Array range queries to count the number of Fibonacci numbers with updates

Given an array arr[] of N integers, the task is to perform the following two queries:

  • query(start, end): Print the number of fibonacci numbers in the subarray from start to end
  • update(i, x): Add x to the array element referenced by array index i, that is: arr[i] = x

Examples:

Input: arr = { 1, 2, 3, 4, 8, 9 }
Query 1: query(start = 0, end = 4)
Query 2: update(i = 3, x = 5)
Query 3: query(start = 0, end = 4)
Output:4
5
Explanation
In Query 1, the subarray [0…4] has 4 fibonacci numbers viz. {1, 2, 3, 8}
In Query 2, the value at index 3 is updated to 5, the array arr now is, {1, 2, 3, 5, 8, 9}
In Query 3, the subarray [0…4] has 5 fibonacci numbers viz. {1, 2, 3, 5, 8}

Approach: To handle both point updates and range queries, a segment tree is optimal for this purpose.

In order to check for Fibonacci numbers, we can build a hash table using dynamic programming containing all the Fibonacci numbers less than or equal to the maximum value arri. We can take MAX which will be used to test a number in O(1) time complexity.



Building the segment tree:

  • The problem is now reduced to the subarray sum using segment tree problem.
  • Now, we can build the segment tree where a leaf node is represented as either 0 (if it is not a prime number) or 1 (if it is a Fibonacci number).
  • The internal nodes of the segment tree equal to the sum of its child nodes, thus a node represent the total Fibonacci numbers in the range from L to R with range [L, R] falling under this node and the sub-tree underneath it.

Handling Queries and Point Updates:

  • Whenever we receive a query from beginning to end, we can query the segment tree for the sum of nodes in range from start to end, which in turn represent the number of Fibonacci numbers in the range start to end.
  • To perform a point update and to update the value at index i to x, we check for the following cases:

    Let the old value of arri be y and the new value be x.

    1. Case 1: Fibonacci: If x and y both are Fibonacci numbers
      Count of Fibonacci numbers in the subarray does not change so we just update array and do not modify the segment tree
    2. Case 2: If x and y both are not Fibonacci numbers
      Count of Fibonacci numbers in the subarray does not change so we just update array and do not modify the segment tree
    3. Case 3: If y is a Fibonacci number but x is not
      Count of Fibonacci numbers in the subarray decreases so we update array and add -1 to every range. The index i which is to be updated is a part of in the segment tree
    4. Case 4: If y is not a Fibonacci number but x is a Fibonacci number
      Count of Fibonacci numbers in the subarray increases so we update array and add 1 to every range. The index i which is to be updated is a part of in the segment tree

Below is the implementation of the above approach:

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// C++ program to find number of fibonacci numbers
// in a subarray and performing updates
  
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 1000
  
// Function to create hash table
// to check Fibonacci numbers
void createHash(set<int>& hash,
                int maxElement)
{
    int prev = 0, curr = 1;
    hash.insert(prev);
    hash.insert(curr);
  
    while (curr <= maxElement) {
        int temp = curr + prev;
        hash.insert(temp);
        prev = curr;
        curr = temp;
    }
}
  
// A utility function to get the middle
// index from corner indexes.
int getMid(int s, int e)
{
    return s + (e - s) / 2;
}
  
// Recursive function to get the number
// of Fibonacci numbers in a given range
/* where
    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 queryFibonacciUtil(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 number of Fibonacci numbers
    // in 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 queryFibonacciUtil(st, ss, mid, qs,
                              qe, 2 * index + 1)
           + queryFibonacciUtil(st, mid + 1, se,
                                qs, qe, 2 * index + 2);
}
  
// Recursive function to update
// the nodes which have the given
// index in their range.
/* where
    st, si, ss and se are same as getSumUtil()
    i --> index of the element to be updated. 
          This index is in input array.
   diff --> Value to be added to all nodes
          which have i in range 
*/
void updateValueUtil(int* st, int ss,
                     int se, int i,
                     int diff, int si)
{
    // Base Case:
    // If the input index lies outside
    // the range of this segment
    if (i < ss || i > se)
        return;
  
    // If the input index is in range
    // of this node, then update the value
    // of the node and its children
    st[si] = st[si] + diff;
    if (se != ss) {
  
        int mid = getMid(ss, se);
        updateValueUtil(st, ss, mid, i,
                        diff, 2 * si + 1);
        updateValueUtil(st, mid + 1, se,
                        i, diff, 2 * si + 2);
    }
}
  
// Function to update a value in the
// input array and segment tree.
// It uses updateValueUtil() to update
// the value in segment tree
void updateValue(int arr[], int* st,
                 int n, int i,
                 int new_val,
                 set<int> hash)
{
    // Check for erroneous input index
    if (i < 0 || i > n - 1) {
        printf("Invalid Input");
        return;
    }
  
    int diff, oldValue;
  
    oldValue = arr[i];
  
    // Update the value in array
    arr[i] = new_val;
  
    // Case 1: Old and new values
    // both are Fibonacci numbers
    if (hash.find(oldValue) != hash.end()
        && hash.find(new_val) != hash.end())
        return;
  
    // Case 2: Old and new values
    // both not Fibonacci numbers
    if (hash.find(oldValue) == hash.end()
        && hash.find(new_val) == hash.end())
        return;
  
    // Case 3: Old value was Fibonacci,
    // new value is non Fibonacci
    if (hash.find(oldValue) != hash.end()
        && hash.find(new_val) == hash.end()) {
        diff = -1;
    }
  
    // Case 4: Old value was non Fibonacci,
    // new_val is Fibonacci
    if (hash.find(oldValue) == hash.end()
        && hash.find(new_val) != hash.end()) {
        diff = 1;
    }
  
    // Update the values of nodes in segment tree
    updateValueUtil(st, 0, n - 1, i, diff, 0);
}
  
// Return number of Fibonacci numbers
// in range from index qs (query start)
// to qe (query end).
// It mainly uses queryFibonacciUtil()
void queryFibonacci(int* st, int n,
                    int qs, int qe)
{
    int FibonacciInRange
        = queryFibonacciUtil(st, 0, n - 1,
                             qs, qe, 0);
  
    cout << "Number of Fibonacci numbers "
         << "in subarray from "
         << qs << " to "
         << qe << " = "
         << FibonacciInRange << "\n";
}
  
// 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, set<int> hash)
{
    // If there is one element in array,
    // check if it is Fibonacci number
    // then store 1 in the segment tree
    // else store 0 and return
    if (ss == se) {
  
        // if arr[ss] is fibonacci number
        if (hash.find(arr[ss]) != hash.end())
            st[si] = 1;
        else
            st[si] = 0;
  
        return st[si];
    }
  
    // If there are more than one elements,
    // then recur for left and right subtrees
    // and store the sum of the
    // two values in this node
    int mid = getMid(ss, se);
    st[si] = constructSTUtil(arr, ss, mid, st,
                             si * 2 + 1, hash)
             + constructSTUtil(arr, mid + 1, se, st,
                               si * 2 + 2, hash);
    return st[si];
}
  
// Function to construct a 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, set<int> hash)
{
    // 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, hash);
  
    // Return the constructed segment tree
    return st;
}
  
// Driver Code
int main()
{
  
    int arr[] = { 1, 2, 3, 4, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // find the largest node value in the array
    int maxEle = *max_element(arr, arr + n);
  
    // Creating a set containing all Fibonacci numbers
    // upto the maximum data value in the array
    set<int> hash;
    createHash(hash, maxEle);
  
    // Build segment tree from given array
    int* st = constructST(arr, n, hash);
  
    // Query 1: Query(start = 0, end = 4)
    int start = 0;
    int end = 4;
    queryFibonacci(st, n, start, end);
  
    // Query 2: Update(i = 3, x = 5),
    // i.e Update a[i] to x
    int i = 3;
    int x = 5;
    updateValue(arr, st, n, i, x, hash);
  
    // uncomment to see array after update
    // for(int i = 0; i < n; i++)
    // cout << arr[i] << " ";
  
    // Query 3: Query(start = 0, end = 4)
    start = 0;
    end = 4;
    queryFibonacci(st, n, start, end);
  
    return 0;
}

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Output:

Number of Fibonacci numbers in subarray from 0 to 4 = 4
Number of Fibonacci numbers in subarray from 0 to 4 = 5

Time Complexity: The time complexity of each query and update is O(log n) and that of building the segment tree is O(n)

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