Find element position in given monotonic sequence
Last Updated :
26 Feb, 2023
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
Naive Approach: Given values of a, b, c, find values of f(n) for every value of n and compare it.
Time Complexity: O(n)
Space Complexity: O(1)
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 <iostream>
#include <math.h>
#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;
}
// Java implementation of the approach
import java.util.*;
class GFG {
static long SMALL_N = 1000000;
static Long LARGE_N = Long.parseUnsignedLong("1000000000000000");
// Function to return the value of f(n) for given values
// of a, b, c, n
static long func(long a, long b, long c, long n)
{
long res = a * n;
long logVlaue = (long)(Math.log(n) / Math.log(2));
res += b * n * logVlaue;
res += c * (n * n * n);
return res;
}
static long getPositionInSeries(long a, long b, long c,
long k)
{
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 ans = 0;
// for efficient searching, use binary search.
while (start <= end) {
long mid = (start + end) / 2;
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
public static void main(String[] args)
{
long a = 2, b = 1, c = 1;
Long k = Long.parseUnsignedLong("12168587437017");
System.out.println(getPositionInSeries(a, b, c, k));
}
}
// This code is contributed by phasing17
# 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
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
static long SMALL_N = 1000000;
static long LARGE_N = 1000000000000000;
// Function to return the value of f(n) for given values
// of a, b, c, n
static long func(long a, long b, long c, long n)
{
long res = a * n;
long logVlaue = (long)(Math.Log(n) / Math.Log(2));
res += b * n * logVlaue;
res += c * (n * n * n);
return res;
}
static long getPositionInSeries(long a, long b, long c,
long k)
{
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 ans = 0;
// for efficient searching, use binary search.
while (start <= end) {
long mid = (start + end) / 2;
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
public static void Main(string[] args)
{
long a = 2, b = 1, c = 1;
long k = 12168587437017;
Console.WriteLine(getPositionInSeries(a, b, c, k));
}
}
// This code is contributed by phasing17
<?php
// PHP implementation of the approach
// from math import log2, floor
$SMALL_N = 1000000;
$LARGE_N = 1000000000000000;
// Function to return the value of f(n)
// for given values of a, b, c, n
function func($a, $b, $c, $n)
{
$res = $a * $n;
$logVlaue = floor(log($n, 2));
$res += $b * $n * $logVlaue;
$res += $c * ($n * $n * $n);
return $res;
}
function getPositionInSeries($a, $b, $c, $k)
{
global $SMALL_N, $LARGE_N;
$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)
{
$mid = (int)(($start + $end) / 2);
$val = func($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
$a = 2;
$b = 1;
$c = 1;
$k = 12168587437017;
print(getPositionInSeries($a, $b, $c, $k));
// This code is contributed by mits
?>
<script>
// Javascript implementation of the approach
const SMALL_N = 1000000;
const LARGE_N = 1000000000000000;
// Function to return the value of f(n)
// for given values of a, b, c, n
function func(a, b, c, n)
{
let res = a * n;
let logVlaue = Math.floor(Math.log(n) /
Math.log(2));
res += b * n * logVlaue;
res += c * (n * n * n);
return res;
}
function getPositionInSeries(a, b, c, k)
{
let 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;
}
let ans = 0;
// For efficient searching, use binary search.
while (start <= end)
{
let mid = parseInt((start + end) / 2);
let 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
let a = 2, b = 1, c = 1;
let k = 12168587437017;
document.write(getPositionInSeries(a, b, c, k));
// This code is contributed by souravmahato348
</script>
Time Complexity: O(log n), since it's a binary search where each time the elements are reduced to half.
Auxiliary Space: O(1), since no extra space has been taken.
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