<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>

//Greedy approach wont work here.a*x^3+b*x^2+c*x+d
//It might not work but the thinking process is still same

//We always prefer to reduce the target so 
//if(upper-target<target) check is performed
//Even in base cases 
//Return after including extra plus
//At the end subtract 1

#define ll long long
class Solution {
public:
    ll logo(ll num,ll x){
        if(num<x)return 0;
        ll cnt = 0;
        while(num!=0){
            num=num/x;
            cnt++;
        }
        return cnt-1;
    }
    ll f(ll x,ll tg,map<ll,ll>& mp){
        if(mp.find(tg)!=mp.end())return mp[tg];
        if(x>tg){
            if(tg==1)return mp[1] =  1;
            return mp[tg]=min(2*tg-1,2*(x-tg));
        }

        ll poke = logo(tg,x);

        ll up = pow(x,poke+1);
        ll down = pow(x,poke);

        ll a = 1e14;
        ll b = 1e14;
        if(2*tg>up)a = f(x,up-tg,mp)+poke+1;
        b = f(x,tg-down,mp)+poke;

        return mp[tg] = min(a,b);
    }
    int leastOpsExpressTarget(int x, int tg) {
        map<ll,ll>mp;
        ll ans = f(x,tg,mp);
        return ans;
    }
};
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