Solution Approach

The solution uses a recursive approach with memoization to find the minimum number of operators needed to express the target value.

The main function dfs(v) computes the minimum operators needed to create value v. Here's how it works:

Base Case: When x >= v

When the value we want to create (v) is less than or equal to x, we have two strategies:

1. Build using ones: Create v by adding 1s together. Since each 1 is created using x/x (1 operator), and we need v of them with v-1 addition operators between them, the total is 2v - 1 operators.

   • Example: If x = 5 and v = 3, we get 5/5 + 5/5 + 5/5 using 5 operators total.

2. Subtract from x: Express v as x - (x - v). This uses 1 subtraction operator plus the operators needed to create x - v. The total is 2 * (x - v) operators (since creating x - v using ones would need 2(x - v) - 1 operators, plus 1 for the subtraction).

The algorithm chooses the minimum of these two options: min(v * 2 - 1, 2 * (x - v)).

Recursive Case: When x < v

For larger values, we need to use powers of x:

1. Find the appropriate power: The algorithm finds the smallest k such that x^k >= v. This is done through a simple loop:

   k = 2 while x**k < v:  k += 1

2. Decision point: Once we have x^k, we can either:

   • Overshoot and subtract: Use x^k - (x^k - v) to get v. This requires k operators to build x^k (since x^k needs k-1 multiplications), plus the operators to build x^k - v.
   • Undershoot and add: Use x^(k-1) + (v - x^(k-1)) to get v. This requires k-1 operators for x^(k-1), plus the operators to build v - x^(k-1).

3. Optimization: The code includes a smart optimization. If x^k - v < v, meaning the overshoot difference is smaller than v itself, it considers both options. Otherwise, it only uses the undershoot approach since the overshoot would require building a larger value recursively.

Memoization with @cache

The @cache decorator stores previously computed results for each value of v. This prevents redundant calculations when the same subproblem appears multiple times during recursion, significantly improving performance.

Final Result

The function returns dfs(target), which gives the minimum number of operators needed to express the target value using only the number x.

The elegance of this solution lies in its recursive decomposition - it transforms the problem of expressing any number into smaller subproblems, leveraging the fact that any target can be efficiently expressed as a combination of powers of x with appropriate additions and subtractions.

Example Walkthrough

Let's walk through finding the minimum operators needed when x = 3 and target = 10.

Initial Call: dfs(10)

Since x = 3 < 10, we enter the recursive case and need to find an appropriate power of 3:

• 3^2 = 9 (less than 10)
• 3^3 = 27 (greater than 10)

So k = 3, meaning 3^3 = 27 is our overshoot power.

Exploring Two Options:

1. Overshoot approach: Express 10 as 27 - 17
   • We need k - 1 = 2 operators to build 3^3 (i.e., 3 * 3 * 3)
   • We need 1 subtraction operator
   • We need dfs(17) operators to build 17
2. Undershoot approach: Express 10 as 9 + 1
   • We need k - 2 = 1 operator to build 3^2 (i.e., 3 * 3)
   • We need 1 addition operator
   • We need dfs(1) operators to build 1

Computing dfs(1):

Since x = 3 >= 1, we use the base case:

• Option 1: Build 1 using ones: 3/3 requires 1 operator
• Option 2: Subtract from 3: 3 - 2 requires 2 * (3 - 1) = 4 operators

Minimum is 1 operator, so dfs(1) = 1.

Computing dfs(17):

Since 3 < 17, we find powers:

• 3^2 = 9 (less than 17)
• 3^3 = 27 (greater than 17)

So k = 3.

Checking overshoot: 27 - 17 = 10. Since 10 < 17, we consider both options:

• Overshoot: 27 - 10 needs 2 + dfs(10) operators
• Undershoot: 9 + 8 needs 1 + dfs(8) operators

For dfs(8):

• Since 3 < 8, we find 3^2 = 9 > 8, so k = 2
• Overshoot: 9 - 1 needs 1 + dfs(1) = 1 + 1 = 2 operators
• Since we only need the minimum and 9 - 8 = 1 < 8, this gives us dfs(8) = 2

Back to dfs(17):

• Overshoot would create a cycle (needs dfs(10) which we're computing)
• Undershoot: 1 + dfs(8) = 1 + 2 = 3 operators

So dfs(17) = 3.

Final Calculation for dfs(10):

1. Overshoot: 2 + dfs(17) = 2 + 3 = 5 operators
2. Undershoot: 1 + dfs(1) = 1 + 1 = 2 operators

The minimum is 2 operators.

Verifying the Result:

The expression 3 * 3 + 3 / 3 equals 9 + 1 = 10 and uses exactly 2 operators (one multiplication and one division, with addition connecting them). This confirms our answer of 2 operators.

Time and Space Complexity

Time Complexity: O(log(target) * log(target)) or O(log²(target))

The recursion depth is bounded by O(log(target)) because at each recursive call, we're working with values that are differences between powers of x and the current value v. The value v gets reduced significantly in each recursive step - either by subtracting a power of x close to it, or by working with x^k - v where x^k is the smallest power greater than v.

At each recursion level, the while loop to find the appropriate power k takes O(log(v)) time in the worst case, where v ≤ target. Since we're using memoization with @cache, each unique value of v is computed only once.

The number of unique subproblems is bounded by O(target) in the worst case, but due to the nature of the problem where we work with powers and differences, the actual number of unique states visited is much smaller, approximately O(log(target)).

Therefore, the overall time complexity is O(log(target) * log(target)) = O(log²(target)).

Space Complexity: O(log(target))

The space complexity consists of two components:

1. The recursion stack depth: O(log(target)) - as the recursion depth is logarithmic in the target value
2. The memoization cache: O(log(target)) - storing results for the unique subproblems encountered

The dominant factor is O(log(target)) for both the recursion stack and the cache storage.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow with Large Powers

When computing x ** power for large values of x and power, the result can quickly exceed Python's practical memory limits or cause performance issues. For instance, if x = 999 and we're looking for a large target, computing 999^10 would result in an astronomically large number.

Solution: Add an early termination condition to prevent unnecessary large power calculations:

@cache def dfs(current_value: int) -> int:  if x >= current_value:  return min(current_value * 2 - 1, 2 * (x - current_value))   power = 2  power_value = x * x # Start with x^2   # Add upper bound check to prevent overflow  while power_value < current_value and power < 40: # 40 is a reasonable upper bound  power += 1  power_value *= x  if power_value > 10**9: # Safeguard against extremely large values  break   # Rest of the logic remains the same

2. Incorrect Base Case Calculation

A common mistake is misunderstanding how operators are counted in the base case. When building a value using ones (x/x), developers often forget that:

• The first x/x uses 1 operator (division)
• Each additional x/x needs 1 division operator PLUS 1 addition operator to connect it

Incorrect implementation:

# Wrong: Forgetting the addition operators between terms option_add_ones = current_value # This only counts divisions!

Correct implementation:

# Right: Count both divisions and additions option_add_ones = current_value * 2 - 1 # v divisions + (v-1) additions

3. Off-by-One Error in Power Calculation

The recursive case needs k-1 operators to build x^k (not k operators), since x itself requires no operators. This is a subtle but critical distinction.

Incorrect:

# Wrong: Counting x as needing 1 operator option_overshoot = power + 1 + dfs(x ** power - current_value)

Correct:

# Right: x^k needs k-1 multiplications option_overshoot = power - 1 + 1 + dfs(x ** power - current_value) # Simplified to: option_overshoot = power + dfs(x ** power - current_value)

4. Missing Edge Case: target = x

When target equals x, the answer should be 0 (no operators needed), but the base case logic might not handle this correctly if not carefully implemented.

Solution: Add an explicit check at the beginning:

def leastOpsExpressTarget(self, x: int, target: int) -> int:  if target == x:  return 0   # Rest of the implementation...

5. Inefficient Repeated Power Calculations

Computing x ** power multiple times in the same recursive call wastes computational resources.

Inefficient:

if x ** power - current_value < current_value:  option_overshoot = power + dfs(x ** power - current_value) # Recomputes x^power  option_undershoot = power - 1 + dfs(current_value - x ** (power - 1)) # Recomputes x^(power-1)

Efficient:

power_value = x ** power prev_power_value = x ** (power - 1)  if power_value - current_value < current_value:  option_overshoot = power + dfs(power_value - current_value)  option_undershoot = power - 1 + dfs(current_value - prev_power_value)