在一组三个向量中找到最接近的三元组？

Question

Given three vectors of double, I want to pair every element in each vector such that the difference between the largest and smallest element in each triple is minimized, and every element of every vector is part of a triple. Right now, I'm using std::lower_bound() :

double closest(vector<double> const& vec, double value){ auto const ret = std::lower_bound(vec.begin(), vec.end(), value); return(*ret); }

int main(){
    vector<double> a, b, c; vector<vector<double>> triples;
    for(auto x : a){
       triples.push_back({x, closest(b, x), closest(c, x)});
    }
}

Pretend a, b, and c here are populated with some values. The problem is, lower_bound() returns the nearest element not less than the argument. I would also like to consider elements less than the argument. Is there a nice way to to this?

Answer 1

My solution was to implement a binary search terminating in a comparison of neighboring elements. Another possible solution is to iterate over the elements of each vector/array, adjusting the index as necessary to minimize the difference (which may compare to binary search with an ideal complexity of $O(\log{n})$?):

int solve(int A[], int B[], int C[], int i, int j, int k) 
{  
        int min_diff, current_diff, max_term; 
  
        min_diff = Integer.MAX_VALUE; 
  
        while (i != -1 && j != -1 && k != -1)  
        { 
            current_diff = abs(max(A[i], max(B[j], C[k]))  
                            - min(A[i], min(B[j], C[k]))); 
  
            if (current_diff < min_diff) 
                min_diff = current_diff; 
  
            max_term = max(A[i], max(B[j], C[k])); 
            if (A[i] == max_term) 
                i -= 1; 
            else if (B[j] == max_term) 
                j -= 1; 
            else
                k -= 1; 
        } 
          
        return min_diff; 
    }