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hasArrayTwoCandidates (A[], ar_size, sum)
1) Sort the array in non-decreasing order.
2) Initialize two index variables to find the candidate
elements in the sorted array.
(a) Initialize first to the leftmost index: l = 0
(b) Initialize second the rightmost index: r = ar_size-1
3) Loop while l < r.
(a) If (A[l]/A[r] == sum) then return 1
(b) Else if( A[l]/A[r] < sum ) then l++
(c) Else r--
4) No candidates in whole array - return 0
Total time complexity will be O(nlog(n)) since sorting is the expensive here which takes nlogn time.
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