Least Operators to Express Number

Hard

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Dynamic Programming

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /). For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

The division operator (/) returns rational numbers.
There are no parentheses placed anywhere.
We use the usual order of operations: multiplication and division happen before addition and subtraction.
It is not allowed to use the unary negation operator (-). For example, "x - x" is a valid expression as it only uses subtraction, but "-x + x" is not because it uses negation.

We would like to write an expression with the least number of operators such that the expression equals the given target. Return the least number of operators used.

Example 1:

Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3.
The expression contains 5 operations.

Example 2:

Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.
The expression contains 8 operations.

Example 3:

Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100.
The expression contains 3 operations.

Constraints:

2 <= x <= 100
1 <= target <= 2 * 10⁸

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for `x` and `target`? Could they be negative, zero, or very large?
If the `target` cannot be expressed using the given operators and `x`, what should the function return? Should I return -1, throw an exception, or something else?
Can I use parentheses to control the order of operations?
Is integer division allowed? If so, how should I handle the rounding?
Is there a minimum number of times 'x' must be used or only a guarantee that it's used at least once?

Brute Force Solution

Approach

The brute force approach to this problem is all about trying every single way to build the target number using a base number and math operators. We explore all possible combinations of multiplication, division, addition, and subtraction. We methodically try all options until we find the combination that needs the fewest operations.

Here's how the algorithm would work step-by-step:

Start by looking at the simplest ways to reach the target number using the base number. For instance, can we get close by just adding or subtracting the base from itself repeatedly?
Next, try including multiplication and division. For each previous attempt, try multiplying or dividing the base number to see if we can get closer to the target.
For every result obtained, branch out and consider all ways to reach the target using more operations. Add or subtract the base number, then multiply or divide, and keep going to explore different combinations.
Keep track of the number of operations needed for each way to reach the target number.
After trying every combination of operations up to a certain level of complexity, pick the combination that has the least operations.

Code Implementation

def least_ops_brute_force(target, base):
    if target == 0:
        return 0
    if base == 0:
        return float('inf') if target != 1 else 1
    if base == 1:
        return abs(target)

    def explore(current_value, operations_count):
        # Check if we've reached the target
        if current_value == target:
            return operations_count

        if operations_count > 12:
            return float('inf')

        min_ops = float('inf')

        # Try addition
        addition_ops = explore(current_value + base, operations_count + 1)
        min_ops = min(min_ops, addition_ops)

        # Try subtraction
        subtraction_ops = explore(current_value - base, operations_count + 1)
        min_ops = min(min_ops, subtraction_ops)

        # Try multiplication
        if abs(current_value) <= abs(target) * 2 and base > 1:
            multiplication_ops = explore(current_value * base, operations_count + 1)
            min_ops = min(min_ops, multiplication_ops)

        # Try division, but avoid division by zero and unnecessary computations
        if current_value != 0 and abs(current_value) >= abs(target) // 2 and base > 1:
            division_ops = explore(current_value / base, operations_count + 1)
            min_ops = min(min_ops, division_ops)

        return min_ops

    # We use this to keep track of the best solution
    return explore(base, 1)

Big(O) Analysis

Time Complexity

O(4^target) – The brute force approach explores all possible combinations of +, -, *, and / operations using the base to reach the target. Each number can be potentially expressed in up to four ways (+base, -base, *base, /base), forming a recursive tree. The depth of the recursion depends on how far the target is from 1 (since the base is repeatedly used to get closer to the target). In the worst case, the number of possible expression trees grows exponentially, proportional to 4 raised to the power of the target, hence O(4^target). This estimation assumes target is significantly larger than base, and the number of recursion steps is proportional to the target itself.

Space Complexity

O(target) – The brute force approach explores all possible combinations of operations, which can lead to a significant number of intermediate results being stored. The depth of the recursive calls and the number of branches explored depend on the target number we are trying to reach. In the worst case, the recursion depth could be proportional to the target number, as we repeatedly add or subtract the base. Therefore, the space complexity can be approximated as O(target) due to the recursion stack and the storage of intermediate results during the exploration of different combinations.

Optimal Solution

Approach

The most efficient way to solve this problem is to figure out the minimum number of operations (+, -, *) to represent the target number using a given base. The core idea is to represent the number in terms of powers of the base, deciding whether to add or subtract each power to get as close as possible to the target.

Here's how the algorithm would work step-by-step:

First, consider representing the target number using only additions of powers of the base. Figure out the minimum number of operations needed for this representation.
Next, consider representing the target number using subtractions too. This often leads to a shorter expression because subtracting might get you closer faster.
The trick is to decide, for each power of the base, whether it's better to add it, subtract it, or neither. Calculate the cost (number of operations) for each choice.
If adding a power of the base gets you too far over the target, you might be better off subtracting a larger power. Therefore, we need to analyze consecutive powers of the base to find optimal representations.
Store the results of these calculations for each number you encounter to avoid redundant calculations. This 'remembering' is key to efficiency.
Finally, return the smallest number of operations needed to reach the target, considering both addition and subtraction options throughout the process.

Code Implementation

def least_ops_express_target(base, target):
    memo = {}

    def solve(current_target):
        if current_target in memo:
            return memo[current_target]

        if current_target == 0:
            return 0
        if current_target == 1:
            return 0

        exponent = 0
        base_power = 1

        while base_power <= current_target:
            base_power *= base
            exponent += 1

        positive_remainder = base_power - current_target
        negative_remainder = current_target - base_power // base

        #Consider edge case where power is just the base itself.
        if exponent == 1:
            positive_cost = positive_remainder
            negative_cost = negative_remainder
        else:
            # Recurse after deciding to add or subtract.
            positive_cost = solve(positive_remainder) + exponent
            negative_cost = solve(negative_remainder) + exponent - 1

        result = min(positive_cost, negative_cost)
        memo[current_target] = result
        return result

    return solve(target)

Big(O) Analysis

Time Complexity

O(log n) – The algorithm's time complexity is driven by the recursion depth, which corresponds to the number of powers of the base needed to reach the target number. Specifically, we are repeatedly dividing (or effectively dividing) the target by the base until we get to zero. This process resembles the logarithm of the target number (n) with base b, where b is the given base. The memoization ensures that each subproblem is solved only once, avoiding exponential blowup. Therefore, the time complexity is approximately O(log_b n), which simplifies to O(log n) since the base is a constant.

Space Complexity

O(log N) – The algorithm utilizes memoization to store the results of calculations for encountered numbers, which helps avoid redundant computations, as described in step 5. The maximum number of unique values that might need to be stored is related to the powers of the base needed to represent the target number N. This quantity is proportional to the logarithm of N with base 'base'. Therefore, the space used by the memoization dictionary is O(log N).

Edge Cases

Case	How to Handle
x = 0	Division by zero is undefined; return -1 or throw an exception to indicate no solution.
target = 0	Consider the case where target is zero and handle it specifically (e.g., x - x requires two operators).
x = 1	When x=1, the number of operators is simply target - 1; implement this as a direct return.
target = x	Zero operators are needed; return 0 immediately.
target significantly larger than x	Ensure that the recursive/iterative calls are bounded to prevent stack overflow or excessive computation time by using memoization with appropriate depth.
target is negative	Handle negative targets by considering using subtraction or negative powers of x.
Both x and target are very large numbers	Be mindful of potential integer overflow during calculations and use appropriate data types, potentially long.
No solution possible within reasonable bounds	Implement a cutoff or maximum recursion depth to prevent infinite loops and return a designated 'no solution' value, like -1.

x = 0

How to Handle:

Division by zero is undefined; return -1 or throw an exception to indicate no solution.

target = 0

How to Handle:

Consider the case where target is zero and handle it specifically (e.g., x - x requires two operators).

x = 1

How to Handle:

When x=1, the number of operators is simply target - 1; implement this as a direct return.

target = x

How to Handle:

Zero operators are needed; return 0 immediately.

target significantly larger than x

How to Handle:

Ensure that the recursive/iterative calls are bounded to prevent stack overflow or excessive computation time by using memoization with appropriate depth.

target is negative

How to Handle:

Handle negative targets by considering using subtraction or negative powers of x.

Both x and target are very large numbers

How to Handle:

Be mindful of potential integer overflow during calculations and use appropriate data types, potentially long.

No solution possible within reasonable bounds

How to Handle:

Implement a cutoff or maximum recursion depth to prevent infinite loops and return a designated 'no solution' value, like -1.