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

Given a starting position ‘k’ and two jump sizes ‘d1’ and ‘d2’, our task is to find the minimum number of jumps needed to reach ‘x’ if it is possible.

At any position P, we are allowed to jump to positions :

  • P + d1 and P – d1
  • P + d2 and P – d2

Examples:

Input : k = 10, d1 = 4, d2 = 6 and x = 8 
Output : 2
1st step 10 + d1 = 14
2nd step 14 - d2 = 8

Input : k = 10, d1 = 4, d2 = 6 and x = 9
Output : -1
-1 indicates it is not possible to reach x.

In the previous article we discussed a strategy to check whether a list of numbers is reachable by K by making jump of two given lengths.

Here, instead of a list of numbers, we are given a single integer x and if it is reachable from k then the task is to find the minimum number of steps or jumps needed.

We will solve this using Breadth first Search:
Approach:

  • Check if ‘x’ is reachable from k. The number x is reachable from k if it satisfies (x – k) % gcd(d1, d2) = 0.
  • If x is reachable :
    1. Maintain a hash table to store the already visited positions.
    2. Apply bfs algorithm starting from the position k.
    3. If you reach position P in ‘stp’ steps, you can reach p+d1 position in ‘stp+1’ steps.
    4. If position P is the required position ‘x’ then steps taken to reach P is the answer

The image below depicts how the algorithm finds out number of steps needed to reach x = 8 with k = 10, d1 = 4 and d2 = 6.
Algo Example

Below is the implementation of the above approach:

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#include <bits/stdc++.h>
using namespace std;
  
// Function to perform BFS traversal to
// find minimum number of step needed
// to reach x from K
int minStepsNeeded(int k, int d1, int d2, int x)
{
    // Calculate GCD of d1 and d2
    int gcd = __gcd(d1, d2);
  
    // If position is not reachable
    // return -1
    if ((k - x) % gcd != 0)
        return -1;
  
    // Queue for BFS
    queue<pair<int, int> > q;
  
    // Hash Table for marking
    // visited positions
    unordered_set<int> visited;
  
    // we need 0 steps to reach K
    q.push({ k, 0 });
  
    // Mark starting position
    // as visited
    visited.insert(k);
  
    while (!q.empty()) {
  
        int s = q.front().first;
  
        // stp is the number of steps
        // to reach position s
        int stp = q.front().second;
  
        if (s == x)
            return stp;
  
        q.pop();
  
        if (visited.find(s + d1) == visited.end()) {
  
            // if position not visited
            // add to queue and mark visited
            q.push({ s + d1, stp + 1 });
  
            visited.insert(s + d1);
        }
  
        if (visited.find(s + d2) == visited.end()) {
            q.push({ s + d2, stp + 1 });
            visited.insert(s + d2);
        }
  
        if (visited.find(s - d1) == visited.end()) {
            q.push({ s - d1, stp + 1 });
            visited.insert(s - d1);
        }
        if (visited.find(s - d2) == visited.end()) {
            q.push({ s - d2, stp + 1 });
            visited.insert(s - d2);
        }
    }
}
  
// Driver Code
int main()
{
    int k = 10, d1 = 4, d2 = 6, x = 8;
  
    cout << minStepsNeeded(k, d1, d2, x);
  
    return 0;
}

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Output:

2


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