Minimum Cost to Make Array Equalindromic
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> having length <code>n</code>.</p>

<p>You are allowed to perform a special move <strong>any</strong> number of times (<strong>including zero</strong>) on <code>nums</code>. In one <strong>special</strong> <strong>move</strong> you perform the following steps <strong>in order</strong>:</p>

<ul>
	<li>Choose an index <code>i</code> in the range <code>[0, n - 1]</code>, and a <strong>positive</strong> integer <code>x</code>.</li>
	<li>Add <code>|nums[i] - x|</code> to the total cost.</li>
	<li>Change the value of <code>nums[i]</code> to <code>x</code>.</li>
</ul>

<p>A <strong>palindromic number</strong> is a positive integer that remains the same when its digits are reversed. For example, <code>121</code>, <code>2552</code> and <code>65756</code> are palindromic numbers whereas <code>24</code>, <code>46</code>, <code>235</code> are not palindromic numbers.</p>

<p>An array is considered <strong>equalindromic</strong> if all the elements in the array are equal to an integer <code>y</code>, where <code>y</code> is a <strong>palindromic number</strong> less than <code>10<sup>9</sup></code>.</p>

<p>Return <em>an integer denoting the <strong>minimum</strong> possible total cost to make </em><code>nums</code><em> <strong>equalindromic</strong> by performing any number of special moves.</em></p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 6
<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 3 which is a palindromic number. The cost of changing the array to [3,3,3,3,3] using 4 special moves is given by |1 - 3| + |2 - 3| + |4 - 3| + |5 - 3| = 6.
It can be shown that changing all elements to any palindromic number other than 3 cannot be achieved at a lower cost.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10,12,13,14,15]
<strong>Output:</strong> 11
<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 11 which is a palindromic number. The cost of changing the array to [11,11,11,11,11] using 5 special moves is given by |10 - 11| + |12 - 11| + |13 - 11| + |14 - 11| + |15 - 11| = 11.
It can be shown that changing all elements to any palindromic number other than 11 cannot be achieved at a lower cost.
</pre>

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

<pre>
<strong>Input:</strong> nums = [22,33,22,33,22]
<strong>Output:</strong> 22
<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 22 which is a palindromic number. The cost of changing the array to [22,22,22,22,22] using 2 special moves is given by |33 - 22| + |33 - 22| = 22.
It can be shown that changing all elements to any palindromic number other than 22 cannot be achieved at a lower cost.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Greedy; Array; Math; Binary Search; Sorting
public class Solution {
    private static long[] ps;
    private int[] nums;

    static {
        ps = new long[2 * (int) 1e5];
        for (int i = 1; i <= 1e5; i++) {
            String s = Integer.toString(i);
            String t1 = new StringBuilder(s).reverse().toString();
            String t2 = new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString();
            ps[2 * i - 2] = Long.parseLong(s + t1);
            ps[2 * i - 1] = Long.parseLong(s + t2);
        }
        Arrays.sort(ps);
    }

    public long minimumCost(int[] nums) {
        this.nums = nums;
        Arrays.sort(nums);
        int i = Arrays.binarySearch(ps, nums[nums.length / 2]);
        i = i < 0 ? -i - 1 : i;
        long ans = 1L << 60;
        for (int j = i - 1; j <= i + 1; j++) {
            if (0 <= j && j < ps.length) {
                ans = Math.min(ans, f(ps[j]));
            }
        }
        return ans;
    }

    private long f(long x) {
        long ans = 0;
        for (int v : nums) {
            ans += Math.abs(v - x);
        }
        return ans;
    }
}