Find n-th term in series 1 2 2 3 3 3 4 4 4 4….

Given series 1 2 2 3 3 3 4 4 4 4 …., find n-th term of the series. The “pattern” is obvious. There is one “1”, two “2”s three “3”s, etc.

Examples:

`Input : n = 5 Output : 3Input :  n = 7Output : 4`

A naive approach is to run two loops one from 1 to n and the other from 1 to i, and for every iteration of the inner loop keep a count, whenever the count reaches n, we can break out of both the loops and i will be our answer.

C++

 `#include ``using` `namespace` `std;` `// Function to find the nth term of the series``int` `term(``int` `n) {``    ``int` `count = 0;``    ``int` `result = 0;` `    ``// Outer loop from 1 to n``    ``for` `(``int` `i = 1; i <= n; ++i) {``        ``// Inner loop from 1 to the current value of i``        ``for` `(``int` `j = 1; j <= i; ++j) {``            ``// Increment the count for each iteration of the inner loop``            ``++count;` `            ``// If the count reaches n, update the result and break out of both loops``            ``if` `(count == n) {``                ``result = i;``                ``break``;``            ``}``        ``}` `        ``// If the count reaches n, break out of the outer loop``        ``if` `(count == n) {``            ``break``;``        ``}``    ``}` `    ``// Return the result``    ``return` `result;``}` `int` `main() {``    ``// Example usage:``    ``int` `n = 5;``    ``cout<

Java

 `public` `class` `SeriesTerm {``    ``// Function to find the nth term of the series``    ``static` `int` `term(``int` `n) {``        ``int` `count = ``0``;``        ``int` `result = ``0``;` `        ``// Outer loop from 1 to n``        ``for` `(``int` `i = ``1``; i <= n; ++i) {``            ``// Inner loop from 1 to the current value of i``            ``for` `(``int` `j = ``1``; j <= i; ++j) {``                ``// Increment the count for each iteration of the inner loop``                ``++count;` `                ``// If the count reaches n, update the result and break out of both loops``                ``if` `(count == n) {``                    ``result = i;``                    ``break``;``                ``}``            ``}` `            ``// If the count reaches n, break out of the outer loop``            ``if` `(count == n) {``                ``break``;``            ``}``        ``}` `        ``// Return the result``        ``return` `result;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int` `n = ``5``;``        ``System.out.println(term(n));``    ``}``}`

Python3

 `# Function to find the nth term of the series``def` `term(n):``    ``count ``=` `0``    ``result ``=` `0` `    ``# Outer loop from 1 to n``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``# Inner loop from 1 to the current value of i``        ``for` `j ``in` `range``(``1``, i ``+` `1``):``            ``# Increment the count for each iteration of the inner loop``            ``count ``+``=` `1` `            ``# If the count reaches n, update the result and break out of both loops``            ``if` `count ``=``=` `n:``                ``result ``=` `i``                ``break` `        ``# If the count reaches n, break out of the outer loop``        ``if` `count ``=``=` `n:``            ``break` `    ``# Return the result``    ``return` `result` `# Example usage:``n ``=` `5``print``(term(n))` `# This code is contributed by Yash Agarwal(yashagarwal2852002)`

C#

 `using` `System;` `class` `Program``{``    ``// Function to find the nth term of the series``    ``static` `int` `Term(``int` `n)``    ``{``        ``int` `count = 0;``        ``int` `result = 0;` `        ``// Outer loop from 1 to n``        ``for` `(``int` `i = 1; i <= n; ++i)``        ``{``            ``// Inner loop from 1 to the current value of i``            ``for` `(``int` `j = 1; j <= i; ++j)``            ``{``                ``// Increment the count for each iteration of the inner loop``                ``++count;` `                ``// If the count reaches n, update the result and break out of both loops``                ``if` `(count == n)``                ``{``                    ``result = i;``                    ``break``;``                ``}``            ``}` `            ``// If the count reaches n, break out of the outer loop``            ``if` `(count == n)``            ``{``                ``break``;``            ``}``        ``}` `        ``// Return the result``        ``return` `result;``    ``}` `    ``static` `void` `Main()``    ``{``        ``int` `n = 5;``        ``Console.WriteLine(Term(n));``    ``}``}`

Javascript

 `// Function to find the nth term of the series``function` `term(n) {``    ``let count = 0;``    ``let result = 0;` `    ``// Outer loop from 1 to n``    ``for` `(let i = 1; i <= n; ++i) {``        ``// Inner loop from 1 to the current value of i``        ``for` `(let j = 1; j <= i; ++j) {``            ``// Increment the count for each iteration of the inner loop``            ``++count;` `            ``// If the count reaches n, update the result and break out of both loops``            ``if` `(count === n) {``                ``result = i;``                ``break``;``            ``}``        ``}` `        ``// If the count reaches n, break out of the outer loop``        ``if` `(count === n) {``            ``break``;``        ``}``    ``}` `    ``// Return the result``    ``return` `result;``}` `// Example usage:``let n = 5;``console.log(term(n));`

Output
```3

```

Time Complexity: O(n), as we will be using a loop to traverse n times.
Auxiliary Space: O(1), as we will not be using any extra space.

Efficient Approach:

An efficient approach will be to note down a small observation which is:
The trick is to find a pattern.
Consider numbering the given sequence as follows:
1 is at position 1
2 is at positions 2, 3
3 is at positions 4, 5, 6
4 is at positions 7, 8, 9, 10
and so on…
Notice that the last positions of the individual values form a sequence.
1, 3, 6, 10, 15, 21…
If we want a formula for the “nth” term, start by looking at it the other way around. In which term does the number “n” first appear? Taking the first term to be the “0th” term. “1” appears in term 0, “2” appears in term 1, “3” appears in term 1+2=3, “4” appears in term 1+2+3= 6, etc. The number “x” first appears in term 1 + 2 + …+ (x- 2)+ (x-1) = x(x-1)/2.
So solving for the n-th term we get n = x*(x-1)/2
Solving it using quadratic equation we get

x = ( ( 1 + sqrt(1+8*n) )/2 )

n in every case is NOT an integer which means the n-th number is not the first of a sequence of the same integer, but it is clear that the n-th integer is the integer value.

C++

 `// CPP program to find the nth term of the series``// 1 2 2 3 3 3 ...``#include ``using` `namespace` `std;` `// function to solve the quadratic equation``int` `term(``int` `n)``{``    ``// calculating the Nth term``    ``int` `x = (((1) + (``double``)``sqrt``(1 + (8 * n))) / 2);``    ``return` `x;``}` `// driver code to check the above function``int` `main()``{``    ``int` `n = 5;``    ``cout << term(n);``    ``return` `0;``}`

Java

 `// Java program to find the nth``// term of the series 1 2 2 3 3 3 ...``import` `java.io.*;` `class` `Series {``    ` `    ``// function to solve the quadratic``    ``// equation``    ``static` `int` `term(``int` `n)``    ``{``        ``// calculating the Nth term``        ``int` `x = (((``1``) + (``int``)Math.sqrt(``1` `+ ``                           ``(``8` `* n))) / ``2``);``        ``return` `x;``    ``}``    ` `    ``// driver code to check the above function``    ``public` `static` `void` `main (String[] args) {``        ``int` `n = ``5``;``        ``System.out.println(term(n));``    ``}``}` `// This code is contributed by Chinmoy Lenka`

Python3

 `# Python program to find the nth term ``# of the series 1 2 2 3 3 3 ...``import` `math ` `# function to solve the quadratic equation``def` `term( n ):` `    ``# calculating the Nth term``    ``x ``=` `(((``1``) ``+` `math.sqrt(``1` `+` `(``8` `*` `n))) ``/` `2``)``    ``return` `x` `# Driver code``n ``=` `5``print``(``int``(term(n)))` `# This code is contributed by Sharad_Bhardwaj.`

C#

 `// C# program to find the nth``// term of the series 1 2 2 3 3 3 ...``using` `System;` `class` `Series``{``    ``// function to solve the quadratic``    ``// equation``    ``static` `int` `term(``int` `n)``    ``{``        ``// calculating the Nth term``        ``int` `x = (((1) + (``int``)Math.Sqrt(1 + (8 * n))) / 2);``        ``return` `x;``    ``}` `    ``// driver code to check the above function``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 5;``        ``Console.WriteLine(term(n));``    ``}``}` `// This code is contributed by vt_m.`

Javascript

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Output
```3

```

Time Complexity: O(log(n)) as sqrt function takes O(log n).
Auxiliary Space: O(1), as we are not using any extra space.

Another Efficient Approach:

Another efficient approach is to find another pattern related to first and last position index of every number.

Pattern:

Consider numbering the given sequence as follows:
1 is at position 1
2 is at positions 2, 3
3 is at positions 4, 5, 6
4 is at positions 7, 8, 9, 10
5 is at positions 11, 12, 13, 14, 15
6 is at positions 16, 17, 18, 19, 20, 21
7 is at positions 22, 23, 24, 25, 26, 27, 28
and so on…

From the above notice that the first or last positions of the individual values are either their sqrt(2*first or last position) or 1+sqrt(2*first or last position), means

X = sqrt(2n) or 1+sqrt(2n)

For the above pattern let’s do the implementation:

C++

 `// CPP program to find the nth term of the series``// 1 2 2 3 3 3 ...``#include ``using` `namespace` `std;` `// function to solve the quadratic equation``int` `term(``int` `n)``{``    ``// Calculating approx. value``    ``int` `x = ``sqrt``(2 * n);` `    ``// Condition to check if the approx. number true or not``    ``if` `(((x * x) - x + 2) <= 2 * n``        ``&& ((x * x) + x) >= 2 * n)``        ``return` `x;``    ``else``        ``return` `x + 1;``}` `// Driver's code``int` `main()``{``    ``int` `n = 5;``    ``cout << term(n);``    ``return` `0;``}` `// This code is contributed by Susobhan Akhuli`

Java

 `import` `java.util.Scanner;` `public` `class` `Main {``    ``// Function to solve the quadratic equation``    ``static` `int` `findNthTerm(``int` `n)``    ``{``        ``// Calculating the approximate value``        ``int` `x = (``int``)Math.sqrt(``2` `* n);` `        ``// Condition to check if the approximate number is``        ``// true or not``        ``if` `(((x * x) - x + ``2``) <= ``2` `* n``            ``&& ((x * x) + x) >= ``2` `* n)``            ``return` `x;``        ``else``            ``return` `x + ``1``;``    ``}` `    ``// Driver's code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Test case``        ``int` `n = ``5``;` `        ``// Displaying the nth term of the series``        ``System.out.println(findNthTerm(n));``    ``}``}`

Python3

 `import` `math` `def` `term(n):``    ``x ``=` `int``(math.sqrt(``2` `*` `n))` `    ``if` `((x ``*` `x) ``-` `x ``+` `2``) <``=` `2` `*` `n ``and` `((x ``*` `x) ``+` `x) >``=` `2` `*` `n:``        ``return` `x``    ``else``:``        ``return` `x ``+` `1` `# Driver's code``n ``=` `5``print``(term(n))`

C#

 `using` `System;` `class` `Program {``    ``// Function to solve the quadratic equation``    ``static` `int` `Term(``int` `n)``    ``{``        ``// Calculating approximate value``        ``int` `x = (``int``)Math.Sqrt(2 * n);` `        ``// Condition to check if the approximate number is``        ``// true or not``        ``if` `(((x * x) - x + 2) <= 2 * n``            ``&& ((x * x) + x) >= 2 * n)``            ``return` `x;``        ``else``            ``return` `x + 1;``    ``}` `    ``// Driver's code``    ``static` `void` `Main()``    ``{``        ``int` `n = 5;``        ``Console.WriteLine(Term(n));``    ``}``}`

Javascript

 `// Function to solve the quadratic equation``function` `term(n) {``    ``// Calculating approx. value``    ``const x = Math.sqrt(2 * n);` `    ``// Condition to check if the approx. number is true or not``    ``if` `(((x * x) - x + 2) <= 2 * n && ((x * x) + x) >= 2 * n) {``        ``return` `Math.floor(x);``    ``} ``else` `{``        ``return` `Math.floor(x) + 1;``    ``}``}` `// Driver's code``const n = 5;``console.log(term(n));`

Output
```3

```

Time Complexity: O(log(n)) as sqrt function takes O(log n).
Auxiliary Space: O(1)

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