Given three sorted arrays A[], B[] and C[], find 3 elements i, j and k from A, B and C respectively such that max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) is minimized. Here abs() indicates absolute value.

Example :

Input: A[] = {1, 4, 10} B[] = {2, 15, 20} C[] = {10, 12} Output: 10 15 10 10 from A, 15 from B and 10 from C Input: A[] = {20, 24, 100} B[] = {2, 19, 22, 79, 800} C[] = {10, 12, 23, 24, 119} Output: 24 22 23 24 from A, 22 from B and 23 from C

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A** Simple Solution** is to run three nested loops to consider all triplets from A, B and C. Compute the value of max(abs(A[i] – B[j]), abs(B[j] – C[k]), abs(C[k] – A[i])) for every triplet and return minimum of all values. Time complexity of this solution is O(n^{3})

A **Better Solution** is to us Binary Search.

1) Iterate over all elements of A[],

a) Binary search for element just smaller than or equal to in B[] and C[], and note the difference.

2) Repeat step 1 for B[] and C[].

3) Return overall minimum.

Time complexity of this solution is O(nLogn)

**Efficient Solution** Let ‘p’ be size of A[], ‘q’ be size of B[] and ‘r’ be size of C[]

1) Start with i=0, j=0 and k=0 (Three index variables for A, B and C respectively) // p, q and r are sizes of A[], B[] and C[] respectively. 2) Do following while i < p and j < q and k < r a) Find min and maximum of A[i], B[j] and C[k] b) Compute diff = max(X, Y, Z) - min(A[i], B[j], C[k]). c) If new result is less than current result, change it to the new result. d) Increment the pointer of the array which contains the minimum.

Note that we increment the pointer of the array which has the minimum, because our goal is to decrease the difference. Increasing the maximum pointer increases the difference. Increase the second maximum pointer can potentially increase the difference.

// C++ program to find 3 elements such that max(abs(A[i]-B[j]), abs(B[j]- // C[k]), abs(C[k]-A[i])) is minimized. #include<bits/stdc++.h> using namespace std; void findClosest(int A[], int B[], int C[], int p, int q, int r) { int diff = INT_MAX; // Initialize min diff // Initialize result int res_i =0, res_j = 0, res_k = 0; // Traverse arrays int i=0,j=0,k=0; while (i < p && j < q && k < r) { // Find minimum and maximum of current three elements int minimum = min(A[i], min(B[j], C[k])); int maximum = max(A[i], max(B[j], C[k])); // Update result if current diff is less than the min // diff so far if (maximum-minimum < diff) { res_i = i, res_j = j, res_k = k; diff = maximum - minimum; } // We can't get less than 0 as values are absolute if (diff == 0) break; // Increment index of array with smallest value if (A[i] == minimum) i++; else if (B[j] == minimum) j++; else k++; } // Print result cout << A[res_i] << " " << B[res_j] << " " << C[res_k]; } // Driver program int main() { int A[] = {1, 4, 10}; int B[] = {2, 15, 20}; int C[] = {10, 12}; int p = sizeof A / sizeof A[0]; int q = sizeof B / sizeof B[0]; int r = sizeof C / sizeof C[0]; findClosest(A, B, C, p, q, r); return 0; }

Output:

10 15 10

Time complexity of this solution is O(p + q + r) where p, q and r are sizes of A[], B[] and C[] respectively.

Thanks to Gaurav Ahirwar for suggesting above solutions.

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