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 : 3 Input : n = 7 Output : 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.
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 <bits/stdc++.h>using namespace std;// function to solve the quadratic equationint term(int n){ // calculating the Nth term int x = (((1) + (double)sqrt(1 + (8 * n))) / 2); return x;}// driver code to check the above functionint 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 equationdef term( n ): # calculating the Nth term x = (((1) + math.sqrt(1 + (8 * n))) / 2) return x# Driver coden = 5print(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. |
PHP
<?php// PHP program to find the// nth term of the series// 1 2 2 3 3 3 ...// function to solve the// quadratic equationfunction term($n){ // calculating the Nth term $x = (((1) + (double)sqrt(1 + (8 * $n))) / 2); return $x;}// Driver Code$n = 5;echo((int)term($n));// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript program to find the nth // term of the series 1 2 2 3 3 3 ...// Function to solve the quadratic equationfunction term(n){ // Calculating the Nth term let x = parseInt(((1) + Math.sqrt(1 + (8 * n))) / 2); return x;}// Driver code let n = 5;document.write(term(n));// This code is contributed by rishavmahato348</script> |
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 <bits/stdc++.h>using namespace std;// function to solve the quadratic equationint 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 codeint main(){ int n = 5; cout << term(n); return 0;}// This code is contributed by Susobhan Akhuli |
3
Time Complexity: O(log(n)) as sqrt function takes O(log n).
Auxiliary Space: O(1)
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