Minimize (max(A[i], B[j], C[k]) - min(A[i], B[j], C[k])) of three different sorted arrays - GeeksforGeeks

18 Mar, 2022
Given three sorted arrays A, B, and C of not necessarily same sizes. Calculate the minimum absolute difference between the maximum and minimum number of any triplet A[i], B[j], C[k] such that they belong to arrays A, B and C respectively, i.e., minimize (max(A[i], B[j], C[k]) – min(A[i], B[j], C[k]))Examples:  
Input : A : [ 1, 4, 5, 8, 10 ]
        B : [ 6, 9, 15 ]
        C : [ 2, 3, 6, 6 ]
Output : 1
Explanation: When we select A[i] = 5
B[j] = 6, C[k] = 6, we get the minimum difference 
as max(A[i], B[j], C[k]) - min(A[i], B[j], C[k]))
= |6-5| = 1 

Input : A = [ 5, 8, 10, 15 ]
        B = [ 6, 9, 15, 78, 89 ]
        C = [ 2, 3, 6, 6, 8, 8, 10 ]
Output : 1
Explanation: When we select A[i] = 10
b[j] = 9, C[k] = 10.
 
Start with the largest elements in each of the arrays A, B & C. Maintain a variable to update the answer during each of the steps to be followed. In every step, the only possible way to decrease the difference is to decrease the maximum element out of the three elements. So traverse to the next largest element in the array containing the maximum element for this step and update the answer variable. Repeat this step until the array containing the maximum element ends.  
C++
Java
Python3
C#
PHP
Javascript
// C++ code for above approach#include<bits/stdc++.h>using namespace std; int solve(int A[], int B[], int C[], int i, int j, int k){        int min_diff, current_diff, max_term;         // calculating min difference from last        // index of lists        min_diff = Integer.MAX_VALUE;         while (i != -1 && j != -1 && k != -1)        {            current_diff = abs(max(A[i], max(B[j], C[k]))                            - min(A[i], min(B[j], C[k])));             // checking condition            if (current_diff < min_diff)                min_diff = current_diff;             // calculating max term from list            max_term = max(A[i], max(B[j], C[k]));             // Moving to smaller value in the            // array with maximum out of three.            if (A[i] == max_term)                i -= 1;            else if (B[j] == max_term)                j -= 1;            else                k -= 1;        }                 return min_diff;    }     // Driver code    int main()    {        int D[] = { 5, 8, 10, 15 };        int E[] = { 6, 9, 15, 78, 89 };        int F[] = { 2, 3, 6, 6, 8, 8, 10 };        int nD = sizeof(D) / sizeof(D[0]);        int nE = sizeof(E) / sizeof(E[0]);        int nF = sizeof(F) / sizeof(F[0]);        cout << solve(D, E, F, nD-1, nE-1, nF-1);                 return 0;    } // This code is contributed by Ravi Maurya.
// Java code for above approachimport java.util.*; class GFG{    static int solve(int[] A, int[] B, int[] C)    {        int i, j, k;                 // assigning the length -1 value        // to each of three variables        i = A.length - 1;        j = B.length - 1;        k = C.length - 1;                 int min_diff, current_diff, max_term;         // calculating min difference        // from last index of lists        min_diff = Math.abs(Math.max(A[i], Math.max(B[j], C[k]))                - Math.min(A[i], Math.min(B[j], C[k])));         while (i != -1 && j != -1 && k != -1)        {            current_diff = Math.abs(Math.max(A[i], Math.max(B[j], C[k]))                        - Math.min(A[i], Math.min(B[j], C[k])));             // checking condition            if (current_diff < min_diff)                min_diff = current_diff;             // calculating max term from list            max_term = Math.max(A[i], Math.max(B[j], C[k]));             // Moving to smaller value in the            // array with maximum out of three.            if (A[i] == max_term)                i -= 1;            else if (B[j] == max_term)                j -= 1;            else                k -= 1;        }        return min_diff;    }     // Driver code    public static void main(String []args)    {            int[] D = { 5, 8, 10, 15 };        int[] E = { 6, 9, 15, 78, 89 };        int[] F = { 2, 3, 6, 6, 8, 8, 10 };        System.out.println(solve(D, E, F));             }} // This code is contributed by rutvik_56.
# python code for above approach. def solve(A, B, C):         # assigning the length -1 value        # to each of three variables        i = len(A) - 1        j = len(B) - 1        k = len(C) - 1         # calculating min difference        # from last index of lists        min_diff = abs(max(A[i], B[j], C[k]) -        min(A[i], B[j], C[k]))         while i != -1 and j != -1 and k != -1:            current_diff = abs(max(A[i], B[j],            C[k]) - min(A[i], B[j], C[k]))             # checking condition            if current_diff < min_diff:                min_diff = current_diff             # calculating max term from list            max_term = max(A[i], B[j], C[k])             # Moving to smaller value in the            # array with maximum out of three.            if A[i] == max_term:                i -= 1            elif B[j] == max_term:                j -= 1            else:                k -= 1        return min_diff # driver code A = [ 5, 8, 10, 15 ]B = [ 6, 9, 15, 78, 89 ]C = [ 2, 3, 6, 6, 8, 8, 10 ]print(solve(A, B, C))
// C# code for above approachusing System; class GFG{    static int solve(int[] A, int[] B, int[] C)    {        int i, j, k;                 // assigning the length -1 value        // to each of three variables        i = A.Length - 1;        j = B.Length - 1;        k = C.Length - 1;                 int min_diff, current_diff, max_term;         // calculating min difference        // from last index of lists        min_diff = Math.Abs(Math.Max(A[i], Math.Max(B[j], C[k]))                - Math.Min(A[i], Math.Min(B[j], C[k])));         while (i != -1 && j != -1 && k != -1)        {            current_diff = Math.Abs(Math.Max(A[i], Math.Max(B[j], C[k]))                        - Math.Min(A[i], Math.Min(B[j], C[k])));             // checking condition            if (current_diff < min_diff)                min_diff = current_diff;             // calculating max term from list            max_term = Math.Max(A[i], Math.Max(B[j], C[k]));             // Moving to smaller value in the            // array with maximum out of three.            if (A[i] == max_term)                i -= 1;            else if (B[j] == max_term)                j -= 1;            else                k -= 1;        }        return min_diff;    }     // Driver code    public static void Main()    {            int[] D = { 5, 8, 10, 15 };        int[] E = { 6, 9, 15, 78, 89 };        int[] F = { 2, 3, 6, 6, 8, 8, 10 };        Console.WriteLine(solve(D, E, F));             }}// This code is contributed by vt_m
<?php// PHP code for above approachfunction solve($A, $B, $C,               $i, $j, $k){    $min_diff;    $current_diff;    $max_term;     // calculating min difference    // from last index of lists    $min_diff = abs(max($A[$i], max($B[$j], $C[$k])) -                    min($A[$i], min($B[$j], $C[$k])));     while ($i != -1 &&           $j != -1 && $k != -1)    {        $current_diff = abs(max($A[$i], max($B[$j], $C[$k])) -                            min($A[$i], min($B[$j], $C[$k])));         // checking condition        if ($current_diff < $min_diff)            $min_diff = $current_diff;         // calculating max term from list        $max_term = max($A[$i],                    max($B[$j], $C[$k]));         // Moving to smaller value in the        // array with maximum out of three.        if ($A[$i] == $max_term)            $i -= 1;        else if ($B[$j] == $max_term)            $j -= 1;        else            $k -= 1;    }         return $min_diff;} // Driver code $D = array (5, 8, 10, 15);$E = array (6, 9, 15, 78, 89);$F = array (2, 3, 6, 6, 8, 8, 10); $nD = sizeof($D) ;$nE = sizeof($E) ;$nF = sizeof($F);echo solve($D, $E, $F,           $nD - 1, $nE - 1,           $nF - 1); // This code is contributed by ajit?>
<script> // javascript program for above approach function solve( A, B,C)    {        let i, j, k;                   // assigning the length -1 value        // to each of three variables        i = A.length - 1;        j = B.length - 1;        k = C.length - 1;                   let min_diff, current_diff, max_term;           // calculating min difference        // from last index of lists        min_diff = Math.abs(Math.max(A[i], Math.max(B[j], C[k]))                  - Math.min(A[i], Math.min(B[j], C[k])));           while (i != -1 && j != -1 && k != -1)        {            current_diff = Math.abs(Math.max(A[i], Math.max(B[j], C[k]))                          - Math.min(A[i], Math.min(B[j], C[k])));               // checking condition            if (current_diff < min_diff)                min_diff = current_diff;               // calculating max term from list            max_term = Math.max(A[i], Math.max(B[j], C[k]));               // Moving to smaller value in the            // array with maximum out of three.            if (A[i] == max_term)                i -= 1;            else if (B[j] == max_term)                j -= 1;            else                k -= 1;        }        return min_diff;    } // Driver Function         let D = [ 5, 8, 10, 15 ];         let E = [ 6, 9, 15, 78, 89 ];         let F = [ 2, 3, 6, 6, 8, 8, 10 ];        document.write(solve(D, E, F));                    // This code is contributed by susmitakundugoaldnga.</script>
Output:  
1
Auxiliary Space: O(1) 
Ravi_Maurya
jit_t
Samyak Jain 9
aatifnazar26
susmitakundugoaldanga
rutvik_56
rishavnitro
Arrays
Searching
Arrays
Searching
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Comments
Old Comments
Arrays in Java
Stack Data Structure (Introduction and Program)
Arrays in C/C++
Multidimensional Arrays in Java
Python | Using 2D arrays/lists the right way
Binary Search
k largest(or smallest) elements in an array
Linear Search
Program to find largest element in an array
Search an element in a sorted and rotated array