MinAbsSumOfTwo ::: Find the minimal absolute value of a sum of two elements.

Task description

Let A be a non-empty zero-indexed array consisting of N integers.
The abs sum of two for a pair of indices (P, Q) is the absolute value |A[P] + A[Q]|, for 0 ≤ P ≤ Q < N.
For example, the following array A:
A[0] = 1 A[1] = 4 A[2] = -3has pairs of indices (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).
The abs sum of two for the pair (0, 0) is A[0] + A[0] = |1 + 1| = 2.
The abs sum of two for the pair (0, 1) is A[0] + A[1] = |1 + 4| = 5.
The abs sum of two for the pair (0, 2) is A[0] + A[2] = |1 + (−3)| = 2.
The abs sum of two for the pair (1, 1) is A[1] + A[1] = |4 + 4| = 8.
The abs sum of two for the pair (1, 2) is A[1] + A[2] = |4 + (−3)| = 1.
The abs sum of two for the pair (2, 2) is A[2] + A[2] = |(−3) + (−3)| = 6.
Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty zero-indexed array A consisting of N integers, returns the minimal abs sum of two for any pair of indices in this array.
For example, given the following array A:
A[0] = 1 A[1] = 4 A[2] = -3the function should return 1, as explained above.
Given array A:
A[0] = -8 A[1] = 4 A[2] = 5 A[3] =-10 A[4] = 3the function should return |(−8) + 5| = 3.
Assume that:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

• expected worst-case time complexity is O(N*log(N));
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

class Solution { public int solution(int[] A) { // write your code in Java SE 8 int len=A.length; int min=0; int sum=0; for(int i=0;i<len;i++){ for(int j=i;j<len;j++){ sum=Math.abs(A[i]+A[j]); if(i==0&&j==0) min=sum; else{ if(min>sum) min=sum; } } } return min; } }

Analysis summary

The following issues have been detected: timeout errors.

Analysis

Detected time complexity:
O(N * N)