You are given a 0-indexed string num representing a non-negative integer.
In one operation, you can pick any digit of num and delete it. Note that if you delete all the digits of num, num becomes 0.
Return the minimum number of operations required to make num special.
An integer x is considered special if it is divisible by 25.
\n
Example 1:
\n\n\nInput: num = "2245047"\nOutput: 2\nExplanation: Delete digits num[5] and num[6]. The resulting number is "22450" which is special since it is divisible by 25.\nIt can be shown that 2 is the minimum number of operations required to get a special number.\n\n
Example 2:
\n\n\nInput: num = "2908305"\nOutput: 3\nExplanation: Delete digits num[3], num[4], and num[6]. The resulting number is "2900" which is special since it is divisible by 25.\nIt can be shown that 3 is the minimum number of operations required to get a special number.\n\n
Example 3:
\n\n\nInput: num = "10"\nOutput: 1\nExplanation: Delete digit num[0]. The resulting number is "0" which is special since it is divisible by 25.\nIt can be shown that 1 is the minimum number of operations required to get a special number.\n\n\n\n
\n
Constraints:
\n\n1 <= num.length <= 100num only consists of digits '0' through '9'.num does not contain any leading zeros.给你一个下标从 0 开始的字符串 num ，表示一个非负整数。
在一次操作中，您可以选择 num 的任意一位数字并将其删除。请注意，如果你删除 num 中的所有数字，则 num 变为 0。
返回最少需要多少次操作可以使 num 变成特殊数字。
如果整数 x 能被 25 整除，则该整数 x 被认为是特殊数字。
\n\n
\n\n
示例 1：
\n\n\n输入：num = \"2245047\"\n输出：2\n解释：删除数字 num[5] 和 num[6] ，得到数字 \"22450\" ，可以被 25 整除。\n可以证明要使数字变成特殊数字，最少需要删除 2 位数字。\n\n
示例 2：
\n\n\n输入：num = \"2908305\"\n输出：3\n解释：删除 num[3]、num[4] 和 num[6] ，得到数字 \"2900\" ，可以被 25 整除。\n可以证明要使数字变成特殊数字，最少需要删除 3 位数字。\n\n
示例 3：
\n\n\n输入：num = \"10\"\n输出：1\n解释：删除 num[0] ，得到数字 \"0\" ，可以被 25 整除。\n可以证明要使数字变成特殊数字，最少需要删除 1 位数字。\n\n\n
\n\n
提示
\n\n1 <= num.length <= 100num 仅由数字 '0' 到 '9' 组成num 不含任何前导零num contains a single zero digit then the answer is at most n - 1.",
"A number is divisible by 25 if its last two digits are 75, 50, 25, or 00.",
"Iterate over all possible pairs of indices i < j such that num[i] * 10 + num[j] is in [00,25,50,75]. Then, set the answer to min(answer, n - i - 2) ."
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