Maximize distinct elements by incrementing/decrementing an element or keeping it same
Given an array arr[] of N elements, the task is to maximize the count of distinct elements in the array, by either of the given operation on each element of the array:
- either increasing the element by 1
- or decreasing the element by 1
- or keeping the element as it is.
Note: No element can be less than or equal to 0.
Examples:
Input: arr = [4, 4, 5, 5, 5, 5, 6, 6]
Output: 5
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [3, 4, 5, 5, 5, 5, 6, 7]. Here distinct elements are 5.Input: arr = [1, 1, 1, 8, 8, 8, 9, 9]
Output: 6
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [1, 1, 2, 7, 8, 8, 9, 10]. Here distinct elements are 6.
Approach: The idea is to sort the given array first, so that the elements can be checked easily, if it is distinct, by comparing with adjacent elements.
- First, sort all elements of the array.
- Initialize variables count and prev to 0. (To store the count of distinct elements and previous element respectively.)
- After that keep a track of the previous element using prev variable.
- Iterate the sorted array.
- Decrease the current element’s value by 1 and check if the previous element is lesser than the decreased value. If it is lesser then increment the count and assign current value to prev.
- If the decreased value of the current element is not greater than the previous element then keep the current element as it is and check if the previous element is lesser than the current element. If it is lesser then increment the count and assign current value to prev.
- If the current value is not greater than the previous element then increment the current value by 1 and check if the previous element is lesser than the incremented current element. If it is lesser then increment the count and assign current value to prev.
- If incremented value of current element is not lesser than previous value then skip that element.
Below is the implementation of the above approach:
C++
// C++ program to Maximize distinct // elements by incrementing/decrementing // an element or keeping it same #include <bits/stdc++.h> using namespace std; // Function that Maximize // the count of distinct // element int max_dist_ele( int arr[], int n) { // sort the array sort(arr, arr + n); int ans = 0; // keeping track of // previous change int prev = 0; for ( int i = 0; i < n; i++) { // check the // decremented value if (prev < (arr[i] - 1)) { ans++; prev = arr[i] - 1; } // check the current // value else if (prev < (arr[i])) { ans++; prev = arr[i]; } // check the // incremented value else if (prev < (arr[i] + 1)) { ans++; prev = arr[i] + 1; } } return ans; } // Driver Code int main() { int arr[] = { 1, 1, 1, 8, 8, 8, 9, 9 }; int n = sizeof (arr) / sizeof (arr[0]); cout << max_dist_ele(arr, n) << endl; return 0; } |
Java
// Java program to Maximize // the count of distinct element import java.util.*; public class GFG { // Function that Maximize // the count of distinct element static int max_dist_ele( int arr[], int n) { // sort the array Arrays.sort(arr); int ans = 0 ; // keeping track of // previous change int prev = 0 ; for ( int i = 0 ; i < n; i++) { // decrement is possible if (prev < (arr[i] - 1 )) { ans++; prev = arr[i] - 1 ; } // remain as it is else if (prev < (arr[i])) { ans++; prev = arr[i]; } // increment is possible else if (prev < (arr[i] + 1 )) { ans++; prev = arr[i] + 1 ; } } return ans; } // Driver Code public static void main(String args[]) { int arr[] = { 1 , 1 , 1 , 8 , 8 , 8 , 9 , 9 }; int n = arr.length; System.out.println(max_dist_ele(arr, n)); } } |
Python3
# Python3 program to Maximize # the count of distinct element def max_dist_ele(arr, n): # Sort the list arr.sort() ans = 0 # Keeping track of # previous change prev = 0 for i in range (n): # Decrement is possible if prev < (arr[i] - 1 ): ans + = 1 ; prev = arr[i] - 1 # Remain as it is elif prev < (arr[i]): ans + = 1 prev = arr[i] # Increment is possible elif prev < (arr[i] + 1 ): ans + = 1 prev = arr[i] + 1 return ans # Driver Code arr = [ 1 , 1 , 1 , 8 , 8 , 8 , 9 , 9 ] n = len (arr) print (max_dist_ele(arr, n)) # This code is contributed by rutvik_56 |
C#
// C# program to maximize the // count of distinct element using System; class GFG{ // Function that maximize the // count of distinct element static int max_dist_ele( int []arr, int n) { // Sort the array Array.Sort(arr); int ans = 0; // Keeping track of // previous change int prev = 0; for ( int i = 0; i < n; i++) { // Decrement is possible if (prev < (arr[i] - 1)) { ans++; prev = arr[i] - 1; } // Remain as it is else if (prev < (arr[i])) { ans++; prev = arr[i]; } // Increment is possible else if (prev < (arr[i] + 1)) { ans++; prev = arr[i] + 1; } } return ans; } // Driver Code public static void Main(String []args) { int []arr = { 1, 1, 1, 8, 8, 8, 9, 9 }; int n = arr.Length; Console.WriteLine(max_dist_ele(arr, n)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program to Maximize distinct // elements by incrementing/decrementing // an element or keeping it same // Function that Maximize // the count of distinct // element function max_dist_ele( arr, n) { // sort the array arr.sort(); var ans = 0; // keeping track of // previous change var prev = 0; for (let i = 0; i < n; i++) { // check the // decremented value if (prev < (arr[i] - 1)) { ans++; prev = arr[i] - 1; } // check the current // value else if (prev < (arr[i])) { ans++; prev = arr[i]; } // check the // incremented value else if (prev < (arr[i] + 1)) { ans++; prev = arr[i] + 1; } } return ans; } // Driver Code var arr = new Array( 1, 1, 1, 8, 8, 8, 9, 9 ); var n = arr.length; document.write(max_dist_ele(arr, n)); // This code is contributed by ukasp. </script> |
6
Time Complexity: O(N*logN)
Auxiliary Space: O(1) as it is using constant space for variables
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