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# Smallest positive integer K such that all array elements can be made equal by incrementing or decrementing by at most K

Given an array arr[] of size N, the task is to find the smallest positive integer K such that incrementing or decrementing each array element by K at most once makes all elements equal. If it is not possible to make all array elements equal, then print -1.

Examples :

Input: arr[] = { 5, 7, 9 }
Output: 2
Explanation:
Incrementing the value of arr[0] by K(= 2) modifies arr[] to { 7, 7, 9 }.
Decrementing the value of arr[2] by K(= 2) modifies arr[] to { 7, 7, 7 }

Input: arr[] = {1, 3, 9}, N = 3
Output: -1

Approach: Follow the steps below to solve the problem :

• Initialize a Set to store all distinct elements present in the array.
• Count the distinct elements in the array, which is equal to the size of the Set, say M.
• If M > 3, then print -1.
• If M = 3, then check if the difference between the largest and the second largest element of the Set is equal to the difference between the second largest and the third largest element of the Set or not. If found to be true, then print the difference. Otherwise, print -1.
• If M = 2, then check if the difference between the largest and the second largest element of the set is even or not. If found to be true, then print the half of their difference. Otherwise, print the difference.
• If M <= 1, then print 0.

Below is the implementation of the above solution :

## C++

 `// C++ program for the above approach``#include ``#include ``using` `namespace` `std;` `// Function to find smallest integer K such that``// incrementing or decrementing each element by``// K at most once makes all elements equal``void` `findMinKToMakeAllEqual(``int` `N, ``int` `A[])``{` `    ``// Store distinct``    ``// array elements``    ``set<``int``> B;` `    ``// Traverse the array, A[]``    ``for` `(``int` `i = 0; i < N; i++)``        ``B.insert(A[i]);` `    ``// Count elements into the set``    ``int` `M = B.size();` `    ``// Iterator to store first``    ``// element of B``    ``set<``int``>::iterator itr = B.begin();` `    ``// If M is greater than 3``    ``if` `(M > 3)``        ``printf``(``"-1"``);` `    ``// If M is equal to 3``    ``else` `if` `(M == 3) {` `        ``// Stores the first``        ``// smallest element``        ``int` `B_1 = *itr;` `        ``// Stores the second``        ``// smallest element``        ``int` `B_2 = *(++itr);` `        ``// Stores the largest element``        ``int` `B_3 = *(++itr);` `        ``// IF difference between B_2 and B_1``        ``// is equal to B_3 and B_2``        ``if` `(B_2 - B_1 == B_3 - B_2)``            ``printf``(``"%d"``, B_2 - B_1);``        ``else``            ``printf``(``"-1"``);``    ``}` `    ``// If M is equal to 2``    ``else` `if` `(M == 2) {` `        ``// Stores the smallest element``        ``int` `B_1 = *itr;` `        ``// Stores the largest element``        ``int` `B_2 = *(++itr);` `        ``// If difference is an even``        ``if` `((B_2 - B_1) % 2 == 0)``            ``printf``(``"%d"``, (B_2 - B_1) / 2);``        ``else``            ``printf``(``"%d"``, B_2 - B_1);``    ``}` `    ``// If M is equal to 1``    ``else``        ``printf``(``"%d"``, 0);``}` `// Driver Code``int` `main()``{` `    ``// Given array``    ``int` `A[] = { 1, 3, 5, 1 };` `    ``// Given size``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Print the required smallest integer``    ``findMinKToMakeAllEqual(N, A);``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;``class` `GFG {``    ` `    ``// Function to find smallest integer K such that``    ``// incrementing or decrementing each element by``    ``// K at most once makes all elements equal``    ``static` `void` `findMinKToMakeAllEqual(``int` `N, ``int` `A[])``    ``{``      ` `        ``// Store distinct``        ``// array elements``        ``Set B = ``new` `HashSet();``      ` `        ``// Traverse the array, A[]``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{``            ``B.add(A[i]);``        ``}``        ` `        ``ArrayList b = ``new` `ArrayList(B);``      ` `        ``// Count elements into the set``        ``int` `M = b.size();``        ``int` `i = ``0``;``      ` `        ``// If M is greater than 3``        ``if` `(M > ``3``)``        ``{    System.out.print(``"-1"``);}``        ` `        ` `        ``// If M is equal to 3``        ``else` `if` `(M == ``3``)``        ``{``          ` `            ``// Stores the first``            ``// smallest element``            ``int` `B_1 = b.get(i++);``            ` `            ``// Stores the second``            ``// smallest element``            ``int` `B_2 =  b.get(i++);``            ` `            ``// Stores the largest element``            ``int` `B_3 = b.get(i++);``            ` `            ``// IF difference between B_2 and B_1``            ``// is equal to B_3 and B_2``            ``if` `(B_2 - B_1 == B_3 - B_2)``            ``{``                ``System.out.print(B_2 - B_1);``            ``}``            ``else``            ``{``                ``System.out.print(``"-1"``);``            ``}``            ` `            ` `        ``}``        ` `        ``// If M is equal to 2``        ``else` `if` `(M == ``2``)``        ``{``          ` `            ``// Stores the smallest element``            ``int` `B_1 = b.get(i++);``            ` `            ``// Stores the largest element``            ``int` `B_2 = b.get(i++);``            ` `            ``// If difference is an even``            ``if` `((B_2 - B_1) % ``2` `== ``0``)``            ``{``                ``System.out.print((B_2 - B_1) / ``2``);``            ``}``            ` `            ``else``            ``{``                ``System.out.print(B_2 - B_1);``            ``}``        ``}``        ``// If M is equal to 1``        ``else``        ``{``            ``System.out.print(``0``);``        ``}``    ``}``    ` `  ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ` `        ``// Given array``        ``int` `A[] = { ``1``, ``3``, ``5``, ``1` `};`` ` `        ``// Given size``        ``int` `N = A.length;``        ` `        ` `        ``// Print the required smallest integer``        ``findMinKToMakeAllEqual(N, A);``    ``}``}` `// This code is contributed by rag2127`

## Python3

 `# Python3 program for the above approach` `# Function to find smallest integer K such``# that incrementing or decrementing each``# element by K at most once makes all``# elements equal``def` `findMinKToMakeAllEqual(N, A):``    ` `    ``# Store distinct``    ``# array elements``    ``B ``=` `{}``    ` `    ``# Traverse the array, A[]``    ``for` `i ``in` `range``(N):``        ``B[A[i]] ``=` `1``        ` `    ``# Count elements into the set``    ``M ``=` `len``(B)``    ` `    ``# Iterator to store first``    ``# element of B``    ``itr, i ``=` `list``(B.keys()), ``0``    ` `    ``# If M is greater than 3``    ``if` `(M > ``3``):``        ``print``(``"-1"``)` `    ``# If M is equal to 3``    ``elif` `(M ``=``=` `3``):``        ` `        ``# Stores the first``        ``# smallest element``        ``B_1, i ``=` `itr[i], i ``+` `1``        ` `        ``# Stores the second``        ``# smallest element``        ``B_2, i ``=` `itr[i], i ``+` `1` `        ``# Stores the largest element``        ``B_3, i ``=` `itr[i], i ``+` `1` `        ``# IF difference between B_2 and B_1``        ``# is equal to B_3 and B_2``        ``if` `(B_2 ``-` `B_1 ``=``=` `B_3 ``-` `B_2):``            ``print``(B_2 ``-` `B_1)``        ``else``:``            ``print``(``"-1"``)` `    ``# If M is equal to 2``    ``elif` `(M ``=``=` `2``):``        ` `        ``# Stores the smallest element``        ``B_1, i ``=` `itr[i], i ``+` `1``        ` `        ``# Stores the largest element``        ``B_2, i ``=` `itr[i], i ``+` `1``        ` `        ``# If difference is an even``        ``if` `((B_2 ``-` `B_1) ``%` `2` `=``=` `0``):``            ``print``((B_2 ``-` `B_1) ``/``/` `2``)``        ``else``:``            ``print``(B_2 ``-` `B_1)` `    ``# If M is equal to 1``    ``else``:``        ``print``(``0``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given array``    ``A ``=` `[ ``1``, ``3``, ``5``, ``1` `]``    ` `    ``# Given size``    ``N ``=` `len``(A)` `    ``# Print the required smallest integer``    ``findMinKToMakeAllEqual(N, A)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {` `  ``// Function to find smallest integer K such that``  ``// incrementing or decrementing each element by``  ``// K at most once makes all elements equal``  ``static` `void` `findMinKToMakeAllEqual(``int` `N, ``int``[] A)``  ``{` `    ``// Store distinct``    ``// array elements``    ``var` `B = ``new` `HashSet<``int``>();` `    ``// Traverse the array, A[]``    ``for` `(``int` `i = 0; i < N; i++) {``      ``B.Add(A[i]);``    ``}` `    ``List<``int``> b = ``new` `List<``int``>(B);` `    ``// Count elements into the set``    ``int` `M = b.Count;``    ``int` `j = 0;` `    ``// If M is greater than 3``    ``if` `(M > 3) {``      ``Console.Write(``"-1"``);``    ``}` `    ``// If M is equal to 3``    ``else` `if` `(M == 3) {` `      ``// Stores the first``      ``// smallest element``      ``int` `B_1 = b[j++];` `      ``// Stores the second``      ``// smallest element``      ``int` `B_2 = b[j++];` `      ``// Stores the largest element``      ``int` `B_3 = b[j++];` `      ``// IF difference between B_2 and B_1``      ``// is equal to B_3 and B_2``      ``if` `(B_2 - B_1 == B_3 - B_2) {``        ``Console.Write(B_2 - B_1);``      ``}``      ``else` `{``        ``Console.Write(``"-1"``);``      ``}``    ``}` `    ``// If M is equal to 2``    ``else` `if` `(M == 2) {` `      ``// Stores the smallest element``      ``int` `B_1 = b[j++];` `      ``// Stores the largest element``      ``int` `B_2 = b[j++];` `      ``// If difference is an even``      ``if` `((B_2 - B_1) % 2 == 0) {``        ``Console.Write((B_2 - B_1) / 2);``      ``}` `      ``else` `{``        ``Console.Write(B_2 - B_1);``      ``}``    ``}``    ``// If M is equal to 1``    ``else` `{``      ``Console.Write(0);``    ``}``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(``string``[] args)``  ``{` `    ``// Given array``    ``int``[] A = { 1, 3, 5, 1 };` `    ``// Given size``    ``int` `N = A.Length;` `    ``// Print the required smallest integer``    ``findMinKToMakeAllEqual(N, A);``  ``}``}` `// This code is contributed by chitranayal.`

## Javascript

 ``

Output

```2

```

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)

Another Approach:

1. Find the minimum and maximum elements in the array.
• Initialize minElement and maxElement as the first element of the array.
• Iterate through the array and update minElement and maxElement accordingly.
• If the current element is less than minElement, update minElement.
• If the current element is greater than maxElement, update maxElement.

2. Check if all elements are already equal.

• If minElement is equal to maxElement, then all elements are already equal, and the smallest positive integer K is 0. Return 0.

3. Check if it’s possible to make all elements equal.

• If the difference between maxElement and minElement is odd, it’s not possible to make all elements equal by incrementing or decrementing them. Return -1.

4. Calculate the value of ‘diff’ as half of the difference between maxElement and minElement.

• diff = (maxElement – minElement) / 2

5. Check if it’s possible to make all elements equal by adding or subtracting ‘diff’.

• Iterate through the array.
• If the current element is not equal to minElement, maxElement, or minElement + diff, it’s not possible to make all elements equal. Return -1.

6. Return ‘diff’ as the smallest positive integer K.

• ‘diff’ is the value by which each array element needs to be incremented or decremented to make all elements equal.

Below is the implementation of the above approach:

## C++

 `#include ``#include ``#include ` `using` `namespace` `std;` `int` `findSmallestK(vector<``int``>& arr) {``    ``int` `n = arr.size();` `    ``// Find the minimum and maximum elements in the array``    ``int` `minElement = *min_element(arr.begin(), arr.end());``    ``int` `maxElement = *max_element(arr.begin(), arr.end());` `    ``// Check if all elements are already equal``    ``if` `(minElement == maxElement) {``        ``return` `0;``    ``}` `    ``// Check if it's possible to make all elements equal``    ``if` `((maxElement - minElement) % 2 != 0) {``        ``return` `-1;``    ``}` `    ``int` `diff = (maxElement - minElement) / 2;` `    ``// Check if it's possible to make all elements equal by adding or subtracting 'diff'``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] != minElement && arr[i] != maxElement && arr[i] != minElement + diff) {``            ``return` `-1;``        ``}``    ``}` `    ``return` `diff;``}` `int` `main() {``    ``vector<``int``> arr = { 5, 7, 9 };` `    ``int` `smallestK = findSmallestK(arr);``    ``cout << smallestK << endl;` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `GFG {` `    ``// Function to find the smallest possible value of 'k'``    ``// such that all elements in the array can be made equal``    ``// by adding or subtracting 'k'``    ``public` `static` `int` `findSmallestK(List arr) {``        ``int` `n = arr.size();` `        ``// Find the minimum and maximum elements in the array``        ``int` `minElement = Collections.min(arr);``        ``int` `maxElement = Collections.max(arr);` `        ``// Check if all elements are already equal``        ``if` `(minElement == maxElement) {``            ``return` `0``;``        ``}` `        ``// Check if it's possible to make all elements equal``        ``if` `((maxElement - minElement) % ``2` `!= ``0``) {``            ``return` `-``1``;``        ``}` `        ``int` `diff = (maxElement - minElement) / ``2``;` `        ``// Check if it's possible to make all elements equal``        ``// by adding or subtracting 'diff'``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(arr.get(i) != minElement && arr.get(i) != maxElement && arr.get(i) != minElement + diff) {``                ``return` `-``1``;``            ``}``        ``}` `        ``return` `diff;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``List arr = ``new` `ArrayList<>(Arrays.asList(``5``, ``7``, ``9``));` `        ``int` `smallestK = findSmallestK(arr);``        ``System.out.println(smallestK);``    ``}``}`

## Python3

 `def` `find_smallest_k(arr):``    ``n ``=` `len``(arr)` `    ``# Find the minimum and maximum elements in the array``    ``min_element ``=` `min``(arr)``    ``max_element ``=` `max``(arr)` `    ``# Check if all elements are already equal``    ``if` `min_element ``=``=` `max_element:``        ``return` `0` `    ``# Check if it's possible to make all elements equal``    ``if` `(max_element ``-` `min_element) ``%` `2` `!``=` `0``:``        ``return` `-``1` `    ``diff ``=` `(max_element ``-` `min_element) ``/``/` `2` `    ``# Check if it's possible to make all elements equal by adding or subtracting 'diff'``    ``for` `i ``in` `range``(n):``        ``if` `arr[i] !``=` `min_element ``and` `arr[i] !``=` `max_element ``and` `arr[i] !``=` `min_element ``+` `diff:``            ``return` `-``1` `    ``return` `diff` `if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``5``, ``7``, ``9``]` `    ``smallest_k ``=` `find_smallest_k(arr)``    ``print``(smallest_k)`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `namespace` `SmallestKEqualization``{``    ``class` `Program``    ``{``        ``static` `int` `FindSmallestK(List<``int``> arr)``        ``{``            ``int` `n = arr.Count;` `            ``// Find the minimum and maximum elements in the list``            ``int` `minElement = arr.Min();``            ``int` `maxElement = arr.Max();` `            ``// Check if all elements are already equal``            ``if` `(minElement == maxElement)``            ``{``                ``return` `0;``            ``}` `            ``// Check if it's possible to make all elements equal``            ``if` `((maxElement - minElement) % 2 != 0)``            ``{``                ``return` `-1;``            ``}` `            ``int` `diff = (maxElement - minElement) / 2;` `            ``// Check if it's possible to make all elements equal by adding or subtracting 'diff'``            ``foreach` `(``int` `num ``in` `arr)``            ``{``                ``if` `(num != minElement && num != maxElement && num != minElement + diff)``                ``{``                    ``return` `-1;``                ``}``            ``}` `            ``return` `diff;``        ``}` `        ``static` `void` `Main(``string``[] args)``        ``{``            ``List<``int``> arr = ``new` `List<``int``> { 5, 7, 9 };` `            ``int` `smallestK = FindSmallestK(arr);``            ``Console.WriteLine(smallestK);` `            ``// Pause execution to see the result``            ``Console.ReadKey();``        ``}``    ``}``}`

Output

```2

```

Time Complexity: O(N)
Auxiliary Space: O(1)