You are given two positive integers x and y.
In one operation, you can do one of the four following operations:
\n\nx by 11 if x is a multiple of 11.x by 5 if x is a multiple of 5.x by 1.x by 1.Return the minimum number of operations required to make x and y equal.
\n
Example 1:
\n\n\nInput: x = 26, y = 1\nOutput: 3\nExplanation: We can make 26 equal to 1 by applying the following operations: \n1. Decrement x by 1\n2. Divide x by 5\n3. Divide x by 5\nIt can be shown that 3 is the minimum number of operations required to make 26 equal to 1.\n\n\n
Example 2:
\n\n\nInput: x = 54, y = 2\nOutput: 4\nExplanation: We can make 54 equal to 2 by applying the following operations: \n1. Increment x by 1\n2. Divide x by 11 \n3. Divide x by 5\n4. Increment x by 1\nIt can be shown that 4 is the minimum number of operations required to make 54 equal to 2.\n\n\n
Example 3:
\n\n\nInput: x = 25, y = 30\nOutput: 5\nExplanation: We can make 25 equal to 30 by applying the following operations: \n1. Increment x by 1\n2. Increment x by 1\n3. Increment x by 1\n4. Increment x by 1\n5. Increment x by 1\nIt can be shown that 5 is the minimum number of operations required to make 25 equal to 30.\n\n\n
\n
Constraints:
\n\n1 <= x, y <= 104给你两个正整数 x 和 y 。
一次操作中，你可以执行以下四种操作之一：
\n\nx 是 11 的倍数，将 x 除以 11 。x 是 5 的倍数，将 x 除以 5 。x 减 1 。x 加 1 。请你返回让 x 和 y 相等的 最少 操作次数。
\n\n
示例 1：
\n\n\n输入：x = 26, y = 1\n输出：3\n解释：我们可以通过以下操作将 26 变为 1 ：\n1. 将 x 减 1\n2. 将 x 除以 5\n3. 将 x 除以 5\n将 26 变为 1 最少需要 3 次操作。\n\n\n
示例 2：
\n\n\n输入：x = 54, y = 2\n输出：4\n解释：我们可以通过以下操作将 54 变为 2 ：\n1. 将 x 加 1\n2. 将 x 除以 11\n3. 将 x 除以 5\n4. 将 x 加 1\n将 54 变为 2 最少需要 4 次操作。\n\n\n
示例 3：
\n\n\n输入：x = 25, y = 30\n输出：5\n解释：我们可以通过以下操作将 25 变为 30 ：\n1. 将 x 加 1\n2. 将 x 加 1\n3. 将 x 加 1\n4. 将 x 加 1\n5. 将 x 加 1\n将 25 变为 30 最少需要 5 次操作。\n\n\n
\n\n
提示：
\n\n1 <= x, y <= 104x larger is to increase it by 1 so if y >= x the answer is y - x.",
"For y < x, x - y is always a candidate answer since we can repeatedly decrease x by one to reach y.",
"We can also increase x and then use the division operations. For example, if x = 10 and y = 1, we can increment x by 1 then divide it by 11.",
"Find an upper bound U on the maximum value of x we will reach an optimal solution. Since all values of x will be in the range [1, U], we can use BFS to find the answer.",
"One possible upper bound on x is U = x + (x - y) . To reach any number strictly greater than U from x, we will need more than x - y operations which is not optimal since we can always reach y in x - y operations."
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