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

• Difficulty Level : Medium
• Last Updated : 29 Jul, 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.

Implementation:

## C++

 `// C++ code for above approach` `#include` `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 = ``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` `    ``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

 `// Java code for above approach` `import` `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.`

## Python3

 `# 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#

 `// C# code for above approach` `using` `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

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## Javascript

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Output

`1`

Time Complexity : O(n), where n is the combined sizes of all input arrays.
Auxiliary Space: O(1), since no extra space has been taken

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