# 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 ` `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; ` `    ``int` `q = ``sizeof` `B / ``sizeof` `B; ` `    ``int` `r = ``sizeof` `C / ``sizeof` `C; ` ` `  `    ``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

 ` `

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.

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