Find three closest elements from given three sorted arrays

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(n³)

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. 

PHP

<?php 
// PHP 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.  
  
function findClosest($A, $B, $C, $p, $q, $r)  
{  
  
    $diff = PHP_INT_MAX; // Initialize min diff  
  
    // Initialize result  
    $res_i = 0; 
    $res_j = 0; 
    $res_k = 0;  
  
    // Traverse arrays  
    $i = 0; 
    $j = 0; 
    $k = 0;  
    while ($i < $p && $j < $q && $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) $i++;  
        else if ($B[$j] == $minimum) $j++;  
        else $k++;  
    }  
  
    // Print result  
    echo $A[$res_i] , " ", $B[$res_j],  
                      " ", $C[$res_k];  
}  
  
// Driver Code 
$A = array(1, 4, 10);  
$B = array(2, 15, 20);  
$C = array(10, 12);  
  
$p = sizeof($A);  
$q = sizeof($B);  
$r = sizeof($C);  
  
findClosest($A, $B, $C, $p, $q, $r);  
  
// This code is contributed by Sach_Code 
?> 

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.

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My Personal Notes

C++

Java

Python3

C#

PHP

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