Find the minimum difference of all possible pairs - c++

Question

Given 3 numeric sorted arrays:
int[] a;
int[] b;
int[] c;
It is necessary to find three numbers whose difference is minimal:
(a_i- b_j)^2 + (b_j - c_k)^2 + (a_i - c_k)^2 -> min
For examle:
int[] a = {7, 10, 12};
int[] b = {3, 4, 6, 9};
int[] c = {1, 2, 5, 8};
Result should be 7 6 8 (or 7 6 5) because (7 - 6)^2 + (6 - 5)^2 + (7 - 5)^2 = 6 = min
What is the best approach to this problem?
I`ve tried to use 3 variables and increment them depending on the minimum term, but unsuccessfully.
Here is my algorithm written in C++:
void printThreeClosest(int[] a, int[] b, int[] c) {
int64_t diff = (a[0] - b[0]) * (a[0] - b[0]) +
(b[0] - c[0]) * (b[0] - c[0]) +
(a[0] - c[0]) * (a[0] - c[0]);
int i_res = 0, j_res = 0, k_res = 0, i = 0, j = 0, k = 0;
while (i < a.size() && j < b.size() && k < c.size()) {
int first_term = (a[i] - b[j]) * (a[i] - b[j]);
int second_term = (b[j] - c[k]) * (b[j] - c[k]);
int third_term = (a[i] - c[k]) * (a[i] - c[k]);
int min_term = std::min(a[i], std::min(b[j], c[k]));
int64_t current_diff = first_term + second_term + third_term;
if (current_diff < diff) {
diff = current_diff;
i_res = i, j_res = j, k_res = k;
}
if (diff == 0) {
break;
}
if (first_term == min_term) {
++k;
} else if (second_term == min_term) {
++i;
} else {
++j;
}
}
std::cout << a[i_res] << " " << b[j_res] << " " << c[k_res] << "\n";
}

Answer 1

IMO, the straight-forward, but O(n^3) solution may be this:
void printThreeClosest(int[] a, int[] b, int[] c) {
int minVal = -1;
int resa, int resb, int resc;
for(int i=0; i<a.length && minVal != 0; i++) {
int x = a[i];
for(int j=0; j<b.length && minVal != 0; j++) {
int y = b[j];
for(int k=0; k<c.length && minVal != 0; k++) {
int z = c[k];
int sumOfSqr = ( ((x-y)*(x-y)) + ((y-z)*(y-z)) + ((z-x)*(z-x)) );
if (minVal < 0) {
minVal = sumOfSqr;
resa=x; resb=y; resc=z;
} else if (sumOfSqr < minVal) {
minVal = sumOfSqr;
resa=x; resb=y; resc=z;
}
}
}
};
std::cout << resa << ' ' << resb << ' ' << resc << std::endl;
}
This solution implemented in Javascript for quick testing is below:
const a = [7, 10, 12];
const b = [3, 4, 6, 9];
const c = [1, 2, 5, 8];
let minVal = -1;
let result = [];
const sumOfSquares = (x, y, z) => ( ((x-y)*(x-y)) + ((y-z)*(y-z)) + ((x-z)*(x-z)) );
for (let i=0; i < a.length && minVal !== 0; i++) {
for (let j=0; j < b.length && minVal !== 0; j++) {
for (let k=0; k < c.length && minVal !== 0; k++) {
const tempSum = sumOfSquares(a[i], b[j], c[k]);
if (minVal < 0) {
minVal = tempSum;
result = [a[i], b[j], c[k]];
} else if (tempSum < minVal) {
minVal = tempSum;
result = [a[i], b[j], c[k]];
}
}
}
}
console.log('result: ', result);