Given a string s, find the sum of the 3 largest unique prime numbers that can be formed using any of its substrings.
Return the sum of the three largest unique prime numbers that can be formed. If fewer than three exist, return the sum of all available primes. If no prime numbers can be formed, return 0.
\n\nA prime number is a natural number greater than 1 with only two factors, 1 and itself.
\n\nA substring is a contiguous sequence of characters within a string.
\n\nNote: Each prime number should be counted only once, even if it appears in multiple substrings. Additionally, when converting a substring to an integer, any leading zeros are ignored.
\n\n\n
Example 1:
\n\nInput: s = "12234"
\n\nOutput: 1469
\n\nExplanation:
\n\n"12234" are 2, 3, 23, 223, and 1223.Example 2:
\n\nInput: s = "111"
\n\nOutput: 11
\n\nExplanation:
\n\n"111" is 11.\n
Constraints:
\n\n1 <= s.length <= 10s consists of only digits.给定一个字符串 s，找出可以由其 子字符串 组成的 3个最大的不同质数 的和。
返回这些质数的 总和 ，如果少于 3 个不同的质数，则返回 所有 不同质数的和。
\n\n质数是大于 1 且只有两个因数的自然数：1和它本身。
\n\n子字符串 是字符串中的一个连续字符序列。
\n\n注意：每个质数即使出现在 多个 子字符串中，也只能计算 一次 。此外，将子字符串转换为整数时，忽略任何前导零。
\n\n\n\n
示例 1：
\n\n输入： s = \"12234\"
\n\n输出： 1469
\n\n解释：
\n\n\"12234\" 的子字符串形成的不同质数为 2 ，3 ，23 ，223 和 1223。示例 2：
\n\n输入： s = \"111\"
\n\n输出： 11
\n\n解释：
\n\n\"111\" 的子字符串形成的不同质数是 11。\n\n
提示：
\n\n1 <= s.length <= 10s 仅由数字组成。s to generate candidate numbers.",
"Check each candidate for primality in O(sqrt(n)) time.",
"Store unique primes, then sum the three largest (or all if fewer than three)."
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