Smallest positive integer K such that all array elements can be made equal by incrementing or decrementing by at most K
Given an array arr[] of size N, the task is to find the smallest positive integer K such that incrementing or decrementing each array element by K at most once makes all elements equal. If it is not possible to make all array elements equal, then print -1.
Examples :
Input: arr[] = { 5, 7, 9 }
Output: 2
Explanation:
Incrementing the value of arr[0] by K(= 2) modifies arr[] to { 7, 7, 9 }.
Decrementing the value of arr[2] by K(= 2) modifies arr[] to { 7, 7, 7 }Input: arr[] = {1, 3, 9}, N = 3
Output: -1
Approach: Follow the steps below to solve the problem :
- Initialize a Set to store all distinct elements present in the array.
- Count the distinct elements in the array, which is equal to the size of the Set, say M.
- If M > 3, then print -1.
- If M = 3, then check if the difference between the largest and the second largest element of the Set is equal to the difference between the second largest and the third largest element of the Set or not. If found to be true, then print the difference. Otherwise, print -1.
- If M = 2, then check if the difference between the largest and the second largest element of the set is even or not. If found to be true, then print the half of their difference. Otherwise, print the difference.
- If M <= 1, then print 0.
Below is the implementation of the above solution :
C++
// C++ program for the above approach #include <iostream> #include <set> using namespace std; // Function to find smallest integer K such that // incrementing or decrementing each element by // K at most once makes all elements equal void findMinKToMakeAllEqual( int N, int A[]) { // Store distinct // array elements set< int > B; // Traverse the array, A[] for ( int i = 0; i < N; i++) B.insert(A[i]); // Count elements into the set int M = B.size(); // Iterator to store first // element of B set< int >::iterator itr = B.begin(); // If M is greater than 3 if (M > 3) printf ( "-1" ); // If M is equal to 3 else if (M == 3) { // Stores the first // smallest element int B_1 = *itr; // Stores the second // smallest element int B_2 = *(++itr); // Stores the largest element int B_3 = *(++itr); // IF difference between B_2 and B_1 // is equal to B_3 and B_2 if (B_2 - B_1 == B_3 - B_2) printf ( "%d" , B_2 - B_1); else printf ( "-1" ); } // If M is equal to 2 else if (M == 2) { // Stores the smallest element int B_1 = *itr; // Stores the largest element int B_2 = *(++itr); // If difference is an even if ((B_2 - B_1) % 2 == 0) printf ( "%d" , (B_2 - B_1) / 2); else printf ( "%d" , B_2 - B_1); } // If M is equal to 1 else printf ( "%d" , 0); } // Driver Code int main() { // Given array int A[] = { 1, 3, 5, 1 }; // Given size int N = sizeof (A) / sizeof (A[0]); // Print the required smallest integer findMinKToMakeAllEqual(N, A); } |
Java
// Java program for the above approach import java.io.*; import java.util.*; class GFG { // Function to find smallest integer K such that // incrementing or decrementing each element by // K at most once makes all elements equal static void findMinKToMakeAllEqual( int N, int A[]) { // Store distinct // array elements Set<Integer> B = new HashSet<Integer>(); // Traverse the array, A[] for ( int i = 0 ; i < N; i++) { B.add(A[i]); } ArrayList<Integer> b = new ArrayList<Integer>(B); // Count elements into the set int M = b.size(); int i = 0 ; // If M is greater than 3 if (M > 3 ) { System.out.print( "-1" );} // If M is equal to 3 else if (M == 3 ) { // Stores the first // smallest element int B_1 = b.get(i++); // Stores the second // smallest element int B_2 = b.get(i++); // Stores the largest element int B_3 = b.get(i++); // IF difference between B_2 and B_1 // is equal to B_3 and B_2 if (B_2 - B_1 == B_3 - B_2) { System.out.print(B_2 - B_1); } else { System.out.print( "-1" ); } } // If M is equal to 2 else if (M == 2 ) { // Stores the smallest element int B_1 = b.get(i++); // Stores the largest element int B_2 = b.get(i++); // If difference is an even if ((B_2 - B_1) % 2 == 0 ) { System.out.print((B_2 - B_1) / 2 ); } else { System.out.print(B_2 - B_1); } } // If M is equal to 1 else { System.out.print( 0 ); } } // Driver code public static void main (String[] args) { // Given array int A[] = { 1 , 3 , 5 , 1 }; // Given size int N = A.length; // Print the required smallest integer findMinKToMakeAllEqual(N, A); } } // This code is contributed by rag2127 |
Python3
# Python3 program for the above approach # Function to find smallest integer K such # that incrementing or decrementing each # element by K at most once makes all # elements equal def findMinKToMakeAllEqual(N, A): # Store distinct # array elements B = {} # Traverse the array, A[] for i in range (N): B[A[i]] = 1 # Count elements into the set M = len (B) # Iterator to store first # element of B itr, i = list (B.keys()), 0 # If M is greater than 3 if (M > 3 ): print ( "-1" ) # If M is equal to 3 elif (M = = 3 ): # Stores the first # smallest element B_1, i = itr[i], i + 1 # Stores the second # smallest element B_2, i = itr[i], i + 1 # Stores the largest element B_3, i = itr[i], i + 1 # IF difference between B_2 and B_1 # is equal to B_3 and B_2 if (B_2 - B_1 = = B_3 - B_2): print (B_2 - B_1) else : print ( "-1" ) # If M is equal to 2 elif (M = = 2 ): # Stores the smallest element B_1, i = itr[i], i + 1 # Stores the largest element B_2, i = itr[i], i + 1 # If difference is an even if ((B_2 - B_1) % 2 = = 0 ): print ((B_2 - B_1) / / 2 ) else : print (B_2 - B_1) # If M is equal to 1 else : print ( 0 ) # Driver Code if __name__ = = '__main__' : # Given array A = [ 1 , 3 , 5 , 1 ] # Given size N = len (A) # Print the required smallest integer findMinKToMakeAllEqual(N, A) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG { // Function to find smallest integer K such that // incrementing or decrementing each element by // K at most once makes all elements equal static void findMinKToMakeAllEqual( int N, int [] A) { // Store distinct // array elements var B = new HashSet< int >(); // Traverse the array, A[] for ( int i = 0; i < N; i++) { B.Add(A[i]); } List< int > b = new List< int >(B); // Count elements into the set int M = b.Count; int j = 0; // If M is greater than 3 if (M > 3) { Console.Write( "-1" ); } // If M is equal to 3 else if (M == 3) { // Stores the first // smallest element int B_1 = b[j++]; // Stores the second // smallest element int B_2 = b[j++]; // Stores the largest element int B_3 = b[j++]; // IF difference between B_2 and B_1 // is equal to B_3 and B_2 if (B_2 - B_1 == B_3 - B_2) { Console.Write(B_2 - B_1); } else { Console.Write( "-1" ); } } // If M is equal to 2 else if (M == 2) { // Stores the smallest element int B_1 = b[j++]; // Stores the largest element int B_2 = b[j++]; // If difference is an even if ((B_2 - B_1) % 2 == 0) { Console.Write((B_2 - B_1) / 2); } else { Console.Write(B_2 - B_1); } } // If M is equal to 1 else { Console.Write(0); } } // Driver code public static void Main( string [] args) { // Given array int [] A = { 1, 3, 5, 1 }; // Given size int N = A.Length; // Print the required smallest integer findMinKToMakeAllEqual(N, A); } } // This code is contributed by chitranayal. |
Javascript
<script> // Javascript program for the above approach // Function to find smallest integer K such that // incrementing or decrementing each element by // K at most once makes all elements equal function findMinKToMakeAllEqual(N, A) { // Store distinct // array elements var B = new Set(); // Traverse the array, A[] for ( var i = 0; i < N; i++) B.add(A[i]); // Count elements into the set var M = B.size; // Iterator to store first // element of B var itr = [...B].sort((a, b) => a - b); // If M is greater than 3 if (M > 3) document.write( "-1" ); // If M is equal to 3 else if (M == 3) { // Stores the first // smallest element var B_1 = itr[0]; // Stores the second // smallest element var B_2 = itr[1]; // Stores the largest element var B_3 = itr[2]; // IF difference between B_2 and B_1 // is equal to B_3 and B_2 if (B_2 - B_1 == B_3 - B_2) document.write(B_2 - B_1); else document.write( "-1" ); } // If M is equal to 2 else if (M == 2) { // Stores the smallest element var B_1 = itr[0]; // Stores the largest element var B_2 =itr[1]; // If difference is an even if ((B_2 - B_1) % 2 == 0) document.write(parseInt((B_2 - B_1) / 2)); else document.write(B_2 - B_1); } // If M is equal to 1 else document.write(0); } // Driver Code // Given array var A = [ 1, 3, 5, 1 ]; // Given size var N = A.length; // Print the required smallest integer findMinKToMakeAllEqual(N, A); // This code is contributed by noob2000 </script> |
2
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)