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Least Operators to Express Number

Number: 1004

Difficulty: Hard

Paid? No

Companies: Snap

Problem Description

Given a positive integer x and a target value, we want to form an expression using the number x and the operators +, -, *, and / (with the usual precedence, and no parentheses) that equals the target. Each time we use an operator (between consecutive x’s) it is counted. We are not allowed to use the unary negation operator. The task is to find the least number of operators needed to create an expression equal to target.

Key Insights

  • Think of the problem as expressing the target in a “base-x” manner.
  • Break the target into sums and differences of powers of x.
  • Use a recursion and memoization (DP) approach: at each step, determine the largest power p = x^m that is <= target.
  • Two strategies are considered: • Use p directly and solve for target − p. • Use the next power (x^(m+1)) and subtract the extra overreach.
  • Special handling is required when target is less than x because there the typical multiplication chain strategy does not work directly.
  • Finally, subtract one from the computed cost since the very first “x” does not require an operator.

Space and Time Complexity

Time Complexity: O(log(target)) in the average case due to reducing the target by a factor close to x per recursive call; however, worst‐case complexity depends on the number of distinct subproblems processed during memoization. Space Complexity: O(log(target)) due to the recursion stack and memoization storage.

Solution

We use a recursive function dp(target) that returns the minimum number of operators needed plus one extra “virtual operator” such that the very first x does not count. For target < x, we use a base case with the formula:   min(2 * target - 1, 2 * (x - target)) which reflects the cost when we form numbers less than x by either summing many terms (each “x/x” gives 1) or by subtracting from x. For target ≥ x:

  1. Find m = floor(log₍ₓ₎(target)) and let p = x^m.
  2. If target equals p, then dp(target) returns m (or 2 if m is zero since “x/x” is used to form 1).
  3. Otherwise, we consider two cases:   a. Express target as p plus the remainder: cost = m + dp(target − p)   b. Express target using a higher power: cost = m + 1 + dp(x^(m+1) − target)
  4. Finally, return the minimum cost between the two strategies. Since the recurrence counts an extra “virtual operator” in the beginning, the final answer is dp(target) − 1.

The algorithm employs recursion with memoization (caching the results for each target) to ensure we do not recompute overlapping subproblems.

Code Solutions

# Python solution using memoization
class Solution:
    def leastOpsExpressTarget(self, x: int, target: int) -> int:
        from functools import lru_cache
        # dp(n): minimum cost (operators count + 1) to express n
        @lru_cache(maxsize=None)
        def dp(target):
            # Base case: for numbers smaller than x
            if target < x:
                # When target < x, cost is min(2*target - 1, 2*(x - target))
                return min(2 * target - 1, 2 * (x - target))
            
            # Determine highest power m such that x^m <= target
            m = 0
            p = 1
            # Calculate p = x^m while p * x <= target, increment m
            while p * x <= target:
                p *= x
                m += 1
            
            # If target equals p exactly, then cost is m (if m > 0) 
            # because it can be represented as x * x * ... * x with (m) multiplications.
            if target == p:
                return m
            
            # Option 1: use power p directly and handle the remainder.
            # Cost: m (cost for power p) + dp(target - p)
            cost1 = m + dp(target - p)
            
            # Option 2: use the next power x^(m+1) and subtract the extra amount.
            # Only consider if using extra power results in fewer operations.
            cost2 = float('inf')
            # p2 = x^(m+1)
            p2 = p * x
            # Cost: (m+1) (to use x^(m+1)) + dp(p2 - target)
            cost2 = m + 1 + dp(p2 - target)
            
            # Return the minimal cost between the two options.
            return min(cost1, cost2)
        
        # Subtract one because the first x does not need an operator.
        return dp(target) - 1

# Example usage:
# sol = Solution()
# print(sol.leastOpsExpressTarget(3, 19))  # Expected output: 5
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