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

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:

Input: N = 10, K = 3
Output: 2
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find the count ` `// of numbers whose difference with ` `// Fibonacci count upto them is atleast K ` ` `  `#include ` `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 = 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 = 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 `

Output:

```2
```

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