Find a sequence of distinct integers whose Harmonic Mean is N
Last Updated :
09 Oct, 2023
Given an integer N, the task is to find a sequence of distinct integers whose Harmonic Mean is N itself. In case such a sequence doesn’t exist, print -1.
Examples:
Input: N = 5
Output: {3, 4, 5, 6, 20}
Explanation: Given sequence is the correct answer because 5/(1/3​+1/4+1/5​+1/6​+1/20)​=5. Note that there are other possible sequences as well. For example – {2, 6, 12, 20, 5}.
Input: N=3
Output: {2, 3, 6}
Approach: To solve the problem follow the below observations:
Observations:
- For N = 1, the sequence {1} satisfies the conditions.
- For N = 2, there is no possible sequence satisfying the given conditions. So, we’ll consider the case for N > 2.
- We know that 1/X(X+1) = 1/X – 1/(X+1). Using this fact, we can see that the sequence {1*2, 2*3, …, (N-1)*N, N} would satisfy the required conditions. (Reason: {1*2, 2*3, …, (N-1)*N, N} = {2, 6, 12, 20…..(N-1)*N, N}.
- The harmonic mean of this sequence:
- N/(1/(1*2) + 1/(2*3) + 1/(3*4) +…+ 1/(N-1)*N) + 1/N)
- N/(1-1/2 + 1/2-1/3 + 1/3-1/4) +…+ 1/(N-1)-1/N + 1/N)
- N/1 = N
- However, for the cases when N can be represented as N=X*(X-1), for some integer X, it can be seen that the above observation fails as the sequence would contain a duplicate.
For example, for N=6 (6=2*3), the sequence would be {2, 6, 12, 20, 30, 6}. Here 6 is repeated, but we want all distinct integers in the sequence.
- Suppose the sequence from observation 1 is {A1, A2, A3….AN}. To avoid repeating numbers when N is even (as N = X*(X-1) will be possible only when N is even), we’ll print the sequence {2, 2*A1, 2*A2,….,2*A(N-2), 2*(N-1)}.
The following steps can be used to solve the problem :
- For N = 1, print 1.
- For N = 2, print -1.
- For odd N, print the sequence {1*2, 2*3, …, (N-1)*N, N}.
- For even N, print the sequence {2, 2*(1*2, 2*3, …, (N-2)(N-1), N-1}.
Following is the code based on the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printSequence(int N)
{
if (N == 2) {
cout << -1;
return;
}
if (N == 1) {
cout << 1;
return;
}
if (N % 2) {
for (int i = 1; i <= N - 1; i++) {
cout << i * (i + 1) << " ";
}
cout << N << "\n";
}
else {
cout << 2 << " ";
for (int i = 1; i <= N - 2; i++) {
cout << 2 * i * (i + 1) << " ";
}
cout << 2 * (N - 1) << "\n";
}
}
int main()
{
int N = 5;
printSequence(N);
}
|
Java
import java.util.*;
public class GFG {
static void printSequence(int N) {
if (N == 2) {
System.out.println(-1);
return;
}
if (N == 1) {
System.out.println(1);
return;
}
if (N % 2 != 0) {
for (int i = 1; i <= N - 1; i++) {
System.out.print(i * (i + 1) + " ");
}
System.out.println(N);
}
else {
System.out.print(2 + " ");
for (int i = 1; i <= N - 2; i++) {
System.out.print(2 * i * (i + 1) + " ");
}
System.out.println(2 * (N - 1));
}
}
public static void main(String[] args) {
int N = 5;
printSequence(N);
}
}
|
Python3
def print_sequence(N):
if N == 2:
print(-1)
return
if N == 1:
print(1)
return
if N % 2 != 0:
for i in range(1, N):
print(i * (i + 1), end=" ")
print(N)
else:
print(2, end=" ")
for i in range(1, N - 1):
print(2 * i * (i + 1), end=" ")
print(2 * (N - 1))
if __name__ == "__main__":
N = 5
print_sequence(N)
|
C#
using System;
public class GFG {
static void PrintSequence(int N)
{
if (N == 2) {
Console.WriteLine(-1);
return;
}
if (N == 1) {
Console.WriteLine(1);
return;
}
if (N % 2 != 0) {
for (int i = 1; i <= N - 1; i++) {
Console.Write(i * (i + 1) + " ");
}
Console.WriteLine(N);
}
else {
Console.Write(2 + " ");
for (int i = 1; i <= N - 2; i++) {
Console.Write(2 * i * (i + 1) + " ");
}
Console.WriteLine(2 * (N - 1));
}
}
static public void Main()
{
int N = 5;
PrintSequence(N);
}
}
|
Javascript
function printSequence(N) {
if (N === 2) {
console.log(-1);
return;
}
if (N === 1) {
console.log(1);
return;
}
if (N % 2) {
for (let i = 1; i <= N - 1; i++) {
process.stdout.write(i * (i + 1) + " ");
}
console.log(N);
}
else {
process.stdout.write(2 + " ");
for (let i = 1; i <= N - 2; i++) {
process.stdout.write(2 * i * (i + 1) + " ");
}
console.log(2 * (N - 1));
}
}
const N = 5;
printSequence(N);
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Please Login to comment...