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# Find K elements whose absolute difference with median of array is maximum

• Difficulty Level : Medium
• Last Updated : 24 Mar, 2023

Given an array arr[] and an integer K, the task is to find the K elements of the array whose absolute difference with median of array is maximum.
Note: If two elements have equal difference then the maximum element is taken into consideration.

Examples:

Input : arr[] = {1, 2, 3, 4, 5}, k = 3
Output : {5, 1, 4}
Explanation :
Median m = 3,
Difference of each array elements from median,
1 ==> diff(1-3) = 2
2 ==> diff(2-3) = 1
3 ==> diff(3-3) = 0
4 ==> diff(4-3) = 1
5 ==> diff(5-3) = 2
First K elements are 5, 1, 4 in this array.

Input: arr[] = {1, 2, 3}, K = 2
Output: {3, 1}

Approach:

• Sort the array and find the median of the array
• Create a difference array to store the difference of each element with the median of the sorted array.
• Highest difference elements will be the corner elements of the array. Therefore, initialize the two pointers as both the corner elements of the array that is 0 and N – 1.
• Finally include the elements of the array one by one with the maximum difference with the median.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find first K ` `// elements whose difference with the  ` `// median of array is maximum ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function for calculating median  ` `double` `findMedian(``int` `a[], ``int` `n)  ` `{ ` `    ``// check for even case  ` `    ``if` `(n % 2 != 0)  ` `       ``return` `(``double``)a[n/2];  ` `       `  `    ``return` `(``double``)(a[(n-1)/2] + a[n/2])/2.0;  ` `}  ` ` `  `// Function to find the K maximum absolute ` `// difference with the median of the array ` `void` `kStrongest(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array. ` `    ``sort(arr, arr + n); ` ` `  `    ``// Store median ` `    ``double` `median = findMedian(arr, n); ` `    ``int` `diff[n]; ` ` `  `    ``// Find and store difference ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``diff[i] = ``abs``(median - arr[i]); ` `    ``} ` ` `  `    ``int` `i = 0, j = n - 1; ` `    ``while` `(k > 0) { ` `         `  `        ``// If diff[i] is greater print it ` `        ``// Else print diff[j] ` `        ``if` `(diff[i] > diff[j]) { ` `            ``cout << arr[i] << ``" "``; ` `            ``i++; ` `        ``} ` `        ``else` `{ ` `            ``cout << arr[j] << ``" "``; ` `            ``j--; ` `        ``} ` `        ``k--; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `k = 3; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``kStrongest(arr, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find first K ` `// elements whose difference with the  ` `// median of array is maximum ` `import` `java.util.*; ` `class` `GFG{ ` `  `  `// Function for calculating median  ` `static` `double` `findMedian(``int` `a[], ``int` `n)  ` `{ ` `    ``// check for even case  ` `    ``if` `(n % ``2` `!= ``0``)  ` `       ``return` `(``double``)a[n / ``2``];  ` `        `  `    ``return` `(``double``)(a[(n - ``1``) / ``2``] +  ` `                    ``a[n / ``2``]) / ``2.0``;  ` `}  ` `  `  `// Function to find the K maximum absolute ` `// difference with the median of the array ` `static` `void` `kStrongest(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array. ` `    ``Arrays.sort(arr); ` `  `  `    ``// Store median ` `    ``double` `median = findMedian(arr, n); ` `    ``int` `[]diff = ``new` `int``[n]; ` `  `  `    ``// Find and store difference ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``diff[i] = (``int``)Math.abs(median - arr[i]); ` `    ``} ` `  `  `    ``int` `i = ``0``, j = n - ``1``; ` `    ``while` `(k > ``0``)  ` `    ``{ ` `          `  `        ``// If diff[i] is greater print it ` `        ``// Else print diff[j] ` `        ``if` `(diff[i] > diff[j]) ` `        ``{ ` `            ``System.out.print(arr[i] + ``" "``); ` `            ``i++; ` `        ``} ` `        ``else`  `        ``{ ` `            ``System.out.print(arr[j] + ``" "``); ` `            ``j--; ` `        ``} ` `        ``k--; ` `    ``} ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `}; ` `    ``int` `k = ``3``; ` `    ``int` `n = arr.length; ` `  `  `    ``kStrongest(arr, n, k); ` `} ` `} ` `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program to find first K ` `# elements whose difference with the ` `# median of array is maximum ` ` `  `# Function for calculating median ` `def` `findMedian(a, n): ` `     `  `    ``# Check for even case ` `    ``if` `(n ``%` `2` `!``=` `0``): ` `        ``return` `a[``int``(n ``/` `2``)] ` `         `  `    ``return` `(a[``int``((n ``-` `1``) ``/` `2``)] ``+`  `            ``a[``int``(n ``/` `2``)]) ``/` `2.0` ` `  `# Function to find the K maximum  ` `# absolute difference with the  ` `# median of the array ` `def` `kStrongest(arr, n, k): ` `     `  `    ``# Sort the array ` `    ``arr.sort() ` `     `  `    ``# Store median ` `    ``median ``=` `findMedian(arr, n) ` `    ``diff ``=` `[``0``] ``*` `(n) ` `     `  `    ``# Find and store difference ` `    ``for` `i ``in` `range``(n): ` `        ``diff[i] ``=` `abs``(median ``-` `arr[i]) ` `         `  `    ``i ``=` `0` `    ``j ``=` `n ``-` `1` `     `  `    ``while` `(k > ``0``): ` `         `  `        ``# If diff[i] is greater print  ` `        ``# it. Else print diff[j] ` `        ``if` `(diff[i] > diff[j]): ` `            ``print``(arr[i], end ``=` `" "``) ` `            ``i ``+``=` `1` `        ``else``: ` `            ``print``(arr[j], end ``=` `" "``) ` `            ``j ``-``=` `1` `         `  `        ``k ``-``=` `1` `     `  `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `] ` `k ``=` `3` `n ``=` `len``(arr) ` ` `  `kStrongest(arr, n, k) ` ` `  `# This code is contributed by sanjoy_62 `

## C#

 `// C# implementation to find first K ` `// elements whose difference with the  ` `// median of array is maximum ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function for calculating median  ` `static` `double` `findMedian(``int` `[]a, ``int` `n)  ` `{ ` `    ``// Check for even case  ` `    ``if` `(n % 2 != 0)  ` `        ``return` `(``double``)a[n / 2];  ` `         `  `    ``return` `(``double``)(a[(n - 1) / 2] +  ` `                    ``a[n / 2]) / 2.0;  ` `}  ` ` `  `// Function to find the K maximum absolute ` `// difference with the median of the array ` `static` `void` `kStrongest(``int` `[]arr, ``int` `n, ` `                                  ``int` `k) ` `{ ` `     `  `    ``// Sort the array. ` `    ``Array.Sort(arr); ` `     `  `    ``int` `i = 0; ` `     `  `    ``// Store median ` `    ``double` `median = findMedian(arr, n); ` `    ``int` `[]diff = ``new` `int``[n]; ` ` `  `    ``// Find and store difference ` `    ``for``(i = 0; i < n; i++)  ` `    ``{ ` `       ``diff[i] = (``int``)Math.Abs(median - arr[i]); ` `    ``} ` ` `  `    ``int` `j = n - 1; ` `    ``i = 0; ` `    ``while` `(k > 0)  ` `    ``{ ` `         `  `        ``// If diff[i] is greater print it ` `        ``// Else print diff[j] ` `        ``if` `(diff[i] > diff[j]) ` `        ``{ ` `            ``Console.Write(arr[i] + ``" "``); ` `            ``i++; ` `        ``} ` `        ``else` `        ``{ ` `            ``Console.Write(arr[j] + ``" "``); ` `            ``j--; ` `        ``} ` `        ``k--; ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 1, 2, 3, 4, 5 }; ` `    ``int` `k = 3; ` `    ``int` `n = arr.Length; ` ` `  `    ``kStrongest(arr, n, k); ` `} ` `} ` ` `  `// This code is contributed by Rohit_ranjan `

## Javascript

 ` `

Output

`5 1 4 `

Time Complexity: O(nlogn), where nlogn is the time complexity required to sort the given array
Auxiliary Space: O(n), extra space used to create a diff array

Another Approach:- In this approach we will se how we can reduce the space complexity to O(N)->O(1)

• There is no need to take difference array as in the above approach
• As we came to know that the difference will be maximum with corner values so no need to store the difference just take 2 pointer on start and end.
• Print the one which have greater difference and move that pointer

Implementation:-

## C++

 `// C++ implementation to find first K ` `// elements whose difference with the  ` `// median of array is maximum ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function for calculating median  ` `double` `findMedian(``int` `a[], ``int` `n)  ` `{ ` `    ``// check for even case  ` `    ``if` `(n % 2 != 0)  ` `       ``return` `(``double``)a[n/2];  ` `       `  `    ``return` `(``double``)(a[(n-1)/2] + a[n/2])/2.0;  ` `}  ` ` `  `// Function to find the K maximum absolute ` `// difference with the median of the array ` `void` `kStrongest(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the array. ` `    ``sort(arr, arr + n); ` ` `  `    ``// Store median ` `    ``double` `median = findMedian(arr, n); ` ` `  `    ``int` `i = 0, j = n - 1; ` `    ``while` `(k) { ` `       `  `          ``//if difference of element at i with K is greater than element at j ` `        ``if` `(``abs``(median-arr[i]) > ``abs``(median-arr[j])) { ` `            ``cout << arr[i] << ``" "``; ` `            ``i++; ` `        ``} ` `        ``else` `{ ` `            ``cout << arr[j] << ``" "``; ` `            ``j--; ` `        ``} ` `        ``k--; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `k = 3; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``kStrongest(arr, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find first K ` `// elements whose difference with the ` `// median of array is maximum ` ` `  `import` `java.util.*; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function for calculating median ` `    ``public` `static` `double` `findMedian(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// check for even case ` `        ``if` `(n % ``2` `!= ``0``) { ` `            ``return` `(``double``)a[n / ``2``]; ` `        ``} ` ` `  `        ``return` `(``double``)(a[(n - ``1``) / ``2``] + a[n / ``2``]) / ``2.0``; ` `    ``} ` ` `  `    ``// Function to find the K maximum absolute ` `    ``// difference with the median of the array ` `    ``public` `static` `void` `kStrongest(``int``[] arr, ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Sort the array. ` `        ``Arrays.sort(arr); ` ` `  `        ``// Store median ` `        ``double` `median = findMedian(arr, n); ` ` `  `        ``int` `i = ``0``; ` `        ``int` `j = n - ``1``; ` `        ``while` `(k != ``0``) { ` ` `  `            ``// if difference of element at i with K is ` `            ``// greater than element at j ` `            ``if` `(Math.abs(median - arr[i]) ` `                ``> Math.abs(median - arr[j])) { ` `                ``System.out.print(arr[i]); ` `                ``System.out.print(``" "``); ` `                ``i++; ` `            ``} ` `            ``else` `{ ` `                ``System.out.print(arr[j]); ` `                ``System.out.print(``" "``); ` `                ``j--; ` `            ``} ` `            ``k--; ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = { ``1``, ``2``, ``3``, ``4``, ``5` `}; ` `        ``int` `k = ``3``; ` `        ``int` `n = arr.length; ` ` `  `        ``kStrongest(arr, n, k); ` `    ``} ` `} ` ` `  `// this code is contributed by bhardwajji`

## Python3

 `# C++ implementation to find first K ` `# elements whose difference with the ` `# median of array is maximum ` ` `  `# Function for calculating median ` `def` `findMedian(a, n): ` `   `  `      ``# check for even case ` `    ``if` `n ``%` `2` `!``=` `0``: ` `        ``return` `a[n``/``/``2``] ` `    ``return` `(a[(n``-``1``)``/``/``2``] ``+` `a[n``/``/``2``])``/``2` ` `  `# Function to find the K maximum absolute ` `# difference with the median of the array ` `def` `kStrongest(arr, n, k): ` `   `  `      ``# sort array ` `    ``arr.sort() ` `     `  `    ``# store median ` `    ``median ``=` `findMedian(arr, n) ` `    ``i ``=` `0` `    ``j ``=` `n ``-` `1` `    ``while` `k: ` `       `  `          ``# if difference of element at i with K is greater than element at j ` `        ``if` `abs``(median``-``arr[i]) > ``abs``(median``-``arr[j]): ` `            ``print``(arr[i], end``=``" "``) ` `            ``i ``+``=` `1` `        ``else``: ` `            ``print``(arr[j], end``=``" "``) ` `            ``j ``-``=` `1` `        ``k ``-``=` `1` ` `  `#driver code ` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``] ` `k ``=` `3` `n ``=` `len``(arr) ` `kStrongest(arr, n, k) `

## C#

 `// C# implementation to find first K ` `// elements whose difference with the  ` `// median of array is maximum ` ` `  `using` `System; ` ` `  `public` `class` `GFG ` `{ ` `    ``// Function for calculating median  ` `    ``static` `double` `FindMedian(``int``[] a, ``int` `n)  ` `    ``{ ` `        ``// check for even case  ` `        ``if` `(n % 2 != 0)  ` `            ``return` `(``double``)a[n / 2];  ` `        ``return` `(``double``)(a[(n - 1) / 2] + a[n / 2]) / 2.0;  ` `    ``}  ` ` `  `    ``// Function to find the K maximum absolute ` `    ``// difference with the median of the array ` `    ``static` `void` `KStrongest(``int``[] arr, ``int` `n, ``int` `k) ` `    ``{ ` `        ``// Sort the array. ` `        ``Array.Sort(arr); ` ` `  `        ``// Store median ` `        ``double` `median = FindMedian(arr, n); ` ` `  `        ``int` `i = 0, j = n - 1; ` `        ``while` `(k > 0) { ` `            ``// if difference of element at i with K is greater than element at j ` `            ``if` `(Math.Abs(median - arr[i]) > Math.Abs(median - arr[j])) { ` `                ``Console.Write(arr[i] + ``" "``); ` `                ``i++; ` `            ``} ` `            ``else` `{ ` `                ``Console.Write(arr[j] + ``" "``); ` `                ``j--; ` `            ``} ` `            ``k--; ` `        ``} ` `    ``} ` `     `  `    ``// Driver Code ` `    ``static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int``[] arr = { 1, 2, 3, 4, 5 }; ` `        ``int` `k = 3; ` `        ``int` `n = arr.Length; ` ` `  `        ``KStrongest(arr, n, k); ` `    ``} ` `} ` ` `  `//this code is contributed by bhardwajji`

## Javascript

 `// JavaScript implementation to find first K ` `// elements whose difference with the ` `// median of array is maximum ` ` `  `function` `findMedian(a, n) { ` `// check for even case ` `if` `(n % 2 !== 0) { ` `return` `a[(Math.floor(n / 2))]; ` `} ` `return` `(a[(n - 1) / 2] + a[n / 2]) / 2.0; ` `} ` ` `  `function` `kStrongest(arr, n, k) { ` `// Sort the array. ` `arr.sort(``function``(a, b) { ` `return` `a - b; ` `}); ` `// Store median ` `var` `median = findMedian(arr, n); ` ` `  `var` `i = 0; ` `var` `j = n - 1; ` `while` `(k !== 0) { ` ` `  `    ``// if difference of element at i with K is ` `    ``// greater than element at j ` `    ``if` `(Math.abs(median - arr[i]) ` `        ``> Math.abs(median - arr[j])) { ` `        ``console.log(arr[i] + ``" "``); ` `        ``i++; ` `    ``} ` `    ``else` `{ ` `        ``console.log(arr[j] + ``" "``); ` `        ``j--; ` `    ``} ` `    ``k--; ` `} ` `} ` ` `  `// Driver Code ` `var` `arr = [ 1, 2, 3, 4, 5 ]; ` `var` `k = 3; ` `var` `n = arr.length; ` ` `  `kStrongest(arr, n, k); ` ` `  `// this code is contributed by shivamsharma215`

Output

`5 1 4 `

Time Complexity:- O(NlogN)
Auxiliary Space- O(1)

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