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Find maximum element among the elements with minimum frequency in given Array

  • Last Updated : 14 Sep, 2021

Given an array arr[] consisting of N integers, the task is to find the maximum element with the minimum frequency.

Examples:

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Input: arr[] = {2, 2, 5, 50, 1}
Output: 50
Explanation:
The element with minimum frequency is {1, 5, 50}. The maximum element among these element is 50.



Input: arr[] = {3, 2, 5, 6, 1}
Output: 6

Approach: The given problem can be solved by storing the frequency of the array element in a HashMap and then find the maximum value having a minimum frequency. Follow the step below to solve the given problem:

  • Store the frequency of each element in a HashMap, say M.
  • Initialize two variables, say maxValue as INT_MIN and minFreq as INT_MAX that store the resultant maximum element and stores the minimum frequency among all the frequencies.
  • Iterate over the map M and perform the following steps:
    • If the frequency of the current element is less than minFreq then update the value of minFreq to the current frequency and the value of maxValue to the current element.
    • If the frequency of the current element is equal to the minFreq and the value of maxValue is less than the current value then then update the value of maxValue to the current element.
  • After completing the above steps, print the value of maxValue as the resultant element.

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 maximum element
// with the minimum frequency
int maxElementWithMinFreq(int* arr, int N)
{
    // Stores the frequency of array
    // elements
    unordered_map<int, int> mp;
 
    // Find the frequency and store
    // in the map
    for (int i = 0; i < N; i++) {
        mp[arr[i]]++;
    }
 
    // Initialize minFreq to the maximum
    // value and minValue to the minimum
    int minFreq = INT_MAX;
    int maxValue = INT_MIN;
 
    // Traverse the map mp
    for (auto x : mp) {
 
        int num = x.first;
        int freq = x.second;
 
        // If freq < minFreq, then update
        // minFreq to freq and maxValue
        // to the current element
        if (freq < minFreq) {
            minFreq = freq;
            maxValue = num;
        }
 
        // If freq is equal to the minFreq
        // and current element > maxValue
        // then update maxValue to the
        // current element
        else if (freq == minFreq
                 && maxValue < num) {
            maxValue = num;
        }
    }
 
    // Return the resultant maximum value
    return maxValue;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 2, 5, 50, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << maxElementWithMinFreq(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum element
// with the minimum frequency
static int maxElementWithMinFreq(int[] arr, int N)
{
   
    // Stores the frequency of array
    // elements
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
 
    // Find the frequency and store
    // in the map
    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);
    }
    }
 
    // Initialize minFreq to the maximum
    // value and minValue to the minimum
    int minFreq = Integer.MAX_VALUE;
    int maxValue = Integer.MIN_VALUE;
 
    // Traverse the map mp
    for (Map.Entry<Integer,Integer> x : mp.entrySet()){
 
        int num = x.getKey();
        int freq = x.getValue();
 
        // If freq < minFreq, then update
        // minFreq to freq and maxValue
        // to the current element
        if (freq < minFreq) {
            minFreq = freq;
            maxValue = num;
        }
 
        // If freq is equal to the minFreq
        // and current element > maxValue
        // then update maxValue to the
        // current element
        else if (freq == minFreq
                 && maxValue < num) {
            maxValue = num;
        }
    }
 
    // Return the resultant maximum value
    return maxValue;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 2, 5, 50, 1 };
    int N = arr.length;
    System.out.print(maxElementWithMinFreq(arr, N));
 
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python 3 program for the above approach
import sys
from collections import defaultdict
 
# Function to find the maximum element
# with the minimum frequency
def maxElementWithMinFreq(arr, N):
 
    # Stores the frequency of array
    # elements
    mp = defaultdict(int)
 
    # Find the frequency and store
    # in the map
    for i in range(N):
        mp[arr[i]] += 1
 
    # Initialize minFreq to the maximum
    # value and minValue to the minimum
    minFreq = sys.maxsize
    maxValue = -sys.maxsize-1
 
    # Traverse the map mp
    for x in mp:
 
        num = x
        freq = mp[x]
 
        # If freq < minFreq, then update
        # minFreq to freq and maxValue
        # to the current element
        if (freq < minFreq):
            minFreq = freq
            maxValue = num
 
        # If freq is equal to the minFreq
        # and current element > maxValue
        # then update maxValue to the
        # current element
        elif (freq == minFreq
              and maxValue < num):
            maxValue = num
 
    # Return the resultant maximum value
    return maxValue
 
# Driver Code
if __name__ == "__main__":
 
    arr = [2, 2, 5, 50, 1]
    N = len(arr)
    print(maxElementWithMinFreq(arr, N))
 
    # This code is contributed by ukasp.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the maximum element
// with the minimum frequency
static int maxElementWithMinFreq(int []arr, int N)
{
   
    // Stores the frequency of array
    // elements
    Dictionary<int, int> mp = new Dictionary<int,int>();
 
    // Find the frequency and store
    // in the map
    for (int i = 0; i < N; i++) {
        if(mp.ContainsKey(arr[i]))
        mp[arr[i]]++;
        else
          mp.Add(arr[i],1);
    }
 
    // Initialize minFreq to the maximum
    // value and minValue to the minimum
    int minFreq = Int32.MaxValue;
    int maxValue = Int32.MinValue;
 
    // Traverse the map mp
    foreach(KeyValuePair<int,int> x in mp) {
 
        int num = x.Key;
        int freq = x.Value;
 
        // If freq < minFreq, then update
        // minFreq to freq and maxValue
        // to the current element
        if (freq < minFreq) {
            minFreq = freq;
            maxValue = num;
        }
 
        // If freq is equal to the minFreq
        // and current element > maxValue
        // then update maxValue to the
        // current element
        else if (freq == minFreq
                 && maxValue < num) {
            maxValue = num;
        }
    }
 
    // Return the resultant maximum value
    return maxValue;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 2, 2, 5, 50, 1 };
    int N = arr.Length;
    Console.Write(maxElementWithMinFreq(arr, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the maximum element
        // with the minimum frequency
        function maxElementWithMinFreq(arr, N) {
            // Stores the frequency of array
            // elements
            let mp = new Map();
 
            // Find the frequency and store
            // in the map
            for (let i = 0; i < N; i++) {
                if (mp.has(arr[i])) {
                    mp.set(arr[i], mp.get(arr[i]) + 1);
                }
                else {
                    mp.set(arr[i], 1);
                }
 
            }
 
            // Initialize minFreq to the maximum
            // value and minValue to the minimum
            let minFreq = Number.MAX_VALUE
            let maxValue = Number.MIN_VALUE;
 
            // Traverse the map mp
            for (let [key, value] of mp) {
 
                let num = key;
                let freq = value;
 
                // If freq < minFreq, then update
                // minFreq to freq and maxValue
                // to the current element
                if (freq < minFreq) {
                    minFreq = freq;
                    maxValue = num;
                }
 
                // If freq is equal to the minFreq
                // and current element > maxValue
                // then update maxValue to the
                // current element
                else if (freq == minFreq
                    && maxValue < num) {
                    maxValue = num;
                }
            }
 
            // Return the resultant maximum value
            return maxValue;
        }
 
        // Driver Code
 
        let arr = [2, 2, 5, 50, 1];
        let N = arr.length;
        document.write(maxElementWithMinFreq(arr, N));
 
// This code is contributed by Potta Lokesh
    </script>
Output: 
50

 

Time Complexity: O(N)
Auxiliary Space: O(N)




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