Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find the number of jumps to reach X in the number line from zero in C++
Suppose we have an integer X. We have to find minimum number of jumps required to reach X from 0. The first jump made can be of length one unit and each successive jump will be exactly one unit longer than the previous jump in length. It is allowed to go either left or right in each jump. So if X = 8, then output is 4. 0 → -1 → 1→ 4 → 8 are possible stages.
If we observe carefully, then we can say that
- If you have always jumped in the right direction, then after n jumps you will be at the point p = 1 + 2 + 3 + … + n
- If we can jump to the left also, at kth jump you will be at point p – 2k.
- If we choose carefully which jump go to the left and which go to the right, after n jumps, you can be at point between n(n+1)/2 and –(n*(n+1) / 2), with the same parity as n(n+1)/2.
Example
#include<iostream>
#include<cmath>
using namespace std;
inline int sumOneToN(int n) {
return (n * (n + 1)) / 2;
}
int jumps(int n) {
n = abs(n);
int ans = 0;
while (sumOneToN(ans) < n or (sumOneToN(ans) - n) & 1)
ans++;
return ans;
}
int main() {
int n = 9;
cout << "Number of jumps: " << jumps(n);
}Output
Number of jumps: 5
Updated on: 19-Dec-2019
160 Views
Advertisements