# Solving Leetcode Interviews in Seconds with AI: Least Operators to Express Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "964" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the approach and the Python code:

*   **Dynamic Programming:** We use dynamic programming to explore the possible values achievable with different numbers of operations. We maintain a `dp` table where `dp[i]` stores the set of achievable values using `i` operations.

*   **Iterative Calculation:** Starting with `dp[0] = {x}`, we iteratively calculate `dp[i]` based on the values in `dp[i-1]` and the four possible operations (+, -, *, /).

*   **Return First Match:** We continue building `dp` until we find a value in any `dp[i]` that equals the `target`. The index `i` then represents the minimum number of operations.

*   **Runtime Complexity:** O(N * T^2) where N is the number of operations (bounded roughly) and T is the target. Note: In practice, the space is limited and it executes much faster. **Storage Complexity:** O(N * T)

	
	# Code
	```python
	def least_ops(x, target):
    """
    Finds the least number of operators to reach the target using x.

    Args:
        x: The base number.
        target: The target value.

    Returns:
        The least number of operators, or -1 if not possible.
    """

    dp = {0: {x}}
    for ops in range(1, 20):  # Iterate up to a reasonable number of operations
        dp[ops] = set()
        for prev_ops in range(ops):
            for val1 in dp[prev_ops]:
                for val2 in dp[ops - prev_ops - 1]:
                    dp[ops].add(val1 + val2)
                    dp[ops].add(val1 - val2)
                    dp[ops].add(val1 * val2)
                    if val2 != 0:
                        dp[ops].add(val1 / val2)

        if target in dp[ops]:
            return ops

    return -1  # Target not reachable within the operation limit
	```