Maximize the number of indices such that element is greater than element to its left
Given an array arr[] of N integers, the task is to maximize the number of indices such that an element is greater than the element to its left, i.e. arr[i+1] > arr[i] after rearranging the array.
Examples:
Input: arr[] = {200, 100, 100, 200}
Output: 2
Explanation:
By arranging the array in following way we have: arr[] = {100, 200, 100, 200}
The possible indices are 0 and 2 such that:
arr[1] > arr[0] (200 > 100)
arr[3] > arr[2] (200 > 100)
Input: arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7}
Output: 7
Explanation:
By arranging the array in following way we have: arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4}
The possible indices are 0, 1, 2, 3, 5, 6 and 7 such that:
arr[1] > arr[0] (5 > 1)
arr[2] > arr[1] (7 > 5)
arr[3] > arr[2] (8 > 7)
arr[4] > arr[3] (9 > 8)
arr[6] > arr[5] (7 > 5)
arr[7] > arr[6] (8 > 7)
arr[8] > arr[7] (8 > 7)
Approach: This problem can be solved using Greedy Approach. Below are the steps:
- To get the maximum number of indices(say i) such that arr[i+1] > arr[i], arrange the elements of the arr[] such that set of all unique element occurs first, then next set of unique elements occurs after the first set till all the elements are arranged.
For Example:
Let arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7}
1st Set = {1, 5, 7, 8, 9}
2nd Set = {5, 7, 8}
3rd Set = {7, 8}
4th Set = {4}
Now the new array will be:
arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4}
- After the above arrangement, the element with the higher value will not be a part of the given condition as it is followed by a number smaller than itself.
- Therefore the total number of pairs satisfying the given condition can be given by:
total_pairs = (number_of_elements – highest_frequency_of_a_number)
Below is the implementation of the above approach:
C++
// C++ program of the above approach #include <bits/stdc++.h> using namespace std; // Function to find the maximum pairs // such that arr[i+1] > arr[i] void countPairs( int arr[], int N) { // To store the frequency of the // element in arr[] unordered_map< int , int > M; // Store the frequency in map M for ( int i = 0; i < N; i++) { M[arr[i]]++; } int maxFreq = 0; // To find the maximum frequency // store in map M for ( auto & it : M) { maxFreq = max(maxFreq, it.second); } // Print the maximum number of // possible pairs cout << N - maxFreq << endl; } // Driver Code int main() { int arr[] = { 1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7 }; int N = sizeof (arr) / sizeof (arr[0]); countPairs(arr, N); return 0; } |
Java
// Java program of the above approach import java.util.*; class GFG{ // Function to find the maximum pairs // such that arr[i+1] > arr[i] static void countPairs( int arr[], int N) { // To store the frequency of the // element in arr[] HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>(); // Store the frequency in map M for ( int i = 0 ; i < N; i++) { if (mp.containsKey(arr[i])){ mp.put(arr[i], mp.get(arr[i])+ 1 ); } else { mp.put(arr[i], 1 ); } } int maxFreq = 0 ; // To find the maximum frequency // store in map M for (Map.Entry<Integer,Integer> it : mp.entrySet()) { maxFreq = Math.max(maxFreq, it.getValue()); } // Print the maximum number of // possible pairs System.out.print(N - maxFreq + "\n" ); } // Driver Code public static void main(String[] args) { int arr[] = { 1 , 8 , 5 , 9 , 8 , 8 , 7 , 7 , 5 , 7 , 7 }; int N = arr.length; countPairs(arr, N); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the above approach # Function to find the maximum pairs # such that arr[i + 1] > arr[i] def countPairs(arr, N) : # To store the frequency of the # element in arr[] M = dict .fromkeys(arr, 0 ); # Store the frequency in map M for i in range (N) : M[arr[i]] + = 1 ; maxFreq = 0 ; # To find the maximum frequency # store in map M for it in M.values() : maxFreq = max (maxFreq,it); # Print the maximum number of # possible pairs print (N - maxFreq); # Driver Code if __name__ = = "__main__" : arr = [ 1 , 8 , 5 , 9 , 8 , 8 , 7 , 7 , 5 , 7 , 7 ]; N = len (arr); countPairs(arr, N); # This code is contributed by AnkitRai01 |
C#
// C# program of the above approach using System; using System.Collections.Generic; class GFG{ // Function to find the maximum pairs // such that arr[i+1] > arr[i] static void countPairs( int []arr, int N) { // To store the frequency of the // element in []arr Dictionary< int , int > mp = new Dictionary< int , int >(); // Store the frequency in map M for ( int i = 0; i < N; i++) { if (mp.ContainsKey(arr[i])){ mp[arr[i]] = mp[arr[i]]+1; } else { mp.Add(arr[i], 1); } } int maxFreq = 0; // To find the maximum frequency // store in map M foreach (KeyValuePair< int , int > it in mp) { maxFreq = Math.Max(maxFreq, it.Value); } // Print the maximum number of // possible pairs Console.Write(N - maxFreq + "\n" ); } // Driver Code public static void Main(String[] args) { int []arr = { 1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7 }; int N = arr.Length; countPairs(arr, N); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Js program of the above approach // Function to find the maximum pairs // such that arr[i+1] > arr[i] function countPairs( arr, N){ // To store the frequency of the // element in arr[] let M = new Map(); // Store the frequency in map M for (let i = 0; i < N; i++) { if (M[arr[i]]) M[arr[i]]++; else M[arr[i]] = 1 } let maxFreq = 0; // To find the maximum frequency // store in map M for (let it in M) { maxFreq = Math.max(maxFreq, M[it]); } // Print the maximum number of // possible pairs document.write(N - maxFreq, '<br>' ); } // Driver Code let a = [ 1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7 ]; let N = a.length; countPairs(a, N); </script> |
7
Time Complexity: O(N), where N is the number of element in the array.
Auxiliary Space: O(N), where N is the number of element in the array.