index of array. In
>>> while loop, Check the sum of a[i],a[j], If sum<k,increment i or else
>>> decrement j. Run the while loop till i<j..
>>>
>>> CODE:
>>>
>>> int arraysum(int a[], int k, int i, int j)
>>> while(i<j)
>>> {
>>> int p=0;
>>> int b[10]; //to store index of selected nos
>>> sum=a[i]+a[j];
>>> if (sum==k)
>>> {
>>> b[p++]=i;b[p++]=j;
>>> }
>>> elseif(sum<k)
>>> i++;
>>> else(sum>k)
>>> j++;
>>> return b;
>>> }
>>>
>>> On May 20, 4:38 am, amit <amitthecoo...@gmail.com> wrote:
>>> > given an array of integers, and an integer k, find out two elements
>>> > from the array whose sum is k in O(n) time. if no such element exists
>>> > output none.
>>>
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>>
>>
>> --
>> Regards
>> Gunjan Sharma
>> B.Tech IV year CSE
>>
>> Contact No- +91 9997767077
>>
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>
>
>
> --
> -Aakash Johari
> (IIIT Allahabad)
>
>
>
>
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regards
Apoorve Mohan
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