Hello Everyone,
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.
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 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++ 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;
}
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).