In this Max Min HackerRank solution, You will be given a list of integers, , and a single integer . You must create an array of length from elements of such that its unfairness is minimized. Call that array . Unfairness of an array is calculated as
Where:
– max denotes the largest integer in
– min denotes the smallest integer in
Example
Pick any two elements, say .
Testing for all pairs, the solution provides the minimum unfairness.
Note: Integers in may not be unique.
Function Description
Complete the maxMin function in the editor below.
maxMin has the following parameter(s):
- int k: the number of elements to select
- int arr[n]:: an array of integers
Returns
- int: the minimum possible unfairness
Input Format
The first line contains an integer , the number of elements in array .
The second line contains an integer .
Each of the next lines contains an integer where .
Constraints
Sample Input
Sample Input #01
10
4
1
2
3
4
10
20
30
40
100
200
Sample Output
Sample Output #01
3
Explanation
Explanation #01
Here; selecting theintegers, unfairness equals
max(1,2,3,4) - min(1,2,3,4) = 4 - 1 = 3Max Min HackerRank Solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in PHP
function maxMin($k, $arr) {
rsort($arr);
$unfair = $arr[0] - $arr[$k-1] ;
$count = count($arr);
for($i=0; $i<=$count-$k; $i++){
$min = $arr[$i] - $arr[$i + ($k - 1)];
if($min < $unfair){
$unfair = $min;
}
}
return $unfair;
}Problem Solution in JavaScript
function maxMin(k, arr) {
arr.sort((a, b) => a - b);
const n = arr.length;
if (k >= n) return arr[n - 1] - arr[0];
let unfairness = arr[k - 1] - arr[0];
for (let i = 1; i < n; i++) {
if (arr[k - 1 + i] - arr[i] < unfairness) {
unfairness = arr[k - 1 + i] - arr[i];
}
}
return unfairness;
}Problem Solution in Python
def maxMin(k, arr):
# Write your code here
arr.sort()
res = []
index = 0
print(arr)
for i in range(len(arr)):
if(i+k<=len(arr)):
res.append(arr[i+k-1]-arr[i])
return min(res)Problem Solution in Java
public static int maxMin(int k, List<Integer> arr) {
// Write your code here
List<Integer> res = new ArrayList<Integer>();
Collections.sort(arr);
for(int i=0; i<arr.size(); i++)
{
if(i+k<=arr.size()){
int value = arr.get(i+k-1)-arr.get(i);
res.add(value);
}
}
return Collections.min(res);
}Problem Solution in C++
int maxMin(int k, vector<int> arr) {
sort(arr.begin(), arr.end());
int min = numeric_limits<int>::max();
for (int i = 0; i < arr.size() - k + 1; i++)
{
if (arr[k - 1 + i] - arr[i] < min)
min = arr[k - 1 + i] - arr[i];
}
return min;
}