Check if it is possible to reach a number by making jumps of two given length in Python

We need to find the minimum number of steps to reach position q from starting position p by making jumps of length d1 or d2 in either direction (left or right).

The key insight is that we can only reach a target if the difference between start and end positions is divisible by the GCD of the two jump distances.

Algorithm Steps

To solve this problem, we follow these steps ?

• Calculate GCD of d1 and d2
• Check if (p - q) is divisible by GCD - if not, return -1
• Use BFS to find minimum steps
• Track visited positions to avoid cycles
• Try all four possible moves: +d1, +d2, -d1, -d2

Example

Let's implement the solution with p = 5, q = 10, d1 = 4, d2 = 3 ?

from math import gcd
from collections import deque

def solve(p, d1, d2, q):
    gcd_res = gcd(d1, d2)
    if (p - q) % gcd_res != 0:
        return -1
    
    que = deque()
    visited = set()
    que.appendleft([p, 0])
    visited.add(p)
    
    while len(que) > 0:
        pair = que.pop()
        point, step = pair[0], pair[1]
        
        if point == q:
            return step
            
        if point + d1 not in visited:
            que.appendleft([point + d1, step + 1])
            visited.add(point + d1)
            
        if point + d2 not in visited:
            que.appendleft([point + d2, step + 1])
            visited.add(point + d2)
            
        if point - d1 not in visited:
            que.appendleft([point - d1, step + 1])
            visited.add(point - d1)
            
        if point - d2 not in visited:
            que.appendleft([point - d2, step + 1])
            visited.add(point - d2)

# Test the function
p = 5
q = 10
d1 = 4
d2 = 3
result = solve(p, d1, d2, q)
print(f"Minimum steps to reach from {p} to {q}: {result}")

Minimum steps to reach from 5 to 10: 3

How It Works

The solution path for our example is ?

• Start at position 5
• Jump right 4 units ? position 9 (step 1)
• Jump right 4 units ? position 13 (step 2)
• Jump left 3 units ? position 10 (step 3)

Alternative Test Case

Let's test an impossible case ?

from math import gcd
from collections import deque

def solve(p, d1, d2, q):
    gcd_res = gcd(d1, d2)
    if (p - q) % gcd_res != 0:
        return -1
    
    que = deque()
    visited = set()
    que.appendleft([p, 0])
    visited.add(p)
    
    while len(que) > 0:
        pair = que.pop()
        point, step = pair[0], pair[1]
        
        if point == q:
            return step
            
        if point + d1 not in visited:
            que.appendleft([point + d1, step + 1])
            visited.add(point + d1)
            
        if point + d2 not in visited:
            que.appendleft([point + d2, step + 1])
            visited.add(point + d2)
            
        if point - d1 not in visited:
            que.appendleft([point - d1, step + 1])
            visited.add(point - d1)
            
        if point - d2 not in visited:
            que.appendleft([point - d2, step + 1])
            visited.add(point - d2)

# Test impossible case: GCD of 4 and 6 is 2, but (0-1) % 2 != 0
p = 0
q = 1
d1 = 4
d2 = 6
result = solve(p, d1, d2, q)
print(f"Result for impossible case: {result}")

Result for impossible case: -1

Conclusion

This algorithm uses BFS to find the minimum steps, first checking if the target is reachable using GCD theory. The time complexity is dependent on the search space, which can be limited by setting boundaries to avoid infinite exploration.