# Find element position in given monotonic sequence

Given an integer k and a monotonic increasing sequence:
f(n) = an + bn [log2(n)] + cn^3 where (a = 1, 2, 3, …), (b = 1, 2, 3, …), (c = 0, 1, 2, 3, …)
Here, [log2(n)] means, taking the log to the base 2 and round the value down Thus,
if n = 1, the value is 0.
if n = 2-3, the value is 1.
if n = 4-7, the value is 2.
if n = 8-15, the value is 3.
The task is to find the value n such that f(n) = k, if k doesn’t belong to the sequence then print 0.
Note: Values are in such a way that they can be expressed in 64 bits and the three integers a, b and c do not exceed 100.

Examples:

Input: a = 2, b = 1, c = 1, k = 12168587437017
Output: 23001
f(23001) = 12168587437017

Input: a = 7, b = 3, c = 0, k = 119753085330
Output: 1234567890

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

Naive Approach: Given values of a, b, c, find values of f(n) for every value of n and compare it.

Efficient Approach: Use Binary Search, choose n = (min + max) / 2 where min and max are the minimum and maximum values possible for n then,

• If f(n) < k then increment n.
• If f(n) > k then decrement n.
• If f(n) = k then n is the required answer.
• Repeat the above steps until the required value is found or it is not possible in the sequence.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `#define SMALL_N 1000000 ` `#define LARGE_N  1000000000000000 ` `using` `namespace` `std; ` ` `  `// Function to return the value of f(n) for given values of a, b, c, n ` `long` `long` `func(``long` `long` `a, ``long` `long` `b, ``long` `long` `c, ``long` `long` `n) ` `{ ` `    ``long` `long` `res = a * n; ` `    ``long` `long` `logVlaue = ``floor``(log2(n)); ` `    ``res += b * n * logVlaue; ` `    ``res += c * (n * n * n); ` `    ``return` `res; ` `} ` ` `  `long` `long` `getPositionInSeries(``long` `long` `a, ``long` `long` `b,  ` `                             ``long` `long` `c, ``long` `long` `k) ` `{ ` `    ``long` `long` `start = 1, end = SMALL_N; ` ` `  `    ``// if c is 0, then value of n can be in order of 10^15. ` `    ``// if c!=0, then n^3 value has to be in order of 10^18 ` `    ``// so maximum value of n can be 10^6. ` `    ``if` `(c == 0) { ` `        ``end = LARGE_N; ` `    ``} ` `    ``long` `long` `ans = 0; ` ` `  `    ``// for efficient searching, use binary search. ` `    ``while` `(start <= end) { ` `        ``long` `long` `mid = (start + end) / 2; ` `        ``long` `long` `val = func(a, b, c, mid); ` `        ``if` `(val == k) { ` `            ``ans = mid; ` `            ``break``; ` `        ``} ` `        ``else` `if` `(val > k) { ` `            ``end = mid - 1; ` `        ``} ` `        ``else` `{ ` `            ``start = mid + 1; ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``long` `long` `a = 2, b = 1, c = 1; ` `    ``long` `long` `k = 12168587437017; ` ` `  `    ``cout << getPositionInSeries(a, b, c, k); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python 3 implementation of the approach  ` `from` `math ``import` `log2, floor ` `SMALL_N ``=` `1000000` `LARGE_N ``=` `1000000000000000` ` `  `# Function to return the value of f(n) ` `# for given values of a, b, c, n  ` `def` `func(a, b, c, n) : ` `     `  `    ``res ``=` `a ``*` `n ` `    ``logVlaue ``=` `floor(log2(n)) ` `    ``res ``+``=` `b ``*` `n ``*` `logVlaue ` `    ``res ``+``=` `c ``*` `(n ``*` `n ``*` `n)  ` `    ``return` `res ` ` `  `def` `getPositionInSeries(a, b, c, k) : ` `     `  `    ``start ``=` `1` `    ``end ``=` `SMALL_N ` ` `  `    ``# if c is 0, then value of n  ` `    ``# can be in order of 10^15.  ` `    ``# if c!=0, then n^3 value has ` `    ``# to be in order of 10^18  ` `    ``# so maximum value of n can be 10^6.  ` `    ``if` `(c ``=``=` `0``) :  ` `        ``end ``=` `LARGE_N ` `     `  `    ``ans ``=` `0` ` `  `    ``# for efficient searching,  ` `    ``# use binary search.  ` `    ``while` `(start <``=` `end) :  ` `         `  `        ``mid ``=` `(start ``+` `end) ``/``/` `2` `        ``val ``=` `func(a, b, c, mid) ` `        ``if` `(val ``=``=` `k) : ` `            ``ans ``=` `mid ` `            ``break` `     `  `        ``elif` `(val > k) :  ` `            ``end ``=` `mid ``-` `1` ` `  `        ``else` `:  ` `            ``start ``=` `mid ``+` `1` `         `  `    ``return` `ans;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``a ``=` `2` `    ``b ``=` `1` `    ``c ``=` `1` `    ``k ``=` `12168587437017` ` `  `    ``print``(getPositionInSeries(a, b, c, k))  ` ` `  `# This code is contributed by Ryuga `

## PHP

 ` ``\$k``)  ` `            ``\$end` `= ``\$mid` `- 1; ` ` `  `        ``else` `            ``\$start` `= ``\$mid` `+ 1; ` `    ``} ` `    ``return` `\$ans``;  ` `} ` ` `  `// Driver code  ` `\$a` `= 2; ` `\$b` `= 1; ` `\$c` `= 1; ` `\$k` `= 12168587437017; ` ` `  `print``(getPositionInSeries(``\$a``, ``\$b``, ``\$c``, ``\$k``));  ` ` `  `// This code is contributed by mits  ` `?> `

Output:

```23001
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, Mithun Kumar