Minimum distance between any most frequent and least frequent element of an array

Given an integer array arr[] of size N, the task is to find the minimum distance between any most and least frequent element of the given array.

Examples:

Input: arr[] = {1, 1, 2, 3, 2, 3, 3}
Output: 1
Explanation: The least frequent elements are 1 and 2 which occurs at indexes: 0, 1, 2, 4. 
Whereas, the most frequent element is 3 which occurs at indexes: 3, 5, 6.
So the minimum distance is (3-2) = 1.

Input: arr[] = {1, 3, 4, 4, 3, 4}
Output: 2
Explanation: The least frequent element is 1 which occurs at indexes: 0. 
Whereas, the most frequent element is 4 which occurs at indexes: 2, 3, 5. 
So the minimum distance is (2-0) = 2.

Approach: The idea is to find the indices of least and most frequent elements in the array and find the difference between those indices which is minimum. Follow the steps below to solve the problem:

1. Store the frequency of each element in a HashMap.
2. Store the least and most frequent elements in separate Sets.
3. Traverse from the start of the array. If the current element is the least frequent element then update the last index of the least frequent element.
4. Otherwise, if the current element is the most frequent element then calculate the distance between the current and the last index of the least frequent element and update the required minimum distance.
5. Similarly, traverse the array from the end and repeat step 3 and step 4 to find the minimum distance between any most and least frequent element of an array.
6. Print the minimum distance.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum
// distance between any two most
// and least frequent element
void getMinimumDistance(int a[], int n)
{

    // Initialize sets to store the least
    // and the most frequent elements
    set<int> min_set;
    set<int> max_set;

    // Initialize variables to store
    // max and min frequency
    int max = 0, min = INT_MAX;

    // Initialize HashMap to store
    // frequency of each element
    map<int, int> frequency;

    // Loop through the array
    for (int i = 0; i < n; i++) {

        // Store the count of each element
        frequency[a[i]] += 1;
    }

    // Store the least and most frequent
    // elements in the respective sets
    for (int i = 0; i < n; i++) {

        // Store count of current element
        int count = frequency[a[i]];

        // If count is equal
        // to max count
        if (count == max) {

            // Store in max set
            max_set.insert(a[i]);
        }

        // If count is greater
        // then max count
        else if (count > max) {

            // Empty max set
            max_set.clear();

            // Update max count
            max = count;

            // Store in max set
            max_set.insert(a[i]);
        }

        // If count is equal
        // to min count
        if (count == min) {

            // Store in min set
            min_set.insert(a[i]);
        }

        // If count is less
        // then max count
        else if (count < min) {

            // Empty min set
            min_set.clear();

            // Update min count
            min = count;

            // Store in min set
            min_set.insert(a[i]);
        }
    }

    // Initialize a variable to
    // store the minimum distance
    int min_dist = INT_MAX;

    // Initialize a variable to
    // store the last index of
    // least frequent element
    int last_min_found = -1;

    // Traverse array
    for (int i = 0; i < n; i++) {

        // If least frequent element
        if (min_set.find(a[i]) != min_set.end())

            // Update last index of
            // least frequent element
            last_min_found = i;

        // If most frequent element
        if (max_set.find(a[i]) != max_set.end()
            && last_min_found != -1) {

            // Update minimum distance
            if ((i - last_min_found) < min_dist)
                min_dist = i - last_min_found;
        }
    }

    last_min_found = -1;

    // Traverse array from the end
    for (int i = n - 1; i >= 0; i--) {

        // If least frequent element
        if (min_set.find(a[i]) != min_set.end())

            // Update last index of
            // least frequent element
            last_min_found = i;

        // If most frequent element
        if (max_set.find(a[i]) != max_set.end()
            && last_min_found != -1) {

            // Update minimum distance
            if ((last_min_found - i) > min_dist)
                min_dist = last_min_found - i;
        }
    }

    // Print the minimum distance
    cout << (min_dist);
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 1, 2, 3, 2, 3, 3 };

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    getMinimumDistance(arr, N);
}

// This code is contributed by ukasp.

Java

// Java implementation of the approach
import java.util.*;

class GFG {

    // Function to find the minimum
    // distance between any two most
    // and least frequent element
    public static void
    getMinimumDistance(int a[], int n)
    {

        // Initialize sets to store the least
        // and the most frequent elements
        Set<Integer> min_set = new HashSet<>();
        Set<Integer> max_set = new HashSet<>();

        // Initialize variables to store
        // max and min frequency
        int max = 0, min = Integer.MAX_VALUE;

        // Initialize HashMap to store
        // frequency of each element
        HashMap<Integer, Integer> frequency
            = new HashMap<>();

        // Loop through the array
        for (int i = 0; i < n; i++) {

            // Store the count of each element
            frequency.put(
                a[i],
                frequency
                        .getOrDefault(a[i], 0)
                    + 1);
        }

        // Store the least and most frequent
        // elements in the respective sets
        for (int i = 0; i < n; i++) {

            // Store count of current element
            int count = frequency.get(a[i]);

            // If count is equal
            // to max count
            if (count == max) {

                // Store in max set
                max_set.add(a[i]);
            }

            // If count is greater
            // then max count
            else if (count > max) {

                // Empty max set
                max_set.clear();

                // Update max count
                max = count;

                // Store in max set
                max_set.add(a[i]);
            }

            // If count is equal
            // to min count
            if (count == min) {

                // Store in min set
                min_set.add(a[i]);
            }

            // If count is less
            // then max count
            else if (count < min) {

                // Empty min set
                min_set.clear();

                // Update min count
                min = count;

                // Store in min set
                min_set.add(a[i]);
            }
        }

        // Initialize a variable to
        // store the minimum distance
        int min_dist = Integer.MAX_VALUE;

        // Initialize a variable to
        // store the last index of
        // least frequent element
        int last_min_found = -1;

        // Traverse array
        for (int i = 0; i < n; i++) {

            // If least frequent element
            if (min_set.contains(a[i]))

                // Update last index of
                // least frequent element
                last_min_found = i;

            // If most frequent element
            if (max_set.contains(a[i])
                && last_min_found != -1) {

                // Update minimum distance
                min_dist = Math.min(min_dist,
                                    i - last_min_found);
            }
        }

        last_min_found = -1;

        // Traverse array from the end
        for (int i = n - 1; i >= 0; i--) {

            // If least frequent element
            if (min_set.contains(a[i]))

                // Update last index of
                // least frequent element
                last_min_found = i;

            // If most frequent element
            if (max_set.contains(a[i])
                && last_min_found != -1) {

                // Update minimum distance
                min_dist = Math.min(min_dist,
                                    last_min_found - i);
            }
        }

        // Print the minimum distance
        System.out.println(min_dist);
    }

    // Driver Code
    public static void
        main(String[] args)
    {
        // Given array
        int arr[] = { 1, 1, 2, 3, 2, 3, 3 };

        int N = arr.length;

        // Function Call
        getMinimumDistance(arr, N);
    }
}

Python3

# Python3 implementation of the approach
import sys

# Function to find the minimum
# distance between any two most
# and least frequent element
def getMinimumDistance(a, n):

    # Initialize sets to store the least
    # and the most frequent elements
    min_set = {}
    max_set = {}

    # Initialize variables to store
    # max and min frequency
    max, min = 0, sys.maxsize + 1

    # Initialize HashMap to store
    # frequency of each element
    frequency = {}

    # Loop through the array
    for i in range(n):

        # Store the count of each element
        frequency[a[i]] = frequency.get(a[i], 0) + 1

    # Store the least and most frequent
    # elements in the respective sets
    for i in range(n):

        # Store count of current element
        count = frequency[a[i]]

        # If count is equal
        # to max count
        if (count == max):

            # Store in max set
            max_set[a[i]] = 1

        # If count is greater
        # then max count
        elif (count > max):

            # Empty max set
            max_set.clear()

            # Update max count
            max = count

            # Store in max set
            max_set[a[i]] = 1

        # If count is equal
        # to min count
        if (count == min):

            # Store in min set
            min_set[a[i]] = 1

        # If count is less
        # then max count
        elif (count < min):

            # Empty min set
            min_set.clear()

            # Update min count
            min = count

            # Store in min set
            min_set[a[i]] = 1

    # Initialize a variable to
    # store the minimum distance
    min_dist = sys.maxsize + 1

    # Initialize a variable to
    # store the last index of
    # least frequent element
    last_min_found = -1

    # Traverse array
    for i in range(n):

        # If least frequent element
        if (a[i] in min_set):

            # Update last index of
            # least frequent element
            last_min_found = i

        # If most frequent element
        if ((a[i] in max_set) and
            last_min_found != -1):

            # Update minimum distance
            if i-last_min_found < min_dist:
                min_dist = i - last_min_found

    last_min_found = -1

    # Traverse array from the end
    for i in range(n - 1, -1, -1):

        # If least frequent element
        if (a[i] in min_set):

            # Update last index of
            # least frequent element
            last_min_found = i;

        # If most frequent element
        if ((a[i] in max_set) and
            last_min_found != -1):

            # Update minimum distance
            if min_dist > last_min_found - i:
                min_dist = last_min_found - i

    # Print the minimum distance
    print(min_dist)

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [ 1, 1, 2, 3, 2, 3, 3 ]

    N = len(arr)

    # Function Call
    getMinimumDistance(arr, N)

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach

using System;
using System.Collections.Generic;

public class GFG{


    // Function to find the minimum
    // distance between any two most
    // and least frequent element
    public static void
    getMinimumDistance(int[] a, int n)
    {

        // Initialize sets to store the least
        // and the most frequent elements
        HashSet<int> min_set = new HashSet<int>();
        HashSet<int> max_set = new HashSet<int>();

        // Initialize variables to store
        // max and min frequency
        int max = 0, min = Int32.MaxValue;

        // Initialize HashMap to store
        // frequency of each element
        Dictionary<int, int> frequency =
                       new Dictionary<int, int>();

        // Loop through the array
        for (int i = 0; i < n; i++) {

            // Store the count of each element
            if(!frequency.ContainsKey(a[i]))
                frequency.Add(a[i],0);
            frequency[a[i]]++;
        }

        // Store the least and most frequent
        // elements in the respective sets
        for (int i = 0; i < n; i++) {

            // Store count of current element
            int count = frequency[a[i]];

            // If count is equal
            // to max count
            if (count == max) {

                // Store in max set
                max_set.Add(a[i]);
            }

            // If count is greater
            // then max count
            else if (count > max) {

                // Empty max set
                max_set.Clear();

                // Update max count
                max = count;

                // Store in max set
                max_set.Add(a[i]);
            }

            // If count is equal
            // to min count
            if (count == min) {

                // Store in min set
                min_set.Add(a[i]);
            }

            // If count is less
            // then max count
            else if (count < min) {

                // Empty min set
                min_set.Clear();

                // Update min count
                min = count;

                // Store in min set
                min_set.Add(a[i]);
            }
        }

        // Initialize a variable to
        // store the minimum distance
        int min_dist = Int32.MaxValue;

        // Initialize a variable to
        // store the last index of
        // least frequent element
        int last_min_found = -1;

        // Traverse array
        for (int i = 0; i < n; i++) {

            // If least frequent element
            if (min_set.Contains(a[i]))

                // Update last index of
                // least frequent element
                last_min_found = i;

            // If most frequent element
            if (max_set.Contains(a[i])
                && last_min_found != -1) {

                // Update minimum distance
                min_dist = Math.Min(min_dist,
                                    i - last_min_found);
            }
        }

        last_min_found = -1;

        // Traverse array from the end
        for (int i = n - 1; i >= 0; i--) {

            // If least frequent element
            if (min_set.Contains(a[i]))

                // Update last index of
                // least frequent element
                last_min_found = i;

            // If most frequent element
            if (max_set.Contains(a[i])
                && last_min_found != -1) {

                // Update minimum distance
                min_dist = Math.Min(min_dist,
                                    last_min_found - i);
            }
        }

        // Print the minimum distance
        Console.WriteLine(min_dist);
    }

    // Driver Code
    static public void Main ()
    {

        // Given array
        int[] arr = { 1, 1, 2, 3, 2, 3, 3 };
        int N = arr.Length;

        // Function Call
        getMinimumDistance(arr, N);

    }
}

// This code is contributed by patel2127.

Javascript

<script>

// Javascript implementation of the approach

// Function to find the minimum
// distance between any two most
// and least frequent element
function getMinimumDistance(a, n)
{

    // Initialize sets to store the least
    // and the most frequent elements
    var min_set = new Set();
    var max_set = new Set();

    // Initialize variables to store
    // max and min frequency
    var max = 0, min = 1000000000;

    // Initialize HashMap to store
    // frequency of each element
    var frequency = new Map();

    // Loop through the array
    for(var i = 0; i < n; i++)
    {

        // Store the count of each element
        if(frequency.has(a[i]))
            frequency.set(a[i],
            frequency.get(a[i]) + 1)
        else
            frequency.set(a[i], 1)
    }

    // Store the least and most frequent
    // elements in the respective sets
    for(var i = 0; i < n; i++)
    {

        // Store count of current element
        var count = frequency.get(a[i]);

        // If count is equal
        // to max count
        if (count == max)
        {

            // Store in max set
            max_set.add(a[i]);
        }

        // If count is greater
        // then max count
        else if (count > max)
        {

            // Empty max set
            max_set = new Set();

            // Update max count
            max = count;

            // Store in max set
            max_set.add(a[i]);
        }

        // If count is equal
        // to min count
        if (count == min)
        {

            // Store in min set
            min_set.add(a[i]);
        }

        // If count is less
        // then max count
        else if (count < min)
        {

            // Empty min set
            min_set = new Set();

            // Update min count
            min = count;

            // Store in min set
            min_set.add(a[i]);
        }
    }

    // Initialize a variable to
    // store the minimum distance
    var min_dist = 1000000000;

    // Initialize a variable to
    // store the last index of
    // least frequent element
    var last_min_found = -1;

    // Traverse array
    for(var i = 0; i < n; i++)
    {

        // If least frequent element
        if (min_set.has(a[i]))

            // Update last index of
            // least frequent element
            last_min_found = i;

        // If most frequent element
        if (max_set.has(a[i]) &&
            last_min_found != -1)
        {

            // Update minimum distance
            if ((i - last_min_found) < min_dist)
                min_dist = i - last_min_found;
        }
    }

    last_min_found = -1;

    // Traverse array from the end
    for(var i = n - 1; i >= 0; i--)
    {

        // If least frequent element
        if (min_set.has(a[i]))

            // Update last index of
            // least frequent element
            last_min_found = i;

        // If most frequent element
        if (max_set.has(a[i]) &&
            last_min_found != -1)
        {

            // Update minimum distance
            if ((last_min_found - i) > min_dist)
                min_dist = last_min_found - i;
        }
    }

    // Print the minimum distance
    document.write(min_dist);
}

// Driver Code

// Given array
var arr = [ 1, 1, 2, 3, 2, 3, 3 ];
var N = arr.length;

// Function Call
getMinimumDistance(arr, N);

// This code is contributed by itsok

</script>

Output:

1

Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(N)