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++ implementation of the above approach
#include <iostream>
using namespace std;
// returns the minimum time
// required to reach 'X'
long cal_minimum_time(long X)
{
// Stores the minimum time
long t = 0;
long sum = 0;
while (sum < X) {
// increment 't' by 1
t++;
// update the sum
sum = sum + t;
}
return t;
}
// Driver code
int main()
{
long n = 6;
long ans = cal_minimum_time(n);
cout << "The minimum time required is : " << ans ;
return 0;
// This code is contributed by ANKITRAI1
}
// Java implementation of the above approach
class GFG {
// returns the minimum time
// required to reach 'X'
static long cal_minimum_time(long X)
{
// Stores the minimum time
long t = 0;
long sum = 0;
while (sum < X) {
// increment 't' by 1
t++;
// update the sum
sum = sum + t;
}
return t;
}
// Driver code
public static void main(String[] args)
{
long n = 6;
long ans = cal_minimum_time(n);
System.out.println("The minimum time required is : " + ans);
}
}
# Python 3 implementation of the
# above approach
# returns the minimum time
# required to reach 'X'
def cal_minimum_time(X):
# Stores the minimum time
t = 0
sum = 0
while (sum < X):
# increment 't' by 1
t = t + 1
# update the sum
sum = sum + t;
return t;
# Driver code
if __name__ == '__main__':
n = 6
ans = cal_minimum_time(n)
print("The minimum time required is :", ans)
# This code is contributed By
# Surendra_Gangwar
// C# implementation of the above approach
using System;
public class GFG{
// returns the minimum time
// required to reach 'X'
static long cal_minimum_time(long X)
{
// Stores the minimum time
long t = 0;
long sum = 0;
while (sum < X) {
// increment 't' by 1
t++;
// update the sum
sum = sum + t;
}
return t;
}
// Driver code
static public void Main (){
long n = 6;
long ans = cal_minimum_time(n);
Console.WriteLine("The minimum time required is : " + ans);
}
}
<script>
// JavaScript implementation of the above approach
// returns the minimum time
// required to reach 'X'
function cal_minimum_time(X)
{
// Stores the minimum time
let t = 0;
let sum = 0;
while (sum < X) {
// increment 't' by 1
t++;
// update the sum
sum = sum + t;
}
return t;
}
// driver code
let n = 6;
let ans = cal_minimum_time(n);
document.write("The minimum time required is : " + ans);
</script>
<?php
// PHP implementation of the
// above approach
// returns the minimum time
// required to reach 'X'
function cal_minimum_time($X)
{
// Stores the minimum time
$t = 0;
$sum = 0;
while ($sum < $X)
{
// increment 't' by 1
$t++;
// update the sum
$sum = $sum + $t;
}
return $t;
}
// Driver code
$n = 6;
$ans = cal_minimum_time($n);
echo "The minimum time required is : " . $ans;
// This code is contributed
// by ChitraNayal
?>
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].
#include <iostream>
#include <cmath>
// Function to calculate the minimum time
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) {
// If n is a triangular number, return its square root minus 1 divided by 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) {
// Recursively calculate the minimum time for different jump values
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; // Print the minimum time for n
return 0;
}
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));
}
}
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))
using System;
class Program
{
// Function to calculate the minimum time
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)
{
// If n is a triangular number, return its square root minus 1 divided by 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)
{
// Recursively calculate the minimum time for different jump values
time = Math.Min(time, MinTime(n - (int)(jump * (jump + 1) / 2.0)) + jump);
jump++;
}
return (int)Math.Ceiling((double)time); // Explicitly cast to double to resolve ambiguity
}
}
static void Main()
{
int n = 6;
Console.WriteLine(MinTime(n)); // Print the minimum time for n
}
}
// This code is contributed by rambabuguphka
// Define a function to calculate the minimum time required to reach the given target number 'n'.
function minTime(n) {
// Base case: If 'n' is less than or equal to 0, return 0 as no time is needed.
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)) {
// Check if 'n' is a triangular number. If so, return the ceiling of its square root minus 1 divided by 2.
return Math.ceil((Math.sqrt(1 + 8 * n) - 1) / 2);
} else {
// Recursive case: Calculate the minimum time to reach 'n' by considering different possible jumps.
let time = minTime(n - 1) + 1; // Initialize time with the cost of one jump.
let jump = 1;
// Explore different jump possibilities to minimize time.
while (jump * (jump + 1) / 2 <= n) {
time = Math.min(time, minTime(n - Math.floor(jump * (jump + 1) / 2)) + jump);
jump++;
}
return Math.ceil(time); // Return the minimum time, rounded up to the nearest integer.
}
}
// Example usage: Calculate the minimum time for 'n' = 6 and log the result.
let n = 6;
console.log(minTime(n));
Output
3
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).
