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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(IIIT Allahabad)
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