Find the kth number in an array of sums

Question

Given an array A with N elements, I need to find a pair (i, j), so I do not equal j and if we write the sum A [i] + A [j] for all pairs (i, j), then it falls into k-th position.

Example : let N = 4 and arrays A = [1 2 3 4], and if K = 3, then the answer will be 5, since we can clearly see that the total array looks like this: [3, 4,5,5 , 6.7]

I can’t use i and j for all pairs, since N can reach 100,000. Please help solve this problem.

I mean something like this:

int len=N*(N+1)/2;
int sum[len];
int count=0;
for(int i=0;i<N;i++){
   for(int j=i+1;j<N;j++){
        sum[count]=A[i]+A[j]; 
        count++;
   }
}
//Then just find kth element.

We cannot go with this approach.

Answer 1

, http://www.careercup.com/question?id=7457663.

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Answer 2

, , K <= 50: K + 1 . . : , (i, j) , j > K + 1. K : (1, 2), (1, 3), ..., (1, K + 1). , K - .

O(N + K ^ 2) , K + 1, quickselect ( , ). O(N * log N + K ^ 2 * log K).

Answer 3

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//A is the original array, such as A=[1,2,3,4]
//k (an integer) is the element in the 'sum' array to find

N = A.length

//first we find i
i = -1
nl = N
k2 = k
while (k2 >= 0) {
    i++
    nl--
    k2 -= nl
}

//then we find j
j = k2 + nl + i + 1

//now compute the sum at index position k
kSum = A[i] + A[j]

EDIT: . ... k 0. (, OP 1.)

2: . , sum 2D- ( ), ( OP) i j. , [1,2,3,4,5], sum :

3,4,5,6,
5,6,7,
7,8,
9.

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