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
Explanation: 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
Explanation: 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 use 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

    # Traverse 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
?>

Javascript

<script>
     // JavaScript 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)
     {
       var diff = Math.pow(10, 9); // Initialize min diff

       // Initialize result
       var res_i = 0,
         res_j = 0,
         res_k = 0;

       // Traverse arrays
       var i = 0,
         j = 0,
         k = 0;
       while (i < p && j < q && k < r)
       {

         // Find minimum and maximum of current three elements
         var minimum = Math.min(A[i], Math.min(B[j], C[k]));
         var 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
       document.write(A[res_i] + " " + B[res_j] + " " + C[res_k]);
     }

     // Driver program
     var A = [1, 4, 10];
     var B = [2, 15, 20];
     var C = [10, 12];

     var p = A.length;
     var q = B.length;
     var r = C.length;

     findClosest(A, B, C, p, q, r);

     // This code is contributed by rdtank.
   </script>

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.
Auxiliary space: O(1) as constant space is required.

Approach 2: Using Binary Search:

Another approach to solve this problem can be to use binary search along with two pointers.

First, sort all the three arrays A, B, and C. Then, we take three pointers, one for each array. For each i, j, k combination, we calculate the maximum difference using the absolute value formula given in the problem. If the current maximum difference is less than the minimum difference found so far, then we update our result.

Next, we move our pointers based on the value of the maximum element among the current i, j, k pointers. We increment the pointer of the array with the smallest maximum element, hoping to find a smaller difference.

The time complexity of this approach will be O(nlogn) due to sorting, where n is the size of the largest array.

Here’s the code for this approach:

C++

#include<bits/stdc++.h>
using namespace std;

void findClosest(int A[], int B[], int C[], int p, int q, int r)
{

    sort(A, A+p);
    sort(B, B+q);
    sort(C, C+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]));

        // Calculate the maximum difference for the current combination
        int curDiff = abs(maximum - minimum);

        // Update result if current diff is less than the min
        // diff so far
        if (curDiff < diff)
        {
            res_i = i, res_j = j, res_k = k;
            diff = curDiff;
        }

        // If the maximum element of A is the smallest among the three,
        // we move the A pointer forward
        if (A[i] == minimum && A[i] <= B[j] && A[i] <= C[k]) i++;

        // If the maximum element of B is the smallest among the three,
        // we move the B pointer forward
        else if (B[j] == minimum && B[j] <= A[i] && B[j] <= C[k]) j++;

        // If the maximum element of C is the smallest among the three,
        // we move the C pointer forward
        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

import java.util.*;

public class Main {
public static void findClosest(int[] A, int[] B, int[] C, int p, int q, int r) {
      Arrays.sort(A);
    Arrays.sort(B);
    Arrays.sort(C);

    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]));

        // Calculate the maximum difference for the current combination
        int curDiff = Math.abs(maximum - minimum);

        // Update result if current diff is less than the min
        // diff so far
        if (curDiff < diff) {
            res_i = i;
            res_j = j;
            res_k = k;
            diff = curDiff;
        }

        // If the maximum element of A is the smallest among the three,
        // we move the A pointer forward
        if (A[i] == minimum && A[i] <= B[j] && A[i] <= C[k])
            i++;

        // If the maximum element of B is the smallest among the three,
        // we move the B pointer forward
        else if (B[j] == minimum && B[j] <= A[i] && B[j] <= C[k])
            j++;

        // If the maximum element of C is the smallest among the three,
        // we move the C pointer forward
        else
            k++;
    }

    // Print result
    System.out.println(A[res_i] + " " + B[res_j] + " " + C[res_k]);
}

// Driver program
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;

    findClosest(A, B, C, p, q, r);
}
}

Python3

import sys

def find_closest(A, B, C):
    p, q, r = len(A), len(B), len(C)
    A.sort()
    B.sort()
    C.sort()

    diff = sys.maxsize
    res_i, res_j, res_k = 0, 0, 0

    i = j = k = 0
    while i < p and j < q and k < r:
        minimum = min(A[i], min(B[j], C[k]))
        maximum = max(A[i], max(B[j], C[k]))

        cur_diff = abs(maximum - minimum)
        if cur_diff < diff:
            res_i, res_j, res_k = i, j, k
            diff = cur_diff

        if A[i] == minimum and A[i] <= B[j] and A[i] <= C[k]:
            i += 1
        elif B[j] == minimum and B[j] <= A[i] and B[j] <= C[k]:
            j += 1
        else:
            k += 1

    return [A[res_i], B[res_j], C[res_k]]

# Driver program
A = [1, 4, 10]
B = [2, 15, 20]
C = [10, 12]

print(find_closest(A, B, C)) # Output: [4, 10, 10]

C#

using System;

public class Program
{
    public static int[] FindClosest(int[] A, int[] B, int[] C)
    {
        int p = A.Length;
        int q = B.Length;
        int r = C.Length;

        Array.Sort(A);
        Array.Sort(B);
        Array.Sort(C);

        int diff = int.MaxValue;
        int res_i = 0, res_j = 0, res_k = 0;

        int i = 0, j = 0, k = 0;
        while (i < p && j < q && k < r)
        {
            int minimum = Math.Min(A[i], Math.Min(B[j], C[k]));
            int maximum = Math.Max(A[i], Math.Max(B[j], C[k]));

            int curDiff = Math.Abs(maximum - minimum);

            if (curDiff < diff)
            {
                res_i = i;
                res_j = j;
                res_k = k;
                diff = curDiff;
            }

            if (A[i] == minimum && A[i] <= B[j] && A[i] <= C[k])
            {
                i++;
            }
            else if (B[j] == minimum && B[j] <= A[i] && B[j] <= C[k])
            {
                j++;
            }
            else
            {
                k++;
            }
        }

        return new int[] { A[res_i], B[res_j], C[res_k] };
    }

    public static void Main()
    {
        int[] A = { 1, 4, 10 };
        int[] B = { 2, 15, 20 };
        int[] C = { 10, 12 };

        int[] result = FindClosest(A, B, C);
        Console.WriteLine(string.Join(" ", result));
    }
}

Javascript

// Function to find the closest elements from three sorted arrays
function find_closest(A, B, C) {
  // Get the length of the three arrays
  let p = A.length,
    q = B.length,
    r = C.length;

  // Sort the three arrays in non-decreasing order
  A.sort((a, b) => a - b);
  B.sort((a, b) => a - b);
  C.sort((a, b) => a - b);

  // Initialize variables to store the minimum difference and the indices
  // of the closest elements
  let diff = Number.MAX_SAFE_INTEGER;
  let res_i = 0,
    res_j = 0,
    res_k = 0;

  // Initialize indices to traverse the three arrays
  let i = 0,
    j = 0,
    k = 0;

  // Traverse the arrays until we reach the end of any one of them
  while (i < p && j < q && k < r) {
    // Get the minimum and maximum elements from the three arrays
    let minimum = Math.min(A[i], Math.min(B[j], C[k]));
    let maximum = Math.max(A[i], Math.max(B[j], C[k]));

    // Calculate the difference between the maximum and minimum elements
    let cur_diff = Math.abs(maximum - minimum);

    // Update the variables to store the indices of the closest elements
    // if the current difference is less than the minimum difference
    if (cur_diff < diff) {
      res_i = i;
      res_j = j;
      res_k = k;
      diff = cur_diff;
    }

    // Increment the index of the array with the minimum element
    if (A[i] == minimum && A[i] <= B[j] && A[i] <= C[k]) {
      i++;
    } else if (B[j] == minimum && B[j] <= A[i] && B[j] <= C[k]) {
      j++;
    } else {
      k++;
    }
  }

  // Return the closest elements from the three arrays
  return [A[res_i], B[res_j], C[res_k]];
}

// Driver program
let A = [1, 4, 10];
let B = [2, 15, 20];
let C = [10, 12];

// Call the find_closest function with the three arrays and print the result
console.log(find_closest(A, B, C));

OUTPUT:

10 15 10

Time complexity :O(NlogN) 
Auxiliary space: O(1) as constant space is required.

//Thanks to Gaurav Ahirwar for suggesting the above solutions.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.