Program to check we can reach at position n by jumping or not in Python

Suppose there is a number line from 1 to n. At first we are at position 0, jump one step to go 1, then jump two places to reach at position 3, then jump three positions to reach at 6 and so on. We have to check whether maintaining this pattern, we can reach at position n or not.

So, if the input is like n = 21, then the output will be True, because 1+2+3+4+5+6 = 21

Approach

To solve this, we will follow these steps −

• Calculate j := (1 + square root of (1+8*n)) / 2
• If |j - int part of j| <= 0, then
  • return True
• Otherwise return False

How It Works

This problem is essentially checking if n can be expressed as a triangular number. The formula for the k-th triangular number is k*(k+1)/2. We solve the quadratic equation k*(k+1)/2 = n to find if there exists an integer k.

Example

Let us see the following implementation to get better understanding −

from math import sqrt

def solve(n):
    j = (1 + sqrt(1 + 8 * n)) / 2
    if abs(j - int(j)) <= 0:
        return True
    else:
        return False

n = 21
print(solve(n))

The output of the above code is −

True

Alternative Implementation

Here's a cleaner version that checks if j is exactly an integer −

from math import sqrt

def can_reach_position(n):
    # Using quadratic formula to solve k*(k+1)/2 = n
    j = (-1 + sqrt(1 + 8 * n)) / 2
    return j == int(j)

# Test with different values
test_values = [1, 3, 6, 10, 15, 21, 28]

for n in test_values:
    result = can_reach_position(n)
    print(f"n = {n}: {result}")

The output of the above code is −

n = 1: True
n = 3: True
n = 6: True
n = 10: True
n = 15: True
n = 21: True
n = 28: True

Conclusion

This algorithm efficiently determines if a position n can be reached by sequential jumps using the mathematical property of triangular numbers. The solution runs in O(1) time complexity.