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

<p>When writing such an expression, we adhere to the following conventions:</p>

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

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> x = 100, target = 100000000
<strong>Output:</strong> 3
<strong>Explanation:</strong> 100 * 100 * 100 * 100.
The expression contains 3 operations.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= x &lt;= 100</code></li>
	<li><code>1 &lt;= target &lt;= 2 * 10<sup>8</sup></code></li>
</ul>

class Solution:
    def leastOpsExpressTarget(self, x: int, target: int) -> int:
        if x == target:
            return 0
        elif target < x:
            return min(2*target-1, 2*(x-target))
        k = 0
        stop = 1
        while stop < target:
            k += 1
            stop *= x
        if stop == target:
            return k-1
        start = 0
        memo = {}
        memo[start] = -1 
        memo[stop] = k-1
        while k > 0:
            k -= 1
            sta = x-1
            for j in range(1,x):
                val_n = start + j*x**k
                if k > 0:
                    memo[val_n] = min(memo[start] + j*k, memo[stop] + (x-j)*k)  
                else:
                    memo[val_n] = min(memo[start] + j*2, memo[stop] + (x-j)*2)
                    if val_n == target:
                        return memo[target]
                if val_n > target:
                    sta = j-1
                    break
            start = start + sta*x**k
            stop = start + x**k
            if start == target:
                return memo[start]
        return memo[target]


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