Given a positive coordinate ‘X’ and you are at coordinate ‘0’, the task is to find the minimum time required to get to coordinate ‘X’ with the following move :
At time ‘t’, you can either stay at the same position or take a jump of length exactly ‘t’ either to the left or to the right. In other words, you can be at coordinate ‘x – t’, ‘x’ or ‘x + t’ at time ‘t’ where ‘x’ is the current position.
Examples:
Input: 6
Output: 3
At time 1, jump from x = 0 to x = 1 (x = x + 1)
At time 2, jump from x = 1 to x = 3 (x = x + 2)
At time 3, jump from x = 3 to x = 6 (x = x + 3)
So, minimum required time is 3.
Input: 9
Output: 4
At time 1, do not jump i.e x = 0
At time 2, jump from x = 0 to x = 2 (x = x + 2)
At time 3, jump from x = 2 to x = 5 (x = x + 3)
At time 4, jump from x = 5 to x = 9 (x = x + 4)
So, minimum required time is 4.
Approach: The following greedy strategy works:
We just find the minimum ‘t’ such that 1 + 2 + 3 + … + t >= X.
- If (t * (t + 1)) / 2 = X then answer is ‘t’.
- Else if (t * (t + 1)) / 2 > X, then we find (t * (t + 1)) / 2 – X and remove this number from the sequence [1, 2, 3, …, t]. The resulting sequence sums up to ‘X’.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
long cal_minimum_time( long X)
{
long t = 0;
long sum = 0;
while (sum < X) {
t++;
sum = sum + t;
}
return t;
}
int main()
{
long n = 6;
long ans = cal_minimum_time(n);
cout << "The minimum time required is : " << ans ;
return 0;
}
|
Java
class GFG {
static long cal_minimum_time( long X)
{
long t = 0 ;
long sum = 0 ;
while (sum < X) {
t++;
sum = sum + t;
}
return t;
}
public static void main(String[] args)
{
long n = 6 ;
long ans = cal_minimum_time(n);
System.out.println( "The minimum time required is : " + ans);
}
}
|
Python3
def cal_minimum_time(X):
t = 0
sum = 0
while ( sum < X):
t = t + 1
sum = sum + t;
return t;
if __name__ = = '__main__' :
n = 6
ans = cal_minimum_time(n)
print ( "The minimum time required is :" , ans)
|
C#
using System;
public class GFG{
static long cal_minimum_time( long X)
{
long t = 0;
long sum = 0;
while (sum < X) {
t++;
sum = sum + t;
}
return t;
}
static public void Main (){
long n = 6;
long ans = cal_minimum_time(n);
Console.WriteLine( "The minimum time required is : " + ans);
}
}
|
Javascript
<script>
function cal_minimum_time(X)
{
let t = 0;
let sum = 0;
while (sum < X) {
t++;
sum = sum + t;
}
return t;
}
let n = 6;
let ans = cal_minimum_time(n);
document.write( "The minimum time required is : " + ans);
</script>
|
PHP
<?php
function cal_minimum_time( $X )
{
$t = 0;
$sum = 0;
while ( $sum < $X )
{
$t ++;
$sum = $sum + $t ;
}
return $t ;
}
$n = 6;
$ans = cal_minimum_time( $n );
echo "The minimum time required is : " . $ans ;
?>
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Output
The minimum time required is : 3
Time Complexity: O(n), since there is a while loop that runs for n times.
Auxiliary Space: O(1), since no extra space has been taken.
Approach#2: Using math
The problem can be solved using dynamic programming. We can start from the base case where n=0, and build the solution recursively by considering all possible jumps that can be taken at each step. We can use a memoization table to store the minimum time required to reach a point, so that we can avoid recomputing the same subproblems multiple times.
Algorithm
1. Create a memoization table dp of size n+1 and initialize all values to infinity.
2. Set dp[0] to 0.
For i from 1 to n, compute dp[i] as follows:
If i is a triangular number, set dp[i] to the square root of 2*i.
Otherwise, set dp[i] to dp[i-1] + 1.
For each jump size j from 1 to i, compute the time required to reach i-j*(j+1)/2 and add j to it.
If this time is less than dp[i], update dp[i] with this new minimum time.
3. Return dp[n].
C++
#include <iostream>
#include <cmath>
int minTime( int n) {
if (n <= 0) {
return 0;
} else if (( int (std:: sqrt (1 + 8 * n)) - 1) / 2 == int ((std:: sqrt (1 + 8 * n)) - 1) / 2) {
return int (std:: ceil ((std:: sqrt (1 + 8 * n) - 1) / 2));
} else {
int time = minTime(n - 1) + 1;
int jump = 1;
while (jump * (jump + 1) / 2 <= n) {
time = std::min( time , minTime(n - jump * (jump + 1) / 2) + jump);
jump++;
}
return std:: ceil ( time );
}
}
int main() {
int n = 6;
std::cout << minTime(n) << std::endl;
return 0;
}
|
Java
import java.lang.Math;
public class MinTime {
public static int minTime( int n) {
if (n <= 0 ) {
return 0 ;
} else if (( int )(Math.sqrt( 1 + 8 * n) - 1 ) / 2 == ( int )((Math.sqrt( 1 + 8 * n) - 1 ) / 2 )) {
return ( int )Math.ceil((Math.sqrt( 1 + 8 * n) - 1 ) / 2 );
} else {
int time = minTime(n - 1 ) + 1 ;
int jump = 1 ;
while (jump * (jump + 1 ) / 2 <= n) {
time = Math.min(time, minTime(n - ( int )(jump * (jump + 1 ) / 2 )) + jump);
jump++;
}
return ( int )Math.ceil(time);
}
}
public static void main(String[] args) {
int n = 6 ;
System.out.println(minTime(n));
}
}
|
Python3
import math
from math import ceil
def min_time(n):
if n < = 0 :
return 0
elif ( int (math.sqrt( 1 + 8 * n)) - 1 ) / 2 = = int ((math.sqrt( 1 + 8 * n)) - 1 ) / 2 :
return int (ceil((math.sqrt( 1 + 8 * n)) - 1 ) / 2 )
else :
time = min_time(n - 1 ) + 1
jump = 1
while jump * (jump + 1 ) / 2 < = n:
time = min (time, min_time(n - jump * (jump + 1 ) / 2 ) + jump)
jump + = 1
return ceil(time)
n = 6
print (min_time(n))
|
C#
using System;
class Program
{
static int MinTime( int n)
{
if (n <= 0)
{
return 0;
}
else if (( int )(Math.Sqrt(1 + 8 * n)) - 1 / 2 == ( int )((Math.Sqrt(1 + 8 * n)) - 1) / 2)
{
return ( int )Math.Ceiling((Math.Sqrt(1 + 8 * n) - 1) / 2.0);
}
else
{
int time = MinTime(n - 1) + 1;
int jump = 1;
while (jump * (jump + 1) / 2 <= n)
{
time = Math.Min(time, MinTime(n - ( int )(jump * (jump + 1) / 2.0)) + jump);
jump++;
}
return ( int )Math.Ceiling(( double )time);
}
}
static void Main()
{
int n = 6;
Console.WriteLine(MinTime(n));
}
}
|
Javascript
function minTime(n) {
if (n <= 0) {
return 0;
} else if (Math.floor(Math.sqrt(1 + 8 * n) - 1) / 2 === Math.floor((Math.sqrt(1 + 8 * n) - 1) / 2)) {
return Math.ceil((Math.sqrt(1 + 8 * n) - 1) / 2);
} else {
let time = minTime(n - 1) + 1;
let jump = 1;
while (jump * (jump + 1) / 2 <= n) {
time = Math.min(time, minTime(n - Math.floor(jump * (jump + 1) / 2)) + jump);
jump++;
}
return Math.ceil(time);
}
}
let n = 6;
console.log(minTime(n));
|
Time complexity: Computing the time required to reach a point takes O(1) time.
The outer loop runs for n iterations, and the inner loop runs for a maximum of sqrt(2*n) iterations.
Therefore, the time complexity of the algorithm is O(n*sqrt(n)).
Space complexity: We need an array of size n+1 to store the memoization table.
Therefore, the space complexity of the algorithm is O(n).
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