Given an integer K and an array arr[] containing N integers, the task is to find the sum and product of K smallest and K largest fibonacci numbers in the array.
Note: Assume that there are at least K fibonacci numbers in the array.
Examples:
Input: arr[] = {2, 5, 6, 8, 10, 11}, K = 2
Output:
Sum of K-minimum fibonacci numbers is 7
Product of K-minimum fibonacci numbers is 10
Sum of K-maximum fibonacci numbers is 13
Product of K-maximum fibonacci numbers is 40
Explanation :
{2, 5, 8} are the only fibonacci numbers from the array.
{2, 5} are the 2 smallest and {5, 8} are the 2 largest among them.Input: arr[] = {3, 2, 12, 13, 5, 19}, K = 3
Output:
Sum of K-minimum fibonacci numbers is 10
Product of K-minimum fibonacci numbers is 30
Sum of K-maximum fibonacci numbers is 21
Product of K-maximum fibonacci numbers is 195
Approach: The idea is to use hashing to precompute and store the Fibonacci nodes up to the maximum value, in a Set, to make checking easy and efficient (in O(1) time).
- Traverse through the entire array and obtain the maximum value in the list.
- Now, build a hash table containing all the Fibonacci nodes less than or equal to the maximum value of the array.
After performing the above precomputation, traverse the array and insert all the numbers which are fibonacci in two heaps, a min heap and a max heap.
Now, pop out top K elements from the min heap and max heap to compute the sum and product of the K Fibonacci numbers.
Below is the implementation of the above approach:
// C++ program to find the sum and// product of K smallest and K// largest Fibonacci numbers in an array #include <bits/stdc++.h>using namespace std; // Function to create the hash table// to check Fibonacci numbersvoid createHash(set<int>& hash, int maxElement){ // Inserting the first two elements // into the hash int prev = 0, curr = 1; hash.insert(prev); hash.insert(curr); // Computing the remaining // elements using // the previous two elements while (curr <= maxElement) { int temp = curr + prev; hash.insert(temp); prev = curr; curr = temp; }} // Function that calculates the sum// and the product of K smallest and// K largest Fibonacci numbers in an arrayvoid fibSumAndProduct(int arr[], int n, int k){ // Find the maximum value in the array int max_val = *max_element(arr, arr + n); // Creating a hash containing // all the Fibonacci numbers // upto the maximum data value // in the array set<int> hash; createHash(hash, max_val); // Max Heap to store all the // Fibonacci numbers priority_queue<int> maxHeap; // Min Heap to store all the // Fibonacci numbers priority_queue<int, vector<int>, greater<int> > minHeap; // Push all the fibonacci numbers // from the array to the heaps for (int i = 0; i < n; i++) if (hash.find(arr[i]) != hash.end()) { minHeap.push(arr[i]); maxHeap.push(arr[i]); } long long int minProduct = 1, maxProduct = 1, minSum = 0, maxSum = 0; // Finding the K minimum // and the K maximum // elements from the heaps while (k--) { // Calculate the products minProduct *= minHeap.top(); maxProduct *= maxHeap.top(); // Calculate the sum minSum += minHeap.top(); maxSum += maxHeap.top(); // Pop the current // minimum element minHeap.pop(); // Pop the current // maximum element maxHeap.pop(); } cout << "Sum of K-minimum " << "fibonacci numbers is " << minSum << "\n"; cout << "Product of K-minimum " << "fibonacci numbers is " << minProduct << "\n"; cout << "Sum of K-maximum " << "fibonacci numbers is " << maxSum << "\n"; cout << "Product of K-maximum " << "fibonacci numbers is " << maxProduct;} // Driver codeint main(){ int arr[] = { 2, 5, 6, 8, 10, 11 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 2; fibSumAndProduct(arr, N, K); return 0;} |
Sum of K-minimum fibonacci numbers is 7 Product of K-minimum fibonacci numbers is 10 Sum of K-maximum fibonacci numbers is 13 Product of K-maximum fibonacci numbers is 40
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