Given a sorted array arr[] of N distinct positive integers. The task is to generate an array such that the element at each index in the new array is the sum of absolute differences of the corresponding element with all other elements of the given array.
Input: arr[] = [2, 3, 5]
Output: [4, 3, 5]
Explanation:
distance(2) = |2 – 3| + |2 – 5| = 4
distance(3) = |3 – 2| + |3 – 5| = 3
distance(5) = |5 – 2| + |5 – 3| = 5
Therefore, we will return [4, 3, 5]Input: arr[] = [2, 3, 5, 6]
Output: [8, 6, 6, 8]
Explanation:
distance(2) = |2 – 3| + |2 – 5| + |2 – 6|= 8
distance(3) = |3 – 2| + |3 – 5| + |3 – 6|= 6
distance(5) = |5 – 2| + |5 – 3| + |5 – 6|= 6
distance(6) = |6 – 2| + |6 – 3| + |6 – 5|= 8
Therefore, we will return [8, 6, 6, 8]
Naive Approach: The idea is to generate all possible pairs for each element in the array arr[] and add the summation of the absolute difference of the pairs for each element in the new array. Print the array after the above steps for all the elements.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to return the new arrayvector<int> calculate(int* arr, int n){ // Initialize the Arraylist vector<int> ans; // Sum of absolute differences // of element with all elements for(int i = 0; i < n; i++) { // Initialize int sum to 0 int sum = 0; for(int j = 0; j < n; j++) { sum += abs(arr[i] - arr[j]); } // Add the value of sum to ans ans.push_back(sum); } // Return the final ans return ans;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 2, 3, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call vector<int> ans = calculate(arr, n); cout << "["; for(auto itr : ans) cout << itr << ", "; cout << "]"; return 0;}// This code is contributed by jrishabh99 |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to return the new array private static List<Integer> calculate(int[] arr) { // Length of the arraylist int n = arr.length; // Initialize the Arraylist List<Integer> ans = new ArrayList<Integer>(); // Sum of absolute differences // of element with all elements for (int i = 0; i < arr.length; i++) { // Initialize int sum to 0 int sum = 0; for (int j = 0; j < arr.length; j++) { sum += Math.abs(arr[i] - arr[j]); } // Add the value of sum to ans ans.add(sum); } // Return the final ans return ans; } // Driver Code public static void main(String[] args) { // Given array arr[] int[] arr = { 2, 3, 5, 6 }; // Function Call System.out.println(calculate(arr)); } } |
Python3
# Python3 program for the above approach # Function to return the new array # private static List<Integer> def calculate(arr): # Length of the arraylist n = len(arr) # Initialize the Arraylist ans = [] # Sum of absolute differences # of element with all elements for i in range(n): # Initialize sum to 0 sum = 0 for j in range(len(arr)): sum += abs(arr[i] - arr[j]) # Add the value of sum to ans ans.append(sum) # Return the final ans return ans # Driver Code if __name__ == '__main__': # Given array arr[] arr = [ 2, 3, 5, 6 ] # Function call print(calculate(arr)) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{// Function to return the new arrayprivate static List<int> calculate(int[] arr){ // Length of the arraylist int n = arr.Length; // Initialize the Arraylist List<int> ans = new List<int>(); // Sum of absolute differences // of element with all elements for(int i = 0; i < arr.Length; i++) { // Initialize int sum to 0 int sum = 0; for(int j = 0; j < arr.Length; j++) { sum += Math.Abs(arr[i] - arr[j]); } // Add the value of sum to ans ans.Add(sum); } // Return the final ans return ans;}// Driver Codepublic static void Main(string[] args){ // Given array arr[] int[] arr = { 2, 3, 5, 6 }; List<int> tmp = calculate(arr); Console.Write("["); for(int i = 0; i < tmp.Count; i++) { if(i != tmp.Count - 1) { Console.Write(tmp[i] + ", "); } else { Console.Write(tmp[i]); } } Console.Write("]");}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript program for// the above approach // Function to return the new array function calculate(arr) { // Length of the arraylist let n = arr.length; // Initialize the Arraylist let ans = []; // Sum of absolute differences // of element with all elements for (let i = 0; i < arr.length; i++) { // Initialize let sum to 0 let sum = 0; for (let j = 0; j < arr.length; j++) { sum += Math.abs(arr[i] - arr[j]); } // Add the value of sum to ans ans.push(sum); } // Return the final ans return ans; } // Driver Code // Given array arr[] let arr = [ 2, 3, 5, 6 ]; // Function Call document.write(calculate(arr));</script> |
Output:
[8, 6, 6, 8]
Time Complexity: O(N^2)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach the idea is to keep track of the accumulated subtraction of the values to the left and of the sum of values to the right. The sum of the difference of all the pairs for each element is given by:
num_of_elements_to_the_left * current_value -num_of_elements_to_the_right * current_value
- Find the sum of all the elements in the given array(say sum).
- Initialize sub as 0.
- Traverse the given array and for each element do the following:
- Subtract the current value arr[i] from the sum.
- Add the difference of all the pairs for each element using the formula to the resultant array:
sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + sum
- Subtract the current value arr[i] from the sum.
- Print the resultant array after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the // above approach#include <bits/stdc++.h>using namespace std;// Function to return list of// total Distance of each element// from other elements in arrayvector<int> calculate(int arr[], int n){ int sub = 0; int sum = 0; // Initialize the arraylist vector<int> ans; // Keep track of the // accumulated of the // sum of values to right for (int i = n - 1; i >= 0; i--) sum += arr[i]; // Keep track of the // accumulated subtraction // of the values to the left for (int i = 0; i < n; i++) { sum -= arr[i]; // Add the value to the // resultant array ans[] ans.push_back(sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + sum); sub -= arr[i]; } // Return the final // answer return ans;}// Driver Codeint main(){ // Initialize the array int arr[] = {2, 3, 5, 6}; int n = sizeof(arr) / sizeof(arr[0]); vector<int> ans = (calculate(arr, n)); for (int i = 0; i < n; i++) cout << ans[i] << " ";}// This code is contributed by Chitranayal |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to return list of // total Distance of each element // from other elements in array private static List<Integer> calculate(int[] arr) { // Length of the array int n = arr.length; int sub = 0; int sum = 0; // Initialize the arraylist List<Integer> ans = new ArrayList<Integer>(); // Keep track of the accumulated // of the sum of values to right for (int i = n - 1; i >= 0; i--) sum += arr[i]; // Keep track of the accumulated // subtraction of the values to the left for (int i = 0; i < arr.length; i++) { sum -= arr[i]; // Add the value to the resultant // array ans[] ans.add(sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + sum); sub -= arr[i]; } // return the final answer return ans; } // Driver Code public static void main(String[] args) { // Initialize the array int[] arr = { 2, 3, 5, 6 }; // Function Call System.out.println(calculate(arr)); } } |
Python3
# Python3 program for the above approach # Function to return list of # total Distance of each element # from other elements in array def calculate (arr): # Length of the array n = len(arr) sub = 0 Sum = 0 # Initialize the ArrayList ans = [] # Keep track of the accumulated # of the sum of values to right for i in range(n - 1, -1, -1): Sum += arr[i] # Keep track of the accumulated # subtraction of values to left for i in range(len(arr)): Sum -= arr[i] # Add the value to the resultant # array ans[] ans.append(sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + Sum) sub -= arr[i] # Return the final answer return ans# Driver Codeif __name__ == '__main__': # Initialize the array arr = [ 2, 3, 5, 6 ] # Function call print(calculate(arr))# This code is contributed by himanshu77 |
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG{ // Function to return list of // total Distance of each element // from other elements in array private static List<int> calculate(int[] arr) { // Length of the array int n = arr.Length; int sub = 0; int sum = 0; // Initialize the arraylist List<int> ans = new List<int>(); // Keep track of the accumulated // of the sum of values to right for(int i = n - 1; i >= 0; i--) sum += arr[i]; // Keep track of the accumulated // subtraction of the values to the left for(int i = 0; i < arr.Length; i++) { sum -= arr[i]; // Add the value to the resultant // array ans[] ans.Add(sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + sum); sub -= arr[i]; } // return the readonly answer return ans; } // Driver Code public static void Main(String[] args) { // Initialize the array int[] arr = { 2, 3, 5, 6 }; // Function Call Console.Write("[ "); foreach(int i in calculate(arr)) Console.Write(i + ", "); Console.Write("]"); } } // This code is contributed by amal kumar choubey |
Javascript
<script>// Javascript program for the // above approach// Function to return list of// total Distance of each element// from other elements in arrayfunction calculate(arr, n){ var sub = 0; var sum = 0; // Initialize the arraylist var ans = []; // Keep track of the // accumulated of the // sum of values to right for (var i = n - 1; i >= 0; i--) sum += arr[i]; // Keep track of the // accumulated subtraction // of the values to the left for (var i = 0; i < n; i++) { sum -= arr[i]; // Add the value to the // resultant array ans[] ans.push(sub + (i * arr[i]) - ((n - i - 1) * arr[i]) + sum); sub -= arr[i]; } // Return the final // answer return ans;}// Driver Code// Initialize the arrayvar arr = [2, 3, 5, 6]var n = arr.length;var ans = (calculate(arr, n));for (var i = 0; i < n; i++) document.write( ans[i] + " ");// This code is contributed by famously.</script> |
Output:
[8, 6, 6, 8]
Time Complexity: O(N)
Space Complexity: O(N)
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