Given an array arr[] of size N, the task is to find the minimum count of elements required to be inserted into the array such that the absolute difference of all possible pairs exists in the array.
Examples:
Input: arr[] = { 3, 5 }
Output: 3
Explanation:
Inserting 2 into the array modifies arr[] to { 2, 3, 5 }
Inserting 1 into the array modifies arr[] to { 1, 2, 3, 5 }
Inserting 4 into the array modifies arr[] to { 1, 2, 3, 4, 5 }
Since absolute difference of all possible pairs from the array are present in the array, the required output is 3.Input: arr[] = { 2, 4 }
Output: 0
Explanation:
Since absolute difference of all possible pairs array are already present in the array, the required output is 0.
Naive Approach: The simplest approach to solve this problem is to find the absolute difference of every possible pair from the array and check if the difference obtained is present in the array or not. If it is not present, then insert the obtained difference. Otherwise, print the count of elements inserted into the array.
Time Complexity:O(N * X), where X is the maximum element in the array.
Auxiliary Space: O(X)
Efficient Approach: The above approach can be optimized based on the following observations:
Since the absolute difference of all possible pairs of final array must be present in the array.
Therefore, the final array must be in the form of X, 2 * X, 3 * X, 4 * X, …From the above sequence of the final array, it can be observed the value of X must be the GCD of the given array.
Follow the steps below to solve the problem:
- Traverse the array and calculate the GCD of the given array say, X.
- Find the largest element of the array say, Max.
- Total count of elements in the array by inserting the absolute difference of all possible pairs is (Max) / X.
- Therefore, the total count of elements inserted into the array is equal to ((Max / X) – N).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the GCD of an arrayint findGcd(int arr[], int N){ // Stores GCD of an array int gcd = arr[0]; // Traverse the array for (int i = 1; i < N; i++) { // Update gcd gcd = __gcd(gcd, arr[i]); } return gcd;}// Function to find minimum count of elements// inserted into the array such that absolute// difference of pairs present in the arrayvoid findMax(int arr[], int N){ // Stores gcd of the array int gcd = findGcd(arr, N); // Stores the largest elemetn // in the array int Max = INT_MIN; // Traverse the array for (int i = 0; i < N; i++) { // Update Max Max = max(Max, arr[i]); } // Stores minimum count of elements inserted // into the array such that absolute difference // of pairs present in the array int ans = (Max / gcd) - N; cout << ans;}// Driver Codeint main(){ // Given array int arr[] = { 3, 5 }; // Size of the array int N = (sizeof(arr) / (sizeof(arr[0]))); // Function Call findMax(arr, N);} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Recursive function to return gcd of a and bstatic int gcdd(int a, int b){ // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return gcdd(a - b, b); return gcdd(a, b - a);} // Function to find the GCD of an arraystatic int findGcd(int arr[], int N){ // Stores GCD of an array int gcd = arr[0]; // Traverse the array for (int i = 1; i < N; i++) { // Update gcd gcd = gcdd(gcd, arr[i]); } return gcd;}// Function to find minimum count of elements// inserted into the array such that absolute// difference of pairs present in the arraystatic void findMax(int arr[], int N){ // Stores gcd of the array int gcd = findGcd(arr, N); // Stores the largest elemetn // in the array int Max = Integer.MIN_VALUE; // Traverse the array for (int i = 0; i < N; i++) { // Update Max Max = Math.max(Max, arr[i]); } // Stores minimum count of elements inserted // into the array such that absolute difference // of pairs present in the array int ans = (Max / gcd) - N; System.out.println(ans);} // Driver codepublic static void main(String[] args){ // Given array int arr[] = { 3, 5 }; // Size of the array int N = arr.length; // Function Call findMax(arr, N);}}// This code is contributed by susmitakundugoaldanga. |
Python3
# Python3 program for the above approachimport mathimport sys# Function to find the GCD of an arraydef findGcd(arr, N): # Stores GCD of an array gcd = arr[0] # Traverse the array for i in range(1, N): # Update gcd gcd = math.gcd(gcd, arr[i]) return gcd# Function to find minimum count of elements# inserted into the array such that absolute# difference of pairs present in the arraydef findMax(arr, N): # Stores gcd of the array gcd = findGcd(arr, N) # Stores the largest elemetn # in the array Max = -sys.maxsize - 1 # Traverse the array for i in range(N): # Update Max Max = max(Max, arr[i]) # Stores minimum count of elements inserted # into the array such that absolute difference # of pairs present in the array ans = (Max // gcd) - N print(ans)# Driver Codeif __name__ == "__main__": # Given array arr = [3, 5] # Size of the array N = len(arr) # Function Call findMax(arr, N)# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;class GFG{ // Recursive function to return gcd of a and b static int gcdd(int a, int b) { // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // base case if (a == b) return a; // a is greater if (a > b) return gcdd(a - b, b); return gcdd(a, b - a); } // Function to find the GCD of an array static int findGcd(int[] arr, int N) { // Stores GCD of an array int gcd = arr[0]; // Traverse the array for (int i = 1; i < N; i++) { // Update gcd gcd = gcdd(gcd, arr[i]); } return gcd; } // Function to find minimum count of elements // inserted into the array such that absolute // difference of pairs present in the array static void findMax(int[] arr, int N) { // Stores gcd of the array int gcd = findGcd(arr, N); // Stores the largest elemetn // in the array int Max = Int32.MinValue; // Traverse the array for (int i = 0; i < N; i++) { // Update Max Max = Math.Max(Max, arr[i]); } // Stores minimum count of elements inserted // into the array such that absolute difference // of pairs present in the array int ans = (Max / gcd) - N; Console.WriteLine(ans); } // Driver code static public void Main() { // Given array int[] arr = new int[] { 3, 5 }; // Size of the array int N = arr.Length; // Function Call findMax(arr, N); }}// This code is contributed by Dharanendra L V. |
Javascript
<script>// Javascript program for the above approach// Recursive function to return gcd of a and bfunction gcdd(a, b){ // Everything divides 0 if (a == 0) return b; if (b == 0) return a; // Base case if (a == b) return a; // a is greater if (a > b) return gcdd(a - b, b); return gcdd(a, b - a);}// Function to find the GCD of an arrayfunction findGcd(arr, N){ // Stores GCD of an array var gcd = arr[0]; // Traverse the array for(i = 1; i < N; i++) { // Update gcd gcd = gcdd(gcd, arr[i]); } return gcd;}// Function to find minimum count of elements// inserted into the array such that absolute// difference of pairs present in the arrayfunction findMax(arr, N){ // Stores gcd of the array var gcd = findGcd(arr, N); // Stores the largest elemetn // in the array var Max = Number.MIN_VALUE; // Traverse the array for(i = 0; i < N; i++) { // Update Max Max = Math.max(Max, arr[i]); } // Stores minimum count of elements inserted // into the array such that absolute difference // of pairs present in the array var ans = (Max / gcd) - N; document.write(ans);}// Driver code// Given arrayvar arr = [ 3, 5 ];// Size of the arrayvar N = arr.length;// Function CallfindMax(arr, N);// This code is contributed by umadevi9616</script> |
3
Time Complexity:O(N * Log(X)), where X is the largest element in the array.
Auxiliary Space: O(1)
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