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 ]

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++

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// C++ implementation to Sort the integers in
// array according to their distance from given
// element K present in the array
#include <bits/stdc++.h>
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[0].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[0]);
  
    distanceSort(arr, K, n);
  
    return 0;
}

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Python3

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# 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

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Output:

35 0 10 1101 14

Time Complexity: O(N)

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Improved By : SURENDRA_GANGWAR