Question

Given an array A[1..n] of positive integers and given a number x, find if
any two numbers in this array sum upto x. That is, are there i,j such
that A[i]+A[j] = x? Give an O(nlogn) algorithm for this. Suppose now
that A was already sorted, can you obtain O(n) algorithm?

Answer 1

Algorithm in O(nlogn) time for A[i]+A[j] = x:

findpair(A[] ,x)

1. sort Array A by using haep sort . it wiil have O(nlogn) complexity.

2, set low=0

3, set high= arr.size-1

4. while(low <high)

        If (A[low] + A[high] == x)  then return 1

Else if ( A[low] + A[high] < x ) then low++

Else high++

End IF

end while

5. end function

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This algorithm has complexity of O(mlogn+n) =O(nlogn)

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. Suppose now that A was already sorted, can you obtain O(n) algorithm?

Ans: yes as A is already sorted. There will be only a while loop as shown in the algorithm above. so it will the complexity of On).