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:
- Store the frequency of each element in a HashMap.
- Store the least and most frequent elements in separate Sets.
- 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.
- 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.
- 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.
- Print the minimum distance.
Below is the implementation of the above approach:
// 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 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 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# 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. |
<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> |
1
Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(N)