Algorithm to solve a Hacker earth problem - algorithm

Question

I have been working on a Hackerearth Problem. Here is the problem statement:
We have three variables a, b and c. We need to convert a to b and following operations are allowed:
1. Can decrement by 1.
2. Can decrement by 2.
3. Can multiply by c.
Minimum steps required to convert a to b.
Here is the algorithm I came up with:
Increment count to 0.
Loop through till a === b:
1. Perform (x = a * c), (y = a - 1) and (z = a - 2).
2. Among x, y and z, choose the one whose absolute difference with b is the least.
3. Update the value of a to the value chosen among x, y and z.
4. Increment the count by 1.
I can get pass the basic test case but all my advance cases are failing. I guess my logic is correct but due to the complexity it seems to fail.
Can someone suggest a more optimized solution.
Edit 1
Sample Code
function findMinStep(arr) {
let a = parseInt(arr[0]);
let b = parseInt(arr[1]);
let c = parseInt(arr[2]);
let numOfSteps = 0;
while(a !== b) {
let multiply = Math.abs(b - (a * c));
let decrement = Math.abs(b - (a - 1));
let doubleDecrement = Math.abs(b - (a - 2));
let abs = Math.min(multiply, decrement, doubleDecrement);
if(abs === multiply) a = a * c;
else if(abs === decrement) a -= 1;
else a -= 2;
numOfSteps += 1;
}
return numOfSteps.toString()
}
Sample Input: a = 3, b = 10, c = 2
Explanation: Multiply 3 with 2 to get 6, subtract 1 from 6 to get 5, multiply 5 with 2 to get 10.
Reason for tagging both Python and JS: Comfortable with both but I am not looking for code, just an optimized algorithm and analytical thinking.
Edit 2:
function findMinStep(arr) {
let a = parseInt(arr[0]);
let b = parseInt(arr[1]);
let c = parseInt(arr[2]);
let depth = 0;
let queue = [a, 'flag'];
if(a === b ) return 0
if(a > b) {
let output = Math.floor((a - b) / 2);
if((a - b) % 2) return output + 1;
return output
}
while(true) {
let current = queue.shift();
if(current === 'flag') {
depth += 1;
queue.push('flag');
continue;
}
let multiple = current * c;
let decrement = current - 1;
let doubleDecrement = current -2;
if (multiple !== b) queue.push(multiple);
else return depth + 1
if (decrement !== b) queue.push(decrement);
else return depth + 1
if (doubleDecrement !== b) queue.push(doubleDecrement);
else return depth + 1
}
}
Still times out. Any more suggestions?
Link for the question for you reference.

Answer 1

BFS
A greedy approach won't work here.
However it is already on the right track. Consider the graph G, where each node represents a value and each edge represents one of the operations and connects two values that are related by that operation (e.g.: 4 and 3 are connected by "subtract 1"). Using this graph, we can easily perform a BFS-search to find the shortest path:
def a_to_b(a, b, c):
visited = set()
state = {a}
depth = 0
while b not in state:
visited |= state
state = {v - 1 for v in state if v - 1 not in visited} | \
{v - 2 for v in state if v - 2 not in visited} | \
{v * c for v in state if v * c not in visited}
depth += 1
return 1
This query systematically tests all possible combinations of operations until it reaches b by testing stepwise. I.e. generate all values that can be reached with a single operation from a, then test all values that can be reached with two operations, etc., until b is among the generated values.
In depth analysis
(Assuming c >= 0, but can be generalized)
So far for the standard-approach that works with little analysis. This approach has the advantage that it works for any problem of this kind and is easy to implement. However it isn't very efficient and will reach it's limits fairly fast, once the numbers grow. So instead I'll show a way to analyze the problem in depth and gain a (far) more performant solution:
In a first step this answer will analyze the problem:
We need operations -->op such that a -->op b and -->op is a sequence of
subtract 1
subtract 2
multiply by c
First of all, what happens if we first subtract and afterwards multiply?
(a - x) * c = a * c - x * c
Next what happens, if we first multiply and afterwards subtract?
a * c - x'
Positional systems
Well, there's no simplifying transformation for this. But we've got the basic pieces to analyze more complicated chains of operations. Let's see what happens when we chain subtractions and multiplications alternatingly:
(((a - x) * c - x') * c - x'') * c - x'''=
((a * c - x * c - x') * c - x'') * c - x''' =
(a * c^2 - x * c^2 - x' * c - x'') * c - x''' =
a * c^3 - x * c^3 - x' * c^2 - x'' * c - x'''
Looks familiar? We're one step away from defining the difference between a and b in a positional system base c:
a * c^3 - x * c^3 - x' * c^2 - x'' * c - x''' = b
x * c^3 + x' * c^2 + x'' * c + x''' = a * c^3 - b
Unfortunately the above is still not quite what we need. All we can tell is that the LHS of the equation will always be >=0. In general, we first need to derive the proper exponent n (3 in the above example), s.t. it is minimal, nonnegative and a * c^n - b >= 0. Solving this for the individual coefficients (x, x', ...), where all coefficients are non-negative is a fairly trivial task.
We can show two things from the above:
if a < b and a < 0, there is no solution
solving as above and transforming all coefficients into the appropriate operations leads to the optimal solution
Proof of optimality
The second statement above can be proven by induction over n.
n = 0: In this case a - b < c, so there is only one -->op
n + 1: let d = a * c^(n + 1) - b. Let d' = d - m * c^(n + 1), where m is chosen, such that d' is minimal and nonnegative. Per induction-hypothesis d' can be generated optimally via a positional system. Leaving a difference of exactly m * c^n. This difference can not be covered more efficiently via lower-order terms than by m / 2 subtractions.
Algorithm (The TLDR-part)
Consider a * c^n - b as a number base c and try to find it's digits. The final number should have n + 1 digits, where each digit represents a certain number of subtractions. Multiple subtractions are represented by a single digit by addition of the subtracted values. E.g. 5 means -2 -2 -1. Working from the most significant to the least significant digit, the algorithm operates as follows:
perform the subtractions as specified by the digit
if the current digit is was the last, terminate
multiply by c and repeat from 1. with the next digit
E.g.:
a = 3, b = 10, c = 2
choose n = 2
a * c^n - b = 3 * 4 - 10 = 2
2 in binary is 010
steps performed: 3 - 0 = 3, 3 * 2 = 6, 6 - 1 = 5, 5 * 2 = 10
or
a = 2, b = 25, c = 6
choose n = 2
a * c^n - b = 47
47 base 6 is 115
steps performed: 2 - 1 = 1, 1 * 6 = 6, 6 - 1 = 5, 5 * 6 = 30, 30 - 2 - 2 - 1 = 25
in python:
def a_to_b(a, b, c):
# calculate n
n = 0
pow_c = 1
while a * pow_c - b < 0:
n += 1
pow_c *= 1
# calculate coefficients
d = a * pow_c - b
coeff = []
for i in range(0, n + 1):
coeff.append(d // pow_c) # calculate x and append to terms
d %= pow_c # remainder after eliminating ith term
pow_c //= c
# sum up subtractions and multiplications as defined by the coefficients
return n + sum(c // 2 + c % 2 for c in coeff)