Given an unsorted array of length N. The task is to sort the array, such that abs(a[i]-a[i+1]) < = abs(a[i+1]-a[i+2]) for all 0 < = i< N that is abs(a[0]-a[1]) < = abs(a[1]-a[2]) < = abs(a[2]-a[3]) and so on.
Examples:
Input: arr[] = {7, 4, 9, 9, -1, 9}
Output: {9, 7, 9, 4, 9, -1}
Explanation:
For first two elements the difference is abs(9-7)=2
For next two elements the difference is abs(7-9)=2
For next two elements the difference is abs(9-4)=5
For next two elements the difference is abs(7-4)=3
For next two elements the difference is abs(4-(-1))=5
Hence, difference array is 0, 0, 2, 3, 5.Input: arr[] = {1, 4, 6, 7}
Output: {6, 4, 7, 1}
Explanation:
For first two elements the difference is abs(6-4)=2
For next two elements the difference is abs(4-7)=3
For next two elements the difference is abs(7-1)=6
Hence, difference array is 2, 3, 6.
Approach:
To solve the problem mentioned above we sort the given unsorted array in ascending order. Then run a loop from i = 1 to i < n/2 and push elements alternatively in a stack from first half and second half respectively, that is push a[i] once and a[n-i-1] once until the whole array elements get pushed into the stack.
The main observation of the problem is to check if the length of the given array is odd, then push element at index n/2 additionally in order to have all elements of the array in the stack. Then traverse the whole stack until the stack is not empty, and pop elements out of the stack and print them as a result.
Below is the implementation of the discussed approach:
C++
// C++ implementation to Sort a given// unsorted array of length n// according to the given condition#include <bits/stdc++.h>using namespace std;// Functionvoid solve(int a[], int n){ // sort the array in ascending order sort(a, a + n); // declare a stack data structure stack<int> st; // run a loop from i=0 to i<n/2 and // push elements alternatively to stack for (int i = 0; i < n / 2; i++) { // push elements from // first half of array st.push(a[i]); // push elements from // second half of the array st.push(a[n - i - 1]); } // check if array has odd length, // then push a[n/2] to the stack if (n % 2 == 1) st.push(a[n / 2]); // loop until stack is not empty) while (!st.empty()) { int x = st.top(); printf("%d ", x); // pop the topmost // element from the stack st.pop(); }}// Driver codeint main(){ // declaration of unsorted array int a[] = { 7, 4, 9, 9, -1, 9 }; // given size of unsorted array int n = sizeof(a) / sizeof(a[0]); solve(a, n); return 0;} |
Java
// Java implementation to Sort a given// unsorted array of length n// according to the given conditionimport java.util.*;class GFG{// Functionstatic void solve(int a[], int n){ // sort the array in ascending order Arrays.sort(a); // declare a stack data structure Stack<Integer> st = new Stack<Integer>(); // run a loop from i=0 to i<n/2 and // push elements alternatively to stack for (int i = 0; i < n / 2; i++) { // push elements from // first half of array st.add(a[i]); // push elements from // second half of the array st.add(a[n - i - 1]); } // check if array has odd length, // then push a[n/2] to the stack if (n % 2 == 1) st.add(a[n / 2]); // loop until stack is not empty) while (!st.isEmpty()) { int x = st.peek(); System.out.printf("%d ", x); // pop the topmost // element from the stack st.pop(); }}// Driver codepublic static void main(String[] args){ // declaration of unsorted array int a[] = { 7, 4, 9, 9, -1, 9 }; // given size of unsorted array int n = a.length; solve(a, n);}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 implementation to sort a # given unsorted array of length n# according to the given condition# Functiondef solve(a, n): # Sort the array in ascending # order a.sort() # Declare a list used as a # stack data structure st = [] # Run a loop from i=0 to i<n/2 and # push elements alternatively to stack for i in range(n // 2): # Push elements from # first half of array st.append(a[i]) # Push elements from # second half of the array st.append(a[n - i - 1]) # Check if array has odd length, # then push a[n/2] to the stack if(n % 2 == 1): st.append(a[n // 2]) # Loop until stack is not empty while(st != []): x = st[-1] print(x, end = " ") # Pop the topmost element # from the stack st.pop()# Driver codeif __name__ == '__main__': # Declaration of unsorted array a = [ 7, 4, 9, 9, -1, 9 ] # Given size of unsorted array n = len(a) solve(a, n) # This code is contributed by Shivam Singh |
C#
// C# implementation to Sort a given// unsorted array of length n// according to the given conditionusing System;using System.Collections;class GFG{// Functionstatic void solve(int[] a, int n){ // Sort the array in ascending order Array.Sort(a); // Declare a stack data structure Stack st = new Stack(); // Run a loop from i=0 to i<n/2 and // push elements alternatively to stack for(int i = 0; i < n / 2; i++) { // Push elements from // first half of array st.Push(a[i]); // Push elements from // second half of the array st.Push(a[n - i - 1]); } // Check if array has odd length, // then push a[n/2] to the stack if (n % 2 == 1) st.Push(a[n / 2]); // Loop until stack is not empty) while (st.Count != 0) { Console.Write(st.Peek() + " "); // Pop the topmost // element from the stack st.Pop(); }}// Driver codepublic static void Main(String[] args){ // Declaration of unsorted array int[] a = { 7, 4, 9, 9, -1, 9 }; // Given size of unsorted array int n = a.Length; solve(a, n);}}// This code is contributed by jrishabh99 |
Javascript
<script> // Javascript implementation to Sort a given // unsorted array of length n // according to the given condition // Function function solve(a, n) { // Sort the array in ascending order a.sort(function(a, b){return a - b}); // Declare a stack data structure let st = []; // Run a loop from i=0 to i<n/2 and // push elements alternatively to stack for(let i = 0; i < parseInt(n / 2, 10); i++) { // Push elements from // first half of array st.push(a[i]); // Push elements from // second half of the array st.push(a[n - i - 1]); } // Check if array has odd length, // then push a[n/2] to the stack if (n % 2 == 1) st.push(a[parseInt(n / 2, 10)]); // Loop until stack is not empty) while (st.length != 0) { document.write(st[st.length - 1] + " "); // Pop the topmost // element from the stack st.pop(); } } // Declaration of unsorted array let a = [ 7, 4, 9, 9, -1, 9 ]; // Given size of unsorted array let n = a.length; solve(a, n);// This code is contributed by vaibhavrabadiya117.</script> |
9 7 9 4 9 -1
Time complexity: O(n*logn) as we are sorting the array
Auxiliary space: O(n) as we are using extra space for stack st
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