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Count of numbers whose difference with Fibonacci count upto them is atleast K

  • Last Updated : 02 Jun, 2021

Prerequisites: Binary Search
Given two positive integers N and K, the task is to count all the numbers that satisfy the following conditions: 
If the number is num
 

  • num ≤ N.
  • abs(num – count) ≥ K where count is the count of fibonacci numbers upto num.

Examples: 
 

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Input: N = 10, K = 3 
Output:
Explanation: 
9 and 10 are the valid numbers which satisfy the given conditions. 
For 9, the difference between 9 and fibonacci numbers upto 9 (0, 1, 2, 3, 5, 8) is i.e. 9 – 6 = 3. 
For 10, the difference between 9 and fibonacci numbers upto 10 (0, 1, 2, 3, 5, 8) is i.e. 10 – 6 = 4.
Input: N = 30, K = 7 
Output: 11 
 



 

Observation: On observing carefully, the function which is the difference of the number and count of fibonacci numbers upto that number is a monotonically increasing function for a particular K. Also, if a number X is a valid number then X + 1 will also be a valid number.
Proof: 
 

  • Let the function Ci denotes the count of fibonacci numbers upto number i.
  • Now, for the number X + 1 the difference is X + 1 – CX + 1 which is greater than or equal to the difference X – CX for the number X, i.e. (X + 1 – CX + 1) ≥ (X – CX).
  • Thus, if (X – CX) ≥ S, then (X + 1 – CX + 1) ≥ S.

Approach: Therefore, from the above observation, the idea is to use hashing to precompute and store the Fibonacci nodes up to the maximum value and create a prefix array using the prefix sum array concept where every index stores the number of Fibonacci numbers less than ‘i’ to make checking easy and efficient (in O(1) time).
Now, we can use binary search to find the minimum valid number X, as all the numbers in range [X, N] are valid. Therefore, the answer would be N – X + 1.
Below is the implementation of the above approach:
 
 

C++




// C++ program to find the count
// of numbers whose difference with
// Fibonacci count upto them is atleast K
 
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000005;
 
// fibUpto[i] denotes the count of
// fibonacci numbers upto i
int fibUpto[MAX + 1];
 
// Function to compute all the Fibonacci
// numbers and update fibUpto array
void compute(int sz)
{
    bool isFib[sz + 1];
    memset(isFib, false, sizeof(isFib));
 
    // Store the first two Fibonacci numbers
    int prev = 0, curr = 1;
    isFib[prev] = isFib[curr] = true;
 
    // Compute the Fibonacci numbers
    // and store them in isFib array
    while (curr <= sz) {
        int temp = curr + prev;
        isFib[temp] = true;
        prev = curr;
        curr = temp;
    }
 
    // Compute fibUpto array
    fibUpto[0] = 1;
    for (int i = 1; i <= sz; i++) {
        fibUpto[i] = fibUpto[i - 1];
        if (isFib[i])
            fibUpto[i]++;
    }
}
 
// Function to return the count
// of valid numbers
int countOfNumbers(int N, int K)
{
 
    // Compute fibUpto array
    compute(N);
 
    // Binary search to find the minimum
    // number that follows the condition
    int low = 1, high = N, ans = 0;
    while (low <= high) {
        int mid = (low + high) >> 1;
 
        // Check if the number is
        // valid, try to reduce it
        if (mid - fibUpto[mid] >= K) {
            ans = mid;
            high = mid - 1;
        }
        else
            low = mid + 1;
    }
 
    // Ans is the minimum valid number
    return (ans ? N - ans + 1 : 0);
}
 
// Driver Code
int main()
{
    int N = 10, K = 3;
 
    cout << countOfNumbers(N, K);
}

Java




// Java program to find the count
// of numbers whose difference with
// Fibonacci count upto them is atleast K
import java.util.*;
 
class GFG{
  
static int MAX = 1000005;
  
// fibUpto[i] denotes the count of
// fibonacci numbers upto i
static int []fibUpto = new int[MAX + 1];
  
// Function to compute all the Fibonacci
// numbers and update fibUpto array
static void compute(int sz)
{
    boolean []isFib = new boolean[sz + 1];
  
    // Store the first two Fibonacci numbers
    int prev = 0, curr = 1;
    isFib[prev] = isFib[curr] = true;
  
    // Compute the Fibonacci numbers
    // and store them in isFib array
    while (curr <= sz) {
        int temp = curr + prev;
        if(temp <= sz)
            isFib[temp] = true;
        prev = curr;
        curr = temp;
    }
  
    // Compute fibUpto array
    fibUpto[0] = 1;
    for (int i = 1; i <= sz; i++) {
        fibUpto[i] = fibUpto[i - 1];
        if (isFib[i])
            fibUpto[i]++;
    }
}
  
// Function to return the count
// of valid numbers
static int countOfNumbers(int N, int K)
{
  
    // Compute fibUpto array
    compute(N);
  
    // Binary search to find the minimum
    // number that follows the condition
    int low = 1, high = N, ans = 0;
    while (low <= high) {
        int mid = (low + high) >> 1;
  
        // Check if the number is
        // valid, try to reduce it
        if (mid - fibUpto[mid] >= K) {
            ans = mid;
            high = mid - 1;
        }
        else
            low = mid + 1;
    }
  
    // Ans is the minimum valid number
    return (ans>0 ? N - ans + 1 : 0);
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 10, K = 3;
  
    System.out.print(countOfNumbers(N, K));
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python 3 program to find the count
# of numbers whose difference with
# Fibonacci count upto them is atleast K
 
MAX = 1000005
 
# fibUpto[i] denotes the count of
# fibonacci numbers upto i
fibUpto = [0]*(MAX + 1)
 
# Function to compute all the Fibonacci
# numbers and update fibUpto array
def compute(sz):
 
    isFib = [False]*(sz + 1)
    # Store the first two Fibonacci numbers
    prev = 0
    curr = 1
    isFib[prev] = True
    isFib[curr] = True
 
    # Compute the Fibonacci numbers
    # and store them in isFib array
    while (curr <=sz):
        temp = curr + prev
        if(temp<=sz):
            isFib[temp] = True
        prev = curr
        curr = temp
 
    # Compute fibUpto array
    fibUpto[0] = 1
    for i in range( 1,sz+1):
        fibUpto[i] = fibUpto[i - 1]
        if (isFib[i]):
            fibUpto[i]+=1
 
# Function to return the count
# of valid numbers
def countOfNumbers(N, K):
 
    # Compute fibUpto array
    compute(N)
 
    # Binary search to find the minimum
    # number that follows the condition
    low , high, ans = 1, N, 0
    while (low <= high):
        mid = (low + high) >> 1
 
        # Check if the number is
        # valid, try to reduce it
        if (mid - fibUpto[mid] >= K):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1
 
    # Ans is the minimum valid number
    if(ans):
        return (N - ans + 1)
    return 0
     
# Driver Code
if __name__ == "__main__":
     
    N = 10
    K = 3
 
    print(countOfNumbers(N, K))
 
# This code is contributed by chitranayal

C#




// C# program to find the count
// of numbers whose difference with
// Fibonacci count upto them is atleast K
using System;
  
class GFG{
   
static int MAX = 1000005;
   
// fibUpto[i] denotes the count of
// fibonacci numbers upto i
static int []fibUpto = new int[MAX + 1];
   
// Function to compute all the Fibonacci
// numbers and update fibUpto array
static void compute(int sz)
{
    bool []isFib = new bool[sz + 1];
   
    // Store the first two Fibonacci numbers
    int prev = 0, curr = 1;
    isFib[prev] = isFib[curr] = true;
   
    // Compute the Fibonacci numbers
    // and store them in isFib array
    while (curr <= sz) {
        int temp = curr + prev;
        if(temp <= sz)
            isFib[temp] = true;
        prev = curr;
        curr = temp;
    }
   
    // Compute fibUpto array
    fibUpto[0] = 1;
    for (int i = 1; i <= sz; i++) {
        fibUpto[i] = fibUpto[i - 1];
        if (isFib[i])
            fibUpto[i]++;
    }
}
   
// Function to return the count
// of valid numbers
static int countOfNumbers(int N, int K)
{
   
    // Compute fibUpto array
    compute(N);
   
    // Binary search to find the minimum
    // number that follows the condition
    int low = 1, high = N, ans = 0;
    while (low <= high) {
        int mid = (low + high) >> 1;
   
        // Check if the number is
        // valid, try to reduce it
        if (mid - fibUpto[mid] >= K) {
            ans = mid;
            high = mid - 1;
        }
        else
            low = mid + 1;
    }
   
    // Ans is the minimum valid number
    return (ans>0 ? N - ans + 1 : 0);
}
   
// Driver Code
public static void Main()
{
    int N = 10, K = 3;
   
    Console.WriteLine(countOfNumbers(N, K));
}
}
  
// This code is contributed by Mohitkumar29

Javascript




<script>
// Javascript program to find the count
// of numbers whose difference with
// Fibonacci count upto them is atleast K
let MAX = 1000005;
 
// fibUpto[i] denotes the count of
// fibonacci numbers upto i
let fibUpto = new Array(MAX+1);
 
// Function to compute all the Fibonacci
// numbers and update fibUpto array
function compute(sz)
{
    let isFib = new Array(sz + 1);
    
    // Store the first two Fibonacci numbers
    let prev = 0, curr = 1;
    isFib[prev] = isFib[curr] = true;
    
    // Compute the Fibonacci numbers
    // and store them in isFib array
    while (curr <= sz) {
        let temp = curr + prev;
        if(temp <= sz)
            isFib[temp] = true;
        prev = curr;
        curr = temp;
    }
    
    // Compute fibUpto array
    fibUpto[0] = 1;
    for (let i = 1; i <= sz; i++) {
        fibUpto[i] = fibUpto[i - 1];
        if (isFib[i])
            fibUpto[i]++;
    }
}
 
// Function to return the count
// of valid numbers
function countOfNumbers(N,K)
{
    // Compute fibUpto array
    compute(N);
    
    // Binary search to find the minimum
    // number that follows the condition
    let low = 1, high = N, ans = 0;
    while (low <= high) {
        let mid = (low + high) >> 1;
    
        // Check if the number is
        // valid, try to reduce it
        if (mid - fibUpto[mid] >= K) {
            ans = mid;
            high = mid - 1;
        }
        else
            low = mid + 1;
    }
    
    // Ans is the minimum valid number
    return (ans>0 ? N - ans + 1 : 0);
}
 
// Driver Code
let N = 10, K = 3;
document.write(countOfNumbers(N, K));
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
2

 




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