Maximum Occurrence in a Given Range

Given an array of n integers in non-decreasing order. Find the number of occurrences of the most frequent value within a given range.

Examples:

Input : arr[] = {-5, -5, 2, 2, 2, 2, 3, 7, 7, 7}
        Query 1: start = 0, end = 9
        Query 2: start = 4, end = 9
Output : 4
         3
Explanation:  
Query 1: '2' occurred the most number of times
with a frequency of 4 within given range.
Query 2: '7' occurred the most number of times
with a frequency of 3 within given range.

Segment Trees can be used to solve this problem efficiently.
Refer here for the implementation of segment trees

The key idea behind this problem is that the given array is in non-decreasing order which means that all occurrences of a number are consecutively placed in the array as the array is in sorted order.
A segment tree can be constructed where each node would store the maximum count of its respective range [i, j]. For that we will build the frequency array and call RMQ (Range Maximum Query) on this array. For e.g.

arr[] =  {-5, -5, 2, 2, 2, 2, 3, 7, 7, 7}
freq_arr[] = {2, 2, 4, 4, 4, 4, 1, 3, 3, 3}
where, freq_arr[i] = frequency(arr[i])

Now there are two cases to be considered,

Case 1: The value of the numbers at index i and j for the given range are same, i.e. arr[i] = arr[j].

Solving this case is very easy. Since arr[i] = arr[j], all numbers between these indices are same (since the array is non-decreasing). Hence answer for this case is simply count of all numbers between i and j (inclusive both) i.e. (j – i + 1)

For e.g.

arr[] =  {-5, -5, 2, 2, 2, 2, 3, 7, 7, 7}
if the given query range is [3, 5], answer would 
be (5 - 3 + 1) = 3, as 2 occurs 3 times within 
given range

Case 2: The value of the numbers at index i and j for the given range are different, i.e. arr[i] != arr[j]
If arr[i] != arr[j], then there exists an index k where arr[i] = arr[k] and arr[i] != arr[k + 1]. This may be a case of partial overlap where some occurrences of a particular number lie in the leftmost part of the given range and some lie just before range starts. Here simply calling RMQ would result into an incorrect answer.

For e.g.

arr[] =  {-5, -5, 2, 2, 2, 2, 3, 7, 7, 7}
freq_arr[] = {2, 2, 4, 4, 4, 4, 1, 3, 3, 3}
if the given query is [4, 9], calling RMQ on 
freq_arr[] will give us 4 as answer which 
is incorrect as some occurrences of 2 are 
lying outside the range. Correct answer 
is 3.

Similar situation can happen at the rightmost part of the given range where some occurrences of a particular number lies inside the range and some lies just after the range ends.
Hence for this case, inside the given range we have to count the leftmost same numbers upto some index say i and rightmost same numbers from index say j to the end of the range. And then calling RMQ (Range Maximum Query) between indices i and j and taking maximum of all these three.

For e.g.

arr[] =  {-5, -5, 2, 2, 2, 2, 3, 7, 7, 7}
freq_arr[] = {2, 2, 4, 4, 4, 4, 1, 3, 3, 3}
if the given query is [4, 7], counting leftmost
same numbers i.e 2 which occurs 2 times inside 
the range and rightmost same numbers i.e. 3 
which occur only 1 time and RMQ on [6, 6] is 
1. Hence maximum would be 2.

Below is the implementation of the above approach

C++

// C++ Program to find the occurrence
// of the most frequent number within
// a given range
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to get the middle index
// from corner indexes.
int getMid(int s, int e) { return s + (e - s) / 2; }
 
/* A recursive function to get the maximum value in
    a given range of array indexes. The following
    are parameters for this function.
 
    st --> Pointer to segment tree
    index --> Index of current node in the segment
            tree. Initially 0 is passed as root is
            always at index 0
    ss & se --> Starting and ending indexes of the
                segment represented by current node,
                i.e., st[index]
    qs & qe --> Starting and ending indexes of query
                range */
int RMQUtil(int* st, int ss, int se, int qs, int qe,
                                        int index)
{
    // If segment of this node is a part of given range,
    // then return the min of the segment
    if (qs <= ss && qe >= se)
        return st[index];
 
    // If segment of this node is outside the
    // given range
    if (se < qs || ss > qe)
        return 0;
 
    // If a part of this segment overlaps
    // with the given range
    int mid = getMid(ss, se);
    return max(RMQUtil(st, ss, mid, qs, qe, 2 * index + 1),
            RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2));
}
 
// Return minimum of elements in range from
// index qs (query start) to
// qe (query end). It mainly uses RMQUtil()
int RMQ(int* st, int n, int qs, int qe)
{
    // Check for erroneous input values
    if (qs < 0 || qe > n - 1 || qs > qe) {
        printf("Invalid Input");
        return -1;
    }
 
    return RMQUtil(st, 0, n - 1, qs, qe, 0);
}
 
// A recursive function that constructs Segment Tree
// for array[ss..se]. si is index of current node in
// segment tree st
int constructSTUtil(int arr[], int ss, int se, int* st,
                                            int si)
{
    // If there is one element in array, store it in
    // current node of segment tree and return
    if (ss == se) {
        st[si] = arr[ss];
        return arr[ss];
    }
 
    // If there are more than one elements, then
    // recur for left and right subtrees and store
    // the minimum of two values in this node
    int mid = getMid(ss, se);
    st[si] = max(constructSTUtil(arr, ss, mid, st, si * 2 + 1),
                constructSTUtil(arr, mid + 1, se, st, si * 2 + 2));
    return st[si];
}
 
/* Function to construct segment tree from given
array. This function allocates memory for segment
tree and calls constructSTUtil() to fill the
allocated memory */
int* constructST(int arr[], int n)
{
    // Allocate memory for segment tree
 
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
 
    // Maximum size of segment tree
    int max_size = 2 * (int)pow(2, x) - 1;
 
    int* st = new int[max_size];
 
    // Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0);
 
    // Return the constructed segment tree
    return st;
}
 
int maximumOccurrence(int arr[], int n, int qs, int qe)
{
    // Declaring a frequency array
    int freq_arr[n + 1];
 
    // Counting frequencies of all array elements.
    unordered_map<int, int> cnt;
    for (int i = 0; i < n; i++)
        cnt[arr[i]]++;
 
    // Creating frequency array by replacing the
    // number in array to the number of times it
    // has appeared in the array
    for (int i = 0; i < n; i++)
        freq_arr[i] = cnt[arr[i]];
 
    // Build segment tree from this frequency array
    int* st = constructST(freq_arr, n);
 
    int maxOcc; // to store the answer
 
    // Case 1: numbers are same at the starting
    // and ending index of the query
    if (arr[qs] == arr[qe])
        maxOcc = (qe - qs + 1);
 
    // Case 2: numbers are different
    else {
        int leftmost_same = 0, righmost_same = 0;
 
        // Partial Overlap Case of a number with some
        // occurrences lying inside the leftmost
        // part of the range and some just before the
        // range starts
        while (qs > 0 && qs <= qe && arr[qs] == arr[qs - 1]) {
            qs++;
            leftmost_same++;
        }
 
        // Partial Overlap Case of a number with some
        // occurrences lying inside the rightmost part of
        // the range and some just after the range ends
        while (qe >= qs && qe < n - 1 && arr[qe] == arr[qe + 1]) {
            qe--;
            righmost_same++;
        }
        // Taking maximum of all three
        maxOcc = max({leftmost_same, righmost_same,
                                RMQ(st, n, qs, qe)});
    }
    return maxOcc;
}
 
// Driver Code
int main()
{
    int arr[] = { -5, -5, 2, 2, 2, 2, 3, 7, 7, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int qs = 0; // Starting index of query range
    int qe = 9; // Ending index of query range
 
    // Print occurrence of most frequent number
    // within given range
    cout << "Maximum Occurrence in range is = "
        << maximumOccurrence(arr, n, qs, qe) << endl;
 
    qs = 4; // Starting index of query range
    qe = 9; // Ending index of query range
 
    // Print occurrence of most frequent number
    // within given range
    cout << "Maximum Occurrence in range is = "
        << maximumOccurrence(arr, n, qs, qe) << endl;
 
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.HashMap;
 
 class GFG {
     
    // A utility function to get the middle index
    // from corner indexes.
    public static int getMid(int s, int e) {
        return s + (e - s) / 2;
    }
 
    /* A recursive function to get the maximum value in
       a given range of array indexes. The following
       are parameters for this function.
 
       st --> Pointer to segment tree
       index --> Index of current node in the segment
               tree. Initially 0 is passed as root is
               always at index 0
       ss & se --> Starting and ending indexes of the
                   segment represented by current node,
                   i.e., st[index]
       qs & qe --> Starting and ending indexes of query
                   range */
  public static int RMQUtil(int[] st, int ss, int se, int qs, int qe,
                         int index) {
        // If segment of this node is a part of given range,
        // then return the min of the segment
        if (qs <= ss && qe >= se)
            return st[index];
 
        // If segment of this node is outside the
        // given range
        if (se < qs || ss > qe)
            return 0;
 
        // If a part of this segment overlaps
        // with the given range
        int mid = getMid(ss, se);
        return Math.max(RMQUtil(st, ss, mid, qs, qe, 2 * index + 1),
                RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2));
    }
 
    // Return minimum of elements in range from
    // index qs (query start) to
    // qe (query end). It mainly uses RMQUtil()
   public static int RMQ(int[] st, int n, int qs, int qe) {
        // Check for erroneous input values
        if (qs < 0 || qe > n - 1 || qs > qe) {
            System.out.println("Invalid Input");
            return -1;
        }
 
        return RMQUtil(st, 0, n - 1, qs, qe, 0);
    }
 
    // A recursive function that constructs Segment Tree
    // for array[ss..se]. si is index of current node in
    // segment tree st
   public static int constructSTUtil(int arr[], int ss, int se, int[] st,
                                            int si) {
        // If there is one element in array, store it in
        // current node of segment tree and return
        if (ss == se) {
            st[si] = arr[ss];
            return arr[ss];
        }
 
        // If there are more than one elements, then
        // recur for left and right subtrees and store
        // the minimum of two values in this node
        int mid = getMid(ss, se);
        st[si] = Math.max(constructSTUtil(arr, ss, mid, st, si * 2 + 1),
                constructSTUtil(arr, mid + 1, se, st, si * 2 + 2));
        return st[si];
    }
   /* Function to construct segment tree from given
array. This function allocates memory for segment
tree and calls constructSTUtil() to fill the
allocated memory */
public static int[] constructST(int arr[], int n)
{
    // Allocate memory for segment tree
  
    // Height of segment tree
    int x = (int)(Math.ceil(Math.log(n) / Math.log(2)));
  
    // Maximum size of segment tree
    int max_size = 2 * (int)Math.pow(2, x) - 1;
  
    int[] st = new int[max_size];
  
    // Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0);
  
    // Return the constructed segment tree
    return st;
}
  
public static int maximumOccurrence(int arr[], int n, int qs, int qe)
{
    // Declaring a frequency array
    int freq_arr[] = new int[n + 1];
  
    // Counting frequencies of all array elements.
    HashMap<Integer, Integer> cnt = new HashMap<>();
    for (int i = 0; i < n; i++)
        cnt.put(arr[i], cnt.getOrDefault(arr[i], 0) + 1);
  
    // Creating frequency array by replacing the
    // number in array to the number of times it
    // has appeared in the array
    for (int i = 0; i < n; i++)
        freq_arr[i] = cnt.get(arr[i]);
  
    // Build segment tree from this frequency array
    int[] st = constructST(freq_arr, n);
  
    int maxOcc; // to store the answer
  
    // Case 1: numbers are same at the starting
    // and ending index of the query
    if (arr[qs] == arr[qe])
        maxOcc = (qe - qs + 1);
  
    // Case 2: numbers are different
    else {
        int leftmost_same = 0, righmost_same = 0;
  
        // Partial Overlap Case of a number with some
        // occurrences lying inside the leftmost
        // part of the range and some just before the
        // range starts
        while (qs > 0 && qs <= qe && arr[qs] == arr[qs - 1]) {
            qs++;
            leftmost_same++;
        }
  
        // Partial Overlap Case of a number with some
        // occurrences lying inside the rightmost part of
        // the range and some just after the range ends
        while (qe >= qs && qe < n - 1 && arr[qe] == arr[qe + 1]) {
            qe--;
            righmost_same++;
        }
        // Taking maximum of all three
        maxOcc = Math.max(Math.max(leftmost_same, righmost_same),
                                RMQ(st, n, qs, qe));
    }
    return maxOcc;
}
   //driver code
   public static void main(String[] args) {
        int[] arr = { -5, -5, 2, 2, 2, 2, 3, 7, 7, 7 };
        int n = arr.length;
 
        int qs = 0; // Starting index of query range
        int qe = 9; // Ending index of query range
 
        // Print occurrence of most frequent number
        // within given range
        System.out.println("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));
 
        qs = 4; // Starting index of query range
        qe = 9; // Ending index of query range
 
        // Print occurrence of most frequent number
        // within given range
        System.out.println("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));
    }
 
}

Python3

# Python 3 Program to find the occurrence
# of the most frequent number within
# a given range
from collections import defaultdict
import math
 
# A utility function to get the middle index
# from corner indexes.
def getMid(s,  e):
    return s + (e - s) // 2
 
''' A recursive function to get the maximum value in
    a given range of array indexes. The following
    are parameters for this function.
 
    st --> Pointer to segment tree
    index --> Index of current node in the segment
            tree. Initially 0 is passed as root is
            always at index 0
    ss & se --> Starting and ending indexes of the
                segment represented by current node,
                i.e., st[index]
    qs & qe --> Starting and ending indexes of query
                range '''
def RMQUtil(st,  ss,  se,  qs,  qe, index):
   
    # If segment of this node is a part of given range
    # then return the min of the segment
    if (qs <= ss and qe >= se):
        return st[index]
       
    # If segment of this node is outside the
    # given range
    if (se < qs or ss > qe):
        return 0
 
    # If a part of this segment overlaps
    # with the given range
    mid = getMid(ss, se)
    return max(RMQUtil(st, ss, mid, qs, qe, 2 * index + 1),
               RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2))
 
 
# Return minimum of elements in range from
# index qs (query start) to
# qe (query end). It mainly uses RMQUtil()
def RMQ(st,  n,  qs,  qe):
 
    # Check for erroneous input values
    if (qs < 0 or qe > n - 1 or qs > qe):
        prf("Invalid Input")
        return -1
 
    return RMQUtil(st, 0, n - 1, qs, qe, 0)
 
# A recursive function that constructs Segment Tree
# for array[ss..se]. si is index of current node in
# segment tree st
def constructSTUtil(arr,  ss,  se,  st,
                    si):
   
    # If there is one element in array, store it in
    # current node of segment tree and return
    if (ss == se):
        st[si] = arr[ss]
        return arr[ss]
       
    # If there are more than one elements, then
    # recur for left and right subtrees and store
    # the minimum of two values in this node
    mid = getMid(ss, se)
    st[si] = max(constructSTUtil(arr, ss, mid, st, si * 2 + 1),
                 constructSTUtil(arr, mid + 1, se, st, si * 2 + 2))
    return st[si]
 
''' Function to construct segment tree from given
array. This function allocates memory for segment
tree and calls constructSTUtil() to fill the
allocated memory '''
def constructST(arr, n):
 
    # Allocate memory for segment tree
    # Height of segment tree
    x = (math.ceil(math.log2(n)))
     
    # Maximum size of segment tree
    max_size = 2 * pow(2, x) - 1
    st = [0]*max_size
 
    # Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0)
     
    # Return the constructed segment tree
    return st
 
def maximumOccurrence(arr,  n,  qs,  qe):
 
    # Declaring a frequency array
    freq_arr = [0]*(n + 1)
     
    # Counting frequencies of all array elements.
    cnt = defaultdict(int)
    for i in range(n):
        cnt[arr[i]] += 1
 
    # Creating frequency array by replacing the
    # number in array to the number of times it
    # has appeared in the array
    for i in range(n):
        freq_arr[i] = cnt[arr[i]]
 
    # Build segment tree from this frequency array
    st = constructST(freq_arr, n)
    maxOcc = 0  # to store the answer
    # Case 1: numbers are same at the starting
    # and ending index of the query
    if (arr[qs] == arr[qe]):
        maxOcc = (qe - qs + 1)
         
    # Case 2: numbers are different
    else:
        leftmost_same = 0
        righmost_same = 0
         
        # Partial Overlap Case of a number with some
        # occurrences lying inside the leftmost
        # part of the range and some just before the
        # range starts
        while (qs > 0 and qs <= qe and arr[qs] == arr[qs - 1]):
            qs += 1
            leftmost_same += 1
 
        # Partial Overlap Case of a number with some
    # occurrences lying inside the rightmost part of
        # the range and some just after the range ends
        while (qe >= qs and qe < n - 1 and arr[qe] == arr[qe + 1]):
            qe -= 1
            righmost_same += 1
 
        # Taking maximum of all three
        maxOcc = max([leftmost_same, righmost_same,
                      RMQ(st, n, qs, qe)])
 
        return maxOcc
 
# Driver Code
if __name__ == "__main__":
 
    arr = [-5, -5, 2, 2, 2, 2, 3, 7, 7, 7]
    n = len(arr)
 
    qs = 0  # Starting index of query range
    qe = 9  # Ending index of query range
 
    # Print occurrence of most frequent number
    # within given range
    print("Maximum Occurrence in range is = ",
          maximumOccurrence(arr, n, qs, qe))
 
    qs = 4  # Starting index of query range
    qe = 9  # Ending index of query range
 
    # Print occurrence of most frequent number
    # within given range
    print("Maximum Occurrence in range is = ",
          maximumOccurrence(arr, n, qs, qe))
 
    # This code is contributed by ukasp.

C#

// C# code addition
 
using System;
using System.Collections.Generic;
 
class GFG
{
    // A utility function to get the middle index
    // from corner indexes.
    public static int GetMid(int s, int e)
    {
        return s + (e - s) / 2;
    }
 
    /* A recursive function to get the maximum value in
    a given range of array indexes. The following
    are parameters for this function.
 
    st --> Pointer to segment tree
    index --> Index of current node in the segment
    tree. Initially 0 is passed as root is
    always at index 0
    ss & se --> Starting and ending indexes of the
                segment represented by current node,
                i.e., st[index]
    qs & qe --> Starting and ending indexes of query
                range */
    public static int RMQUtil(int[] st, int ss, int se, int qs, int qe,
                            int index)
    {
        // If segment of this node is a part of given range,
        // then return the min of the segment
        if (qs <= ss && qe >= se)
            return st[index];
 
        // If segment of this node is outside the
        // given range
        if (se < qs || ss > qe)
            return 0;
 
        // If a part of this segment overlaps
        // with the given range
        int mid = GetMid(ss, se);
        return Math.Max(RMQUtil(st, ss, mid, qs, qe, 2 * index + 1),
                RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2));
    }
 
    // Return minimum of elements in range from
    // index qs (query start) to
    // qe (query end). It mainly uses RMQUtil()
    public static int RMQ(int[] st, int n, int qs, int qe)
    {
        // Check for erroneous input values
        if (qs < 0 || qe > n - 1 || qs > qe)
        {
            Console.WriteLine("Invalid Input");
            return -1;
        }
 
        return RMQUtil(st, 0, n - 1, qs, qe, 0);
    }
 
    // A recursive function that constructs Segment Tree
    // for array[ss..se]. si is index of current node in
    // segment tree st
    public static int ConstructSTUtil(int[] arr, int ss, int se, int[] st,
                                            int si)
    {
        // If there is one element in array, store it in
        // current node of segment tree and return
        if (ss == se)
        {
            st[si] = arr[ss];
            return arr[ss];
        }
 
        // If there are more than one elements, then
        // recur for left and right subtrees and store
        // the minimum of two values in this node
        int mid = GetMid(ss, se);
        st[si] = Math.Max(ConstructSTUtil(arr, ss, mid, st, si * 2 + 1),
                ConstructSTUtil(arr, mid + 1, se, st, si * 2 + 2));
        return st[si];
    }
 
    /* Function to construct segment tree from given
    array. This function allocates memory for segment
    tree and calls constructSTUtil() to fill the
    allocated memory */
    public static int[] ConstructST(int[] arr, int n)
    {
        // Allocate memory for segment tree
 
        // Height of segment tree
        int x =  (int)Math.Ceiling(Math.Log(n, 2));
 
        // Maximum size of segment tree
        int max_size = 2 * (int)Math.Pow(2, x) - 1;
 
        int[] st = new int[max_size];
 
        // Fill the allocated memory st
        ConstructSTUtil(arr, 0, n - 1, st, 0);
 
        // Return the constructed segment tree
        return st;
    }
 
    public static int maximumOccurrence(int[] arr, int n, int qs, int qe)
    {
        // Declaring a frequency array
        int[] freq_arr = new int[n + 1];
 
        // Counting frequencies of all array elements.
        Dictionary<int,int> cnt = new Dictionary<int,int>();
 
        for (int i = 0; i < n; i++){
            if(!cnt.ContainsKey(arr[i]))
           { 
               cnt.Add(arr[i], 0);
           }
           cnt[arr[i]]++;
        }
 
        // Creating frequency array by replacing the
        // number in array to the number of times it
        // has appeared in the array
        for (int i = 0; i < n; i++)
            freq_arr[i] = cnt[arr[i]];
 
        // Build segment tree from this frequency array
        int[] st = ConstructST(freq_arr, n);
 
        int maxOcc; // to store the answer
 
        // Case 1: numbers are same at the starting
        // and ending index of the query
        if (arr[qs] == arr[qe])
            maxOcc = (qe - qs + 1);
 
        // Case 2: numbers are different
        else {
            int leftmost_same = 0, righmost_same = 0;
 
            // Partial Overlap Case of a number with some
            // occurrences lying inside the leftmost
            // part of the range and some just before the
            // range starts
            while (qs > 0 && qs <= qe && arr[qs] == arr[qs - 1]) {
                qs++;
                leftmost_same++;
            }
 
            // Partial Overlap Case of a number with some
            // occurrences lying inside the rightmost part of
            // the range and some just after the range ends
            while (qe >= qs && qe < n - 1 && arr[qe] == arr[qe + 1]) {
                qe--;
                righmost_same++;
            }
            // Taking maximum of all three
            maxOcc = Math.Max(Math.Max(leftmost_same, righmost_same),
                                    RMQ(st, n, qs, qe));
        }
        return maxOcc;
    }
     
   //driver code
   static void Main() {
        int[] arr = { -5, -5, 2, 2, 2, 2, 3, 7, 7, 7 };
        int n = arr.Length;
 
        int qs = 0; // Starting index of query range
        int qe = 9; // Ending index of query range
 
        // Print occurrence of most frequent number
        // within given range
        Console.WriteLine("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));
 
        qs = 4; // Starting index of query range
        qe = 9; // Ending index of query range
 
        // Print occurrence of most frequent number
        // within given range
        Console.WriteLine("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));
    }
 
}
 
// the code is contributed by Nidhi goel.

Javascript

function getMid(s, e) {
    /* A recursive function to get the maximum value in
    a given range of array indexes. The following
    are parameters for this function.
 
    st --> Pointer to segment tree
    index --> Index of current node in the segment
            tree. Initially 0 is passed as root is
            always at index 0
    ss & se --> Starting and ending indexes of the
                segment represented by current node,
                i.e., st[index]
    qs & qe --> Starting and ending indexes of query
                range */
    return s + Math.floor((e - s) / 2);
}
 
function RMQUtil(st, ss, se, qs, qe, index) {
    // If segment of this node is a part of given range,
    // then return the min of the segment
    if (qs <= ss && qe >= se)
        return st[index];
 
    // If segment of this node is outside the
    // given range
    if (se < qs || ss > qe)
        return 0;
 
    // If a part of this segment overlaps
    // with the given range
    let mid = getMid(ss, se);
    return Math.max(RMQUtil(st, ss, mid, qs, qe, 2 * index + 1), RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2));
}
// Return minimum of elements in range from
// index qs (query start) to
// qe (query end). It mainly uses RMQUtil()
function RMQ(st, n, qs, qe) {
 
    // Check for erroneous input values
    if (qs < 0 || qe > n - 1 || qs > qe) {
        console.log("Invalid Input");
        return -1;
    }
    return RMQUtil(st, 0, n - 1, qs, qe, 0);
}
// A recursive function that constructs Segment Tree
// for array[ss..se]. si is index of current node in
// segment tree st
function constructSTUtil(arr, ss, se, st, si) {
    // If there is one element in array, store it in
    // current node of segment tree and return
    if (ss == se) {
        st[si] = arr[ss];
        return arr[ss];
    }
    // If there are more than one elements, then
    // recur for left and right subtrees and store
    // the minimum of two values in this node
    let mid = getMid(ss, se);
    st[si] = Math.max(constructSTUtil(arr, ss, mid, st, si * 2 + 1), constructSTUtil(arr, mid + 1, se, st, si * 2 + 2));
    return st[si];
}
/* Function to construct segment tree from given
array. This function allocates memory for segment
tree and calls constructSTUtil() to fill the
allocated memory */
function constructST(arr, n) {
    // Allocate memory for segment tree
 
    // Height of segment tree
    let x = Math.ceil(Math.log2(n));
    // Maximum size of segment tree
    let max_size = 2 * Math.pow(2, x) - 1;
    let st = new Array(max_size);
    constructSTUtil(arr, 0, n - 1, st, 0);
    return st;
}
 
function maximumOccurrence(arr, n, qs, qe) {
    // Declaring a frequency array
    let freq_arr = new Array(n + 1);
    // Counting frequencies of all array elements.
    let cnt = new Map();
 
    for (let i = 0; i < n; i++)
        cnt.set(arr[i], (cnt.get(arr[i]) || 0) + 1);
    for (let i = 0; i < n; i++)
        freq_arr[i] = cnt.get(arr[i]);
    let st = constructST(freq_arr, n);
    let maxOcc;
    // to store the answer
    if (arr[qs] == arr[qe])
        maxOcc = (qe - qs + 1);
    else {
        let leftmost_same = 0,
            righmost_same = 0;
        // Partial Overlap Case of a number with some
        // occurrences lying inside the leftmost
        // part of the range and some just before the
        // range starts
        while (qs > 0 && qs <= qe && arr[qs] == arr[qs - 1]) {
            qs++;
            leftmost_same++;
        }
        // Partial Overlap Case of a number with some
        // occurrences lying inside the rightmost part of
        // the range and some just after the range ends
        while (qe >= qs && qe < n - 1 && arr[qe] == arr[qe + 1]) {
            qe--;
            righmost_same++;
        }
        maxOcc = Math.max(Math.max(leftmost_same, righmost_same), RMQ(st, n, qs, qe));
    }
    return maxOcc;
}
 
let arr = [-5, -5, 2, 2, 2, 2, 3, 7, 7, 7];
let n = arr.length;
let qs = 0; // Starting index of query range
let qe = 9; // Ending index of query range
console.log("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));
qs = 4; // Starting index of query range
qe = 9; // Ending index of query range
console.log("Maximum Occurrence in range is = " + maximumOccurrence(arr, n, qs, qe));

Output

Maximum Occurrence in range is = 4
Maximum Occurrence in range is = 3

Further Optimization: For the partial overlapping case we have to run a loop to calculate the count of same numbers on both sides. To avoid this loop and perform this operation in O(1), we can store the index of the first occurrence of every number in the given array and hence by doing some precomputation we can find the required count in O(1).

Time Complexity: Time Complexity for tree construction is O(n). Time complexity to query is O(Log n).

Auxiliary Space: O(n)