# Print X array elements closest to the Kth smallest element in the array

Given two integers K, X, and an array arr[] consisting of N distinct elements, the task is to find X elements closest to the Kth smallest element from the given array.

Examples:

Input: arr[] = {1, 2, 3, 4, 10}, K = 3, X = 2
Output: 2 3
Explanation: Kth smallest element present in the given array is 3 and X(= 2) closest array elements to 3 are {2, 3} or {3, 4}.
Therefore, the required output is 2 3.

Input: arr[] = {1, 9, 8, 2, 7, 3, 6, 4, 5, 10, 13, 12, 16, 14, 11, 15}, K = 3, X = 5
Output: 1 2 3 4 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The simplest approach to solve this problem is to sort the array and print X closest elements to the Kth smallest element of the given array using the two-pointers technique.
Time Complexity: O(N * log N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach the idea is to efficiently compute the value of the Kth smallest element of the given array using the Median Selection Algorithm.  Follow the steps below to solve the problem:

• Calculate the Kth smallest, say KthElem of the given array using the Median Selection Algorithm.
• Initialize an array, say diff[] to store the absolute difference of arr[i] and KthElem.
• Create a map, say maps to map each element of the array to the absolute difference of the current element and KthElem.
• Traverse the given array and append arr[i] to maps[abs(KthElem – arr[i])].
• Calculate Xth smallest element, say XthElem of diff[] array using the Median Selection Algorithm to print exactly X closest items.
• Finally, traverse the diff[] array and check if XthElem less than or equal to diff[i] or not. If found to be true then print the element of the array with the help of maps.

Below is the implementation of the above approach:

 `# Python3 program to implement ` `# the above approach ` ` `  `import` `collections ` ` `  ` `  `# Function to swap ` `# two elements of array ` `def` `swap(arr, a, b): ` `    ``temp ``=` `arr[a] ` `    ``arr[a] ``=` `arr[b] ` `    ``arr[b] ``=` `temp ` `     `  `# Function to partition  ` `# the array around x ` `def` `partition(arr, l, r, x): ` `     `  `    ``# Traverse array  ` `    ``# from index l to r ` `    ``for` `i ``in` `range``(l, r): ` `         `  `        ``# partition array ` `        ``# around x ` `        ``if` `arr[i] ``=``=` `x: ` `            ``swap(arr, r, i) ` `            ``break` `       `  `    ``x ``=` `arr[r] ` `    ``i ``=` `l ` `     `  `    ``# Traverse array  ` `    ``# from index l to r  ` `    ``for` `j ``in` `range``(l, r): ` `        ``if` `(arr[j] <``=` `x): ` `            ``swap(arr, i, j) ` `            ``i ``+``=` `1` `    ``swap(arr, i, r) ` `    ``return` `i ` ` `  `# Function to find ` `# median of arr[]  ` `# from index l to l + n ` `def` `findMedian(arr, l, n): ` `    ``lis ``=` `[] ` `    ``for` `i ``in` `range``(l, l ``+` `n): ` `        ``lis.append(arr[i]) ` ` `  `    ``# Sort the array ` `    ``lis.sort() ` ` `  `    ``# Return middle element ` `    ``return` `lis[n ``/``/` `2``] ` `     `  `# Function to get  ` `# the kth smallest element ` `def` `kthSmallest(arr, l, r, k): ` ` `  `    ``# If k is smaller than ` `    ``# number of elements ` `    ``# in array ` `    ``if` `(k > ``0` `and`  `        ``k <``=` `r ``-` `l ``+` `1``): ` ` `  `        ``# Stores count of  ` `        ``# elements in arr[l..r] ` `        ``n ``=` `r ``-` `l ``+` `1` ` `  `        ``# Divide arr[] in groups ` `        ``# of size 5, calculate  ` `        ``# median  of every group ` `        ``# and store it in ` `        ``# median[] array. ` `        ``median ``=` `[] ` ` `  `        ``i ``=` `0` `        ``while` `(i < n ``/``/` `5``): ` `            ``median.append( ` `            ``findMedian(arr,  ` `                ``l ``+` `i ``*` `5``, ``5``)) ` `            ``i ``+``=` `1` ` `  `        ``# For last group with ` `        ``# less than 5 elements ` `        ``if` `(i ``*` `5` `< n): ` `            ``median.append( ` `             ``findMedian(arr,  ` `                ``l ``+` `i ``*` `5``, n ``%` `5``)) ` `            ``i ``+``=` `1` ` `  `         `  `        ``# If median[] has  ` `        ``# only one element ` `        ``if` `i ``=``=` `1``: ` `            ``medOfMed ``=` `median[i ``-` `1``] ` `             `  `        ``# Find median of all medians ` `        ``# using recursive call. ` `        ``else``: ` `            ``medOfMed ``=` `kthSmallest( ` `             ``median, ``0``, i ``-` `1``, i ``/``/` `2``) ` ` `  `        ``# Stores position of pivot ` `        ``# element in sorted array ` `        ``pos ``=` `partition(arr, l, r, ` `                         ``medOfMed) ` ` `  `        ``# If position is same as k ` `        ``if` `(pos ``-` `l ``=``=` `k ``-` `1``): ` `            ``return` `arr[pos] ` `             `  `        ``# If position is more,     ` `        ``if` `(pos ``-` `l > k ``-` `1``):  ` `             `  `            ``# recur for left subarray ` `            ``return` `kthSmallest(arr, l,  ` `                          ``pos ``-` `1``, k) ` ` `  `        ``# Else recur for right subarray ` `        ``return` `kthSmallest(arr, pos ``+` `1``,  ` `                    ``r, k ``-` `pos ``+` `l ``-` `1``) ` ` `  `    ``# If k is more than  ` `    ``# number of elements ` `    ``# in the array ` `    ``return` `999999999999` ` `  `# Function to print  ` `def` `closestElements(arr, k, x): ` ` `  `    ``# Stores size of arr ` `    ``n ``=` `len``(arr) ` `     `  `    ``# Stores kth smallest  ` `    ``# of the given array ` `    ``KthElem ``=` `kthSmallest( ` `            ``arr, ``0``, n ``-` `1``, k) ` `             `  `    ``# Store the value of  ` `    ``# abs(KthElem - arr[i])  ` `    ``diff ``=` `[] ` `     `  `    ``# Create a map to map  ` `    ``# array element to ` `    ``# abs(KthElem - arr[i]) ` `    ``maps ``=` `collections.defaultdict( ` `                              ``list``) ` `    ``for` `elem ``in` `arr: ` `         `  `        ``# Stres the value of  ` `        ``# abs(elem - KthElem) ` `        ``temp ``=` `abs``(elem ``-` `KthElem) ` `         `  `        ``# map array elements ` `        ``maps[temp].append(elem) ` `         `  `        ``# append temp ` `        ``diff.append(temp) ` ` `  `    ``XthElem ``=` `kthSmallest(diff, ``0``,  ` `                        ``n ``-` `1``, x) ` `     `  `    ``# Store X closest elements ` `    ``res ``=` `set``() ` `     `  `    ``# Traverse diff[] array ` `    ``for` `dx ``in` `diff: ` `         `  `        ``# If absolute difference is  ` `        ``# less than or eqaul to XthElem ` `        ``if` `dx <``=` `XthElem: ` `             `  `            ``# Append closest elements  ` `            ``for` `elem ``in` `maps[dx]: ` `                ``if` `len``(res) < x: ` `                  ``res.add(elem) ` `    ``return` `res ` ` `  ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``10``, ``15``] ` `    ``k ``=` `3` `    ``x ``=` `2` `     `  `    ``# Store X closest elements ` `    ``res ``=` `closestElements(arr, k, x) ` `     `  `    ``# Print X closest elements ` `    ``for` `i ``in` `res: ` `        ``print``(i, end ``=``" "``); ` `    `

Output:
```2 3
```

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

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