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

**We strongly recommend you to minimize your browser and try this yourself first.**

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++

// 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; }

## Java

// Java 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. import java.io.*; class GFG { static void findClosest(int A[], int B[], int C[], int p, int q, int r) { int diff = Integer.MAX_VALUE; // 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 = Math.min(A[i], Math.min(B[j], C[k])); int maximum = Math.max(A[i], Math.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 System.out.println(A[res_i] + " " + B[res_j] + " " + C[res_k]); } // Driver code public static void main (String[] args) { int A[] = {1, 4, 10}; int B[] = {2, 15, 20}; int C[] = {10, 12}; int p = A.length; int q = B.length; int r = C.length; // Function calling findClosest(A, B, C, p, q, r); } } // This code is contributed by Ajit.

## Python3

# Python 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. import sys def findCloset(A, B, C, p, q, r): # Initialize min diff diff = sys.maxsize res_i = 0 res_j = 0 res_k = 0 # Travesre Array i = 0 j = 0 k = 0 while(i < p and j < q and k < r): # Find minimum and maximum of # current three elements minimum = min(A[i], min(B[j], C[k])) 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 = i+1 elif B[j] == minimum: j = j+1 else: k = k+1 # Print result print(A[res_i], " ", B[res_j], " ", C[res_k]) # Driver Program A = [1, 4, 10] B = [2, 15, 20] C = [10, 12] p = len(A) q = len(B) r = len(C) findCloset(A,B,C,p,q,r) # This code is contributed by Shrikant13.

## C#

// 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. using System; class GFG { static void findClosest(int []A, int []B, int []C, int p, int q, int r) { // Initialize min diff int diff = int.MaxValue; // 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 = Math.Min(A[i], Math.Min(B[j], C[k])); int maximum = Math.Max(A[i], Math.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 Console.WriteLine(A[res_i] + " " + B[res_j] + " " + C[res_k]); } // Driver code public static void Main () { int []A = {1, 4, 10}; int []B = {2, 15, 20}; int []C = {10, 12}; int p = A.Length; int q = B.Length; int r = C.Length; // Function calling findClosest(A, B, C, p, q, r); } } // This code is contributed // by anuj_67.

**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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

**Practice Tags :**