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 is at the 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 the 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 <bits/stdc++.h>using namespace std;// Function to get sorted array based on// their distance from given integer Kvoid 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 distance 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 codeint main(){ int arr[] = {14, 1101, 10, 35, 0 }, K = 35; int n = sizeof(arr) / sizeof(arr[0]); distanceSort(arr, K, n); return 0;} |
Java
// Java implementation to Sort the integers in// array according to their distance from given// element K present in the arrayimport java.util.*;class GFG{ // Function to get sorted array based on// their distance from given integer K@SuppressWarnings("unchecked")static void distanceSort(int arr[], int K, int n){ // Vector to store respective elements // with their distance from integer K Vector vd[] = new Vector[n]; for(int i = 0; i < n; i++) { vd[i] = new Vector(); } // Find the position of integer K int pos = 0; 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 distance 0 vd[0].add(arr[pos]); // Elements at left side of K while (i >= 0) { vd[pos - i].add(arr[i]); --i; } // Elements at right side of K while (j < n) { vd[j - pos].add(arr[j]); ++j; } // Print the vector content in sorted order for(i = 0; i <= Math.max(pos, n - pos - 1); ++i) { // Sort elements at same distance Collections.sort(vd[i]); // Print elements at distance i from K for (j = 0; j < vd[i].size(); j++) { int element = (int)vd[i].get(j); System.out.print(element + " "); } }} // Driver Codepublic static void main(String s[]){ int arr[] = {14, 1101, 10, 35, 0 }; int K = 35; int n = arr.length; distanceSort(arr, K, n);} }// This code is contributed by rutvik_56 |
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 Kdef 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 distance 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 codeif __name__ == '__main__': arr = [14, 1101, 10, 35, 0] K = 35 n = len(arr) distanceSort(arr, K, n)# This code is contributed by Surendra_Gangwar |
C#
// C# implementation to Sort the integers in// array according to their distance from given// element K present in the arrayusing System;using System.Collections.Generic;class GFG{ // Function to get sorted array based on// their distance from given integer Kstatic void distanceSort(int []arr, int K, int n){ // List to store respective elements // with their distance from integer K List<int> []vd = new List<int>[n]; int i ; for(i = 0; i < n; i++) { vd[i] = new List<int>(); } // Find the position of integer K int pos = 0; for(i = 0; i < n; i++) { if (arr[i] == K) { pos = i; break; } } // Insert the elements with their // distance from K in vector int j = pos + 1; i = pos - 1; // Element at distance 0 vd[0].Add(arr[pos]); // Elements at left side of K while (i >= 0) { vd[pos - i].Add(arr[i]); --i; } // Elements at right side of K while (j < n) { vd[j - pos].Add(arr[j]); ++j; } // Print the vector content in sorted order for(i = 0; i <= Math.Max(pos, n - pos - 1); ++i) { // Sort elements at same distance vd[i].Sort(); // Print elements at distance i from K for (j = 0; j < vd[i].Count; j++) { int element = (int)vd[i][j]; Console.Write(element + " "); } }} // Driver Codepublic static void Main(String []args){ int []arr = {14, 1101, 10, 35, 0}; int K = 35; int n = arr.Length; distanceSort(arr, K, n);} }// This code is contributed by shikhasingrajput |
Javascript
<script>// JavaScript 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 Kfunction distanceSort(arr, K, n){ // Vector to store respective elements // with their distance from integer K let vd = new Array(n); for(let i = 0; i < n; i++) { vd[i] = new Array(); } // Find the position of integer K for(let i = 0; i < n; i++) { if (arr[i] == K) { pos = i break } } // Insert the elements with their // distance from K in vector let i = pos - 1 let j = pos + 1 // Element at distance 0 vd[0].push(arr[pos]) // Elements at left side of K while (i >= 0){ vd[pos - i].push(arr[i]) i -= 1 } // Elements at right side of K while (j < n){ vd[j - pos].push(arr[j]) j += 1 } // Print the vector content in sorted order for(let i = 0;i< Math.max(pos, n - pos - 1) + 1;i++){ // Sort elements at same distance vd[i].sort() // Print elements at distance i from K for(let element of vd[i]) document.write(element," ") }}// Driver codelet arr = [14, 1101, 10, 35, 0]let K = 35let n = arr.lengthdistanceSort(arr, K, n)// This code is contributed by Shinjanpatra</script> |
Output:
35 0 10 1101 14
Time Complexity: O(N)
Auxiliary Space: O(N)
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