Given an array of both positive and negative integers ‘arr[]’ which are sorted. The task is to sort the square of the numbers of the Array.
Examples:
Input : arr[] = {-6, -3, -1, 2, 4, 5}
Output : 1, 4, 9, 16, 25, 36
Input : arr[] = {-5, -4, -2, 0, 1}
Output : 0, 1, 4, 16, 25
Simple solution is to first convert each array element into its square and then apply any “O(nlogn)” sorting algorithm to sort the array elements.
Below is the implementation of the above idea
C++
// C++ program to Sort square of the numbers// of the array#include <bits/stdc++.h>using namespace std;// Function to sort an square arrayvoid sortSquares(int arr[], int n){ // First convert each array elements // into its square for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort an array using "sort STL function " sort(arr, arr + n);}// Driver program to test above functionint main(){ int arr[] = { -6, -3, -1, 2, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Before sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; sortSquares(arr, n); cout << "\nAfter Sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0;} |
Java
// Java program to Sort square of the numbers// of the arrayimport java.util.*;import java.io.*;class GFG { // Function to sort an square array public static void sortSquares(int arr[]) { int n = arr.length; // First convert each array elements // into its square for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort an array using "inbuilt sort function" // in Arrays class. Arrays.sort(arr); } // Driver program to test above function public static void main(String[] args) { int arr[] = { -6, -3, -1, 2, 4, 5 }; int n = arr.length; System.out.println("Before sort "); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); sortSquares(arr); System.out.println(""); System.out.println("After Sort "); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); }} |
Python3
# Python program to Sort square# of the numbers of the array# Function to sort an square arraydef sortSquare(arr, n): # First convert each array # elements into its square for i in range(n): arr[i]= arr[i] * arr[i] arr.sort()# Driver codearr = [-6, -3, -1, 2, 4, 5]n = len(arr)print("Before sort")for i in range(n): print(arr[i], end = " ")print("\n")sortSquare(arr, n)print("After sort")for i in range(n): print(arr[i], end = " ")# This code is contributed by# Shrikant13 |
C#
// C# program to Sort square// of the numbers of the arrayusing System;class GFG { // Function to sort // an square array public static void sortSquares(int[] arr) { int n = arr.Length; // First convert each array // elements into its square for (int i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort an array using // "inbuilt sort function" // in Arrays class. Array.Sort(arr); } // Driver Code public static void Main() { int[] arr = { -6, -3, -1, 2, 4, 5 }; int n = arr.Length; Console.WriteLine("Before sort "); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); sortSquares(arr); Console.WriteLine(""); Console.WriteLine("After Sort "); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); }}// This code is contributed by anuj_67. |
Javascript
<script>// JavaScript program for the above approach // Function to sort an square array function sortSquares(arr) { let n = arr.length; // First convert each array elements // into its square for (let i = 0; i < n; i++) arr[i] = arr[i] * arr[i]; // Sort an array using "inbuilt sort function" // in Arrays class. arr.sort(); }// Driver Code let arr = [ -6, -3, -1, 2, 4, 5 ]; let n = arr.length; document.write("Before sort " + "<br/>"); for (let i = 0; i < n; i++) document.write(arr[i] + " "); sortSquares(arr); document.write("" + "<br/>"); document.write("After Sort " + "<br/>"); for (let i = 0; i < n; i++) document.write(arr[i] + " ");// This code is contributed by chinmoy1997pal.</script> |
Output
Before sort -6 -3 -1 2 4 5 After Sort 1 4 9 16 25 36
Time complexity: O(n log n)
Space Complexity : O(n)
Efficient solution is based on the fact that the given array is already sorted. We do the following two steps.
- Divide the array into two-part “Negative and positive “.
- Use merge function to merge two sorted arrays into a single sorted array.
Below is the implementation of the above idea.
C++
// C++ program to Sort square of the numbers of the array#include <bits/stdc++.h>using namespace std;// function to sort array after doing squares of elementsvoid sortSquares(int arr[], int n){ // first divide array into negative and positive part int K = 0; for (K = 0; K < n; K++) if (arr[K] >= 0) break; // Now do the same process that we learnt // in merge sort to merge two sorted array // here both two half are sorted and we traverse // first half in reverse manner because // first half contains negative elements int i = K - 1; // Initial index of first half int j = K; // Initial index of second half int ind = 0; // Initial index of temp array // store sorted array int temp[n]; while (i >= 0 && j < n) { if (arr[i] * arr[i] < arr[j] * arr[j]) { temp[ind] = arr[i] * arr[i]; i--; } else { temp[ind] = arr[j] * arr[j]; j++; } ind++; } /* Copy the remaining elements of first half */ while (i >= 0) { temp[ind] = arr[i] * arr[i]; i--; ind++; } /* Copy the remaining elements of second half */ while (j < n) { temp[ind] = arr[j] * arr[j]; j++; ind++; } // copy 'temp' array into original array for (int i = 0; i < n; i++) arr[i] = temp[i];}// Driver program to test above functionint main(){ int arr[] = { -6, -3, -1, 2, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Before sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; sortSquares(arr, n); cout << "\nAfter Sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0;} |
Java
// Java program to Sort square of the numbers// of the arrayimport java.util.*;import java.io.*;class GFG { // Function to sort an square array public static void sortSquares(int arr[]) { int n = arr.length; // first divide the array into negative and positive part int k; for (k = 0; k < n; k++) { if (arr[k] >= 0) break; } // Now do the same process that we learnt // in merge sort to merge two sorted arrays // here both two halves are sorted and we traverse // first half in reverse manner because // first half contains negative elements int i = k - 1; // Initial index of first half int j = k; // Initial index of second half int ind = 0; // Initial index of temp array int[] temp = new int[n]; while (i >= 0 && j < n) { if (arr[i] * arr[i] < arr[j] * arr[j]) { temp[ind] = arr[i] * arr[i]; i--; } else { temp[ind] = arr[j] * arr[j]; j++; } ind++; } while (i >= 0) { temp[ind++] = arr[i] * arr[i]; i--; } while (j < n) { temp[ind++] = arr[j] * arr[j]; j++; } // copy 'temp' array into original array for (int x = 0; x < n; x++) arr[x] = temp[x]; } // Driver program to test above function public static void main(String[] args) { int arr[] = { -6, -3, -1, 2, 4, 5 }; int n = arr.length; System.out.println("Before sort "); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); sortSquares(arr); System.out.println(""); System.out.println("After Sort "); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); }} |
Python3
# Python3 program to Sort square of the numbers of the array# function to sort array after doing squares of elementsdef sortSquares(arr, n): # first divide array into negative and positive part K = 0 for K in range(n): if (arr[K] >= 0 ): break # Now do the same process that we learnt # in merge sort to merge to two sorted array # here both two halves are sorted and we traverse # first half in reverse manner because # first half contains negative elements i = K - 1 # Initial index of first half j = K # Initial index of second half ind = 0 # Initial index of temp array # store sorted array temp = [0]*n while (i >= 0 and j < n): if (arr[i] * arr[i] < arr[j] * arr[j]): temp[ind] = arr[i] * arr[i] i -= 1 else: temp[ind] = arr[j] * arr[j] j += 1 ind += 1 ''' Copy the remaining elements of first half ''' while (i >= 0): temp[ind] = arr[i] * arr[i] i -= 1 ind += 1 ''' Copy the remaining elements of second half ''' while (j < n): temp[ind] = arr[j] * arr[j] j += 1 ind += 1 # copy 'temp' array into original array for i in range(n): arr[i] = temp[i]# Driver codearr = [-6, -3, -1, 2, 4, 5 ]n = len(arr)print("Before sort ")for i in range(n): print(arr[i], end =" " ) sortSquares(arr, n)print("\nAfter Sort ")for i in range(n): print(arr[i], end =" " ) # This code is contributed by shubhamsingh10 |
C#
// C# program to Sort square of the numbers// of the arrayusing System;class GFG { // Function to sort an square array public static void sortSquares(int[] arr) { int n = arr.Length; // first divide array into negative and positive part int k; for (k = 0; k < n; k++) { if (arr[k] >= 0) break; } // Now do the same process that we learnt // in merge sort to merge to two sorted array // here both two halves are sorted and we traverse // first half in reverse manner because // first half contains negative elements int i = k - 1; // Initial index of first half int j = k; // Initial index of second half int ind = 0; // Initial index of temp array int[] temp = new int[n]; while (i >= 0 && j < n) { if (arr[i] * arr[i] < arr[j] * arr[j]) { temp[ind] = arr[i] * arr[i]; i--; } else { temp[ind] = arr[j] * arr[j]; j++; } ind++; } while (i >= 0) { temp[ind++] = arr[i] * arr[i]; i--; } while (j < n) { temp[ind++] = arr[j] * arr[j]; j++; } // copy 'temp' array into original array for (int x = 0; x < n; x++) arr[x] = temp[x]; } // Driver code public static void Main(String[] args) { int[] arr = { -6, -3, -1, 2, 4, 5 }; int n = arr.Length; Console.WriteLine("Before sort "); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); sortSquares(arr); Console.WriteLine(""); Console.WriteLine("After Sort "); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to Sort // square of the numbers// of the array // Function to sort an square array function sortSquares(arr) { let n = arr.length; // first dived array into part // negative and positive let k; for (k = 0; k < n; k++) { if (arr[k] >= 0) break; } // Now do the same process that we learn // in merge sort to merge to two sorted array // here both two half are sorted and we traverse // first half in reverse meaner because // first half contain negative element let i = k - 1; // Initial index of first half let j = k; // Initial index of second half let ind = 0; // Initial index of temp array let temp = new Array(n); while (i >= 0 && j < n) { if (arr[i] * arr[i] < arr[j] * arr[j]) { temp[ind] = arr[i] * arr[i]; i--; } else { temp[ind] = arr[j] * arr[j]; j++; } ind++; } while (i >= 0) { temp[ind++] = arr[i] * arr[i]; i--; } while (j < n) { temp[ind++] = arr[j] * arr[j]; j++; } // copy 'temp' array into original array for (let x = 0; x < n; x++) arr[x] = temp[x]; } // Driver program to test above function let arr=[ -6, -3, -1, 2, 4, 5 ]; let n = arr.length; document.write("Before sort <br>"); for (let i = 0; i < n; i++) document.write(arr[i] + " "); sortSquares(arr); document.write("<br>"); document.write("After Sort <br>"); for (let i = 0; i < n; i++) document.write(arr[i] + " "); // This code is contributed by rag2127</script> |
Output
Before sort -6 -3 -1 2 4 5 After Sort 1 4 9 16 25 36
Time complexity: O(n)
space complexity: O(n)
Method 3:
Another efficient solution is based on the two-pointer method as the array is already sorted we can compare the first and last element to check which is bigger and proceed with the result.
Algorithm:
- Initialize left=0 and right=n-1
- if abs(left) >= abs(right) then store square(arr[left])
at the end of result array and increment left pointer - else store square(arr[right]) in the result array and decrement right pointer
- decrement index of result array
Implementation:
C++
// CPP code for the above approach#include <bits/stdc++.h>using namespace std;// Function to sort an square arrayvoid sortSquares(vector<int>& arr, int n){ int left = 0, right = n - 1; int result[n]; // Iterate from n - 1 to 0 for (int index = n - 1; index >= 0; index--) { // Check if abs(arr[left]) is greater // than arr[right] if (abs(arr[left]) > arr[right]) { result[index] = arr[left] * arr[left]; left++; } else { result[index] = arr[right] * arr[right]; right--; } } for (int i = 0; i < n; i++) arr[i] = result[i];}// Driver Codeint main(){ vector<int> arr; arr.push_back(-6); arr.push_back(-3); arr.push_back(-1); arr.push_back(2); arr.push_back(4); arr.push_back(5); int n = 6; cout << "Before sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; sortSquares(arr, n); cout << endl; cout << "After Sort " << endl; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0;}// this code is contributed by Manu Pathria |
Java
// Java program to Sort square of // the numbers of the arrayimport java.util.*;import java.io.*;class GFG{ // Function to sort an square arraypublic static void sortSquares(int arr[]){ int n = arr.length, left = 0, right = n - 1; int result[] = new int[n]; for(int index = n - 1; index >= 0; index--) { if (Math.abs(arr[left]) > arr[right]) { result[index] = arr[left] * arr[left]; left++; } else { result[index] = arr[right] * arr[right]; right--; } } for(int i = 0; i < n; i++) arr[i] = result[i];}// Driver codepublic static void main(String[] args){ int arr[] = { -6, -3, -1, 2, 4, 5 }; int n = arr.length; System.out.println("Before sort "); for(int i = 0; i < n; i++) System.out.print(arr[i] + " "); sortSquares(arr); System.out.println(""); System.out.println("After Sort "); for(int i = 0; i < n; i++) System.out.print(arr[i] + " ");}}// This code is contributed by jinalparmar2382 |
Python3
# Python3 program to Sort square of the numbers of the array# function to sort array after doing squares of elementsdef sortSquares(arr, n): left, right = 0, n - 1 index = n - 1 result = [0 for x in arr] while index >= 0: if abs(arr[left]) >= abs(arr[right]): result[index] = arr[left] * arr[left] left += 1 else: result[index] = arr[right] * arr[right] right -= 1 index -= 1 for i in range(n): arr[i] = result[i]# Driver code arr = [-6, -3, -1, 2, 4, 5 ] n = len(arr) print("Before sort ") for i in range(n): print(arr[i], end =" " ) sortSquares(arr, n) print("\nAfter Sort ") for i in range(n): print(arr[i], end =" " ) |
C#
// C# program to Sort square of // the numbers of the arrayusing System;class GFG{ // Function to sort an square arraypublic static void sortSquares(int [] arr){ int n = arr.Length, left = 0, right = n - 1; int []result = new int[n]; for(int index = n - 1; index >= 0; index--) { if (Math.Abs(arr[left]) > arr[right]) { result[index] = arr[left] * arr[left]; left++; } else { result[index] = arr[right] * arr[right]; right--; } } for(int i = 0; i < n; i++) arr[i] = result[i];}// Driver codepublic static void Main(string[] args){ int []arr = {-6, -3, -1, 2, 4, 5}; int n = arr.Length; Console.WriteLine("Before sort "); for(int i = 0; i < n; i++) Console.Write(arr[i] + " "); sortSquares(arr); Console.WriteLine(""); Console.WriteLine("After Sort "); for(int i = 0; i < n; i++) Console.Write(arr[i] + " ");}}// This code is contributed by Chitranayal |
Javascript
<script> // This function squares each value in an array // and arranges the result in ascending order, using two pointers. // The first pointer points to the first element of the array, and the second // points to the last. On each iteration, each pointer is compared to the other // and the pointer which has the lowest value is shifted closer to the other. function sortSquares(arr) { let n = nums.length, left = 0, right = n -1, result = new Array(n) ; for (let i = n - 1; i >= 0; i--) { // Here, Math.abs() is used to convert any negative numbers to their // integer equivalent... i.e. -4 becomes 4. if (Math.abs(nums[left]) > Math.abs(nums[right])) { result[i] = nums[left] ** 2; left++; } else { result[i] = nums[right] **2; right--; } } return result; } let arr = [-6, -3, -1, 2, 4, 5]; let n = arr.length; document.write("Before sort " + "</br>"); for(let i = 0; i < n; i++) document.write(arr[i] + " "); sortSquares(arr); document.write("</br>"); document.write("After Sort " + "</br>"); for(let i = 0; i < n; i++) document.write(arr[i] + " "); // This code was contributed by rameshtravel07, then was edited by: Lewis Farnworth</script> |
Output
Before sort -6 -3 -1 2 4 5 After Sort 1 4 9 16 25 36
Time complexity: O(n)
Auxiliary Space: O(n)
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