Optimizing Pairing: How Can I Efficiently Compute the Minimum Sum of Absolute Differences in an Array?

Below is one answer. The idea is that if the array is even length the answer is just the sum of adjacent differences in the sorted array; but if it is odd you must “choose” one index to ignore. A naïve way would be to remove every possible element (giving an O(n²) algorithm). However you can pre‐compute the “pairing cost” for the sorted prefix and suffix and then combine them in O(n).

Let the sorted array be A of length n. Define two auxiliary arrays as follows:

• L (for “left”) where for i from 0 to n:
  – when i is even, L[i] is the cost of pairing A[0] … A[i–1] as
    pair (A[0],A[1]), (A[2],A[3]), … (A[i–2],A[i–1]).
  – for i odd define L[i] = L[i–1] (because if there is exactly one leftover at the end you don’t add cost).

Thus

  L[0] = 0,
  L[1] = 0,
  L[i] = L[i–2] + A[i–1] – A[i–2]  (if i is even, i ≥ 2),
  L[i] = L[i–1]         (if i is odd).

• R (for “right”) is similar but computed from the right end. (We assume indices from 0 to n–1; define R[i] as the cost to pair A[i] … A[n–1] “optimally in pairs” when the number of elements from i to n–1 is even. When it is odd it means that if you “shift” the pairing so that one element is waiting then no cost is added for that gap.) One way is to define R[n] = 0 and R[n–1] = 0, and then for i from n–2 down to 0:
  – if n–i is even then
    R[i] = (A[i+1] – A[i]) + R[i+2]
  – if n–i is odd then just set R[i] = R[i+1].

Then, when n is odd, the candidate answer comes by “removing” an element at index j. There are two cases:

1. If you remove an element at an even index j, then the remaining pairing cost is
     Candidate = L[j] + R[j+1].

2. If you remove an element at an odd index j (with j < n–1) then the leftover in the left is “re‐paired” with the first element of the suffix; that is,
     Candidate = L[j–1] + (A[j+1] – A[j–1]) + R[j+2].
    If j is the very last index (j = n–1) then candidate = L[n–1] which has been computed as L[n–2].

Finally, your answer is the minimum candidate.

Below is one complete Python implementation:

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PYTHONCopied
def smallest_sum(arr):
    A = sorted(arr)
    n = len(A)
    
    # Build L: cost to pair the first i elements (for i even) 
    L = [0]*(n+1)
    L[0] = 0
    if n >= 1:
        L[1] = 0  # one element unpaired, no extra cost
    for i in range(2, n+1):
        if i % 2 == 0:
            L[i] = L[i-2] + A[i-1] - A[i-2]
        else:
            L[i] = L[i-1]
    
    # Build R: cost to pair from index i to the end.
    R = [0]*(n+1)
    R[n] = 0
    if n >= 1:
        R[n-1] = 0  # last element unpaired gets no cost
    # We set R[i] so that if the segment A[i:] has even number of elements, then
    # optimal pairing cost is computed.
    for i in range(n-2, -1, -1):
        if (n - i) % 2 == 0:
            R[i] = R[i+2] + A[i+1] - A[i]
        else:
            R[i] = R[i+1]
    
    # Even length array: just pair adjacent numbers.
    if n % 2 == 0:
        return L[n]
    
    ans = float('inf')
    # Try removal of each candidate j
    for j in range(n):
        cand = None
        if j % 2 == 0:
            # remove an element at even index: left side A[0:j] is even-length, right side A[j+1:] is whatever.
            # In the new array the pairing is just L[j] for the left part and R[j+1] for the right part.
            cand = L[j] + R[j+1]
        else:
            # removal at odd index.
            if j < n - 1:
                # then the leftover from left (which is A[j-1] unpaired) is paired with A[j+1]
                cand = L[j-1] + (A[j+1] - A[j-1]) + R[j+2]
            else:
                # if j is the last index
                cand = L[n-1]  # which equals L[n-2]
        if cand < ans:
            ans = cand
    return ans

# Tests
assert smallest_sum([4, 1, 2, 3]) == 2  # |1-2| + |3-4| = 2
assert smallest_sum([1, 3, 3, 4, 5]) == 1  # best pairing: |3-3| + |4-5| = 1

# Example usage
print(smallest_sum([4, 1, 2, 3]))
print(smallest_sum([1, 3, 3, 4, 5]))

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Explanation

1. We sort the array (so that adjacent pairing minimizes differences).
2. The array L precomputes the cost of pairing from the beginning as if we pair the first i elements optimally.
3. The array R does the same from the right side.
4. In the odd‑length case we “simulate” removing element A[j] and then the best pairing cost on the left and right portions can be read from L and R (with a slight adjustment when j is odd because then the leftover element from the left has to be paired with the first element from the right).
5. Finally, we output the minimum over all removal candidates.

This method runs in O(n log n) (for sorting) plus O(n) processing, much faster than the O(n²) brute‐force approach.