# Sort integers in array according to their distance from the element K

Given an array arr[] of N integers and an integer K, the task is to sort these integers according to their distance from given integer K. If more than 1 element are at same distance, print them in increasing order.

Note: Distance between two elements in the array is measured as the difference between their index.

Note : The integer K is always present in array arr[] and is unique.

Examples:

Input: arr[] = {12, 10, 102, 31, 15}, K = 102
Output: 102 10 31 12 15
Explanation:
Elements at their respective distance from K are,
At distance 0: 102
At distance 1: 10, 31 in sorted form.
At distance 2: 12, 15 in sorted form.
Hence, our resultant array is [ 102, 10, 31, 12, 15 ]

Input: arr[] = {14, 1101, 10, 35, 0}, K = 35
Output: 35 0 10 1101 14
Explanation:
Elements at their respective distance from K are,
At distance 0: 35
At distance 1: 10, 0 and in sorted form we have 0, 10.
At distance 2: 1101
At distance 3: 14
Hence, our resultant array is [ 35, 0, 10, 1101, 14 ]

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

Approach :

To solve the problem mentioned above we create an auxiliary vector to store elements at any distance from K. Then find the position of given integer K in the array arr[] and insert the element K at position 0 in the auxiliary vector. Traverse the array in left direction from K and insert those elements in the vector at their distance from K. Repeat the above process for the right side elements of K. Finally, print the array elements from distance 0 in sorted order.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to Sort the integers in ` `// array according to their distance from given ` `// element K present in the array ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get sorted array based on ` `// their distance from given integer K ` `void` `distanceSort(``int` `arr[], ``int` `K, ``int` `n) ` `{ ` `    ``// Vector to store respective elements ` `    ``// with their distance from integer K ` `    ``vector<``int``> vd[n]; ` ` `  `    ``// Find the position of integer K ` `    ``int` `pos; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] == K) { ` `            ``pos = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Insert the elements with their ` `    ``// distance from K in vector ` ` `  `    ``int` `i = pos - 1, j = pos + 1; ` ` `  `    ``// Element at distnce 0 ` `    ``vd.push_back(arr[pos]); ` ` `  `    ``// Elements at left side of K ` `    ``while` `(i >= 0) { ` `        ``vd[pos - i].push_back(arr[i]); ` `        ``--i; ` `    ``} ` ` `  `    ``// Elements at right side of K ` `    ``while` `(j < n) { ` `        ``vd[j - pos].push_back(arr[j]); ` `        ``++j; ` `    ``} ` ` `  `    ``// Print the vector content in sorted order ` `    ``for` `(``int` `i = 0; i <= max(pos, n - pos - 1); ++i) { ` ` `  `        ``// Sort elements at same distance ` `        ``sort(begin(vd[i]), end(vd[i])); ` ` `  `        ``// Print elements at distance i from K ` `        ``for` `(``auto` `element : vd[i]) ` `            ``cout << element << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {14, 1101, 10, 35, 0 }, K = 35; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``distanceSort(arr, K, n); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to Sort the integers in ` `# array according to their distance from given ` `# element K present in the array ` ` `  `# Function to get sorted array based on ` `# their distance from given integer K ` `def` `distanceSort(arr,K,n): ` `    ``# Vector to store respective elements ` `    ``# with their distance from integer K ` `    ``vd ``=` `[[] ``for` `i ``in` `range``(n)] ` ` `  `    ``# Find the position of integer K ` ` `  `    ``for` `i ``in` `range``(n): ` `        ``if` `(arr[i] ``=``=` `K): ` `            ``pos ``=` `i ` `            ``break` ` `  `    ``# Insert the elements with their ` `    ``# distance from K in vector ` ` `  `    ``i ``=` `pos ``-` `1` `    ``j ``=` `pos ``+` `1` ` `  `    ``# Element at distnce 0 ` `    ``vd[``0``].append(arr[pos]) ` ` `  `    ``# Elements at left side of K ` `    ``while` `(i >``=` `0``): ` `        ``vd[pos ``-` `i].append(arr[i]) ` `        ``i ``-``=` `1` ` `  `    ``# Elements at right side of K ` `    ``while` `(j < n): ` `        ``vd[j ``-` `pos].append(arr[j]) ` `        ``j ``+``=` `1` ` `  `    ``# Print the vector content in sorted order ` `    ``for` `i ``in` `range``(``max``(pos, n ``-` `pos ``-` `1``) ``+` `1``): ` ` `  `        ``# Sort elements at same distance ` `        ``vd[i].sort(reverse``=``False``) ` ` `  `        ``# Print elements at distance i from K ` `        ``for` `element ``in` `vd[i]: ` `            ``print``(element,end ``=` `" "``) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=`  `[``14``, ``1101``, ``10``, ``35``, ``0``] ` `    ``K ``=` `35` ` `  `    ``n ``=` `len``(arr) ` ` `  `    ``distanceSort(arr, K, n) ` ` `  `# This code is contributed by Surendra_Gangwar `

Output:

```35 0 10 1101 14
```

Time Complexity: O(N)

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Improved By : SURENDRA_GANGWAR