{ "data": { "question": { "questionId": "3033", "questionFrontendId": "2896", "categoryTitle": "Algorithms", "boundTopicId": 2470809, "title": "Apply Operations to Make Two Strings Equal", "titleSlug": "apply-operations-to-make-two-strings-equal", "content": "
You are given two 0-indexed binary strings s1
and s2
, both of length n
, and a positive integer x
.
You can perform any of the following operations on the string s1
any number of times:
i
and j
, and flip both s1[i]
and s1[j]
. The cost of this operation is x
.i
such that i < n - 1
and flip both s1[i]
and s1[i + 1]
. The cost of this operation is 1
.Return the minimum cost needed to make the strings s1
and s2
equal, or return -1
if it is impossible.
Note that flipping a character means changing it from 0
to 1
or vice-versa.
\n
Example 1:
\n\n\nInput: s1 = "1100011000", s2 = "0101001010", x = 2\nOutput: 4\nExplanation: We can do the following operations:\n- Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".\n- Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".\n- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2.\nThe total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.\n\n\n
Example 2:
\n\n\nInput: s1 = "10110", s2 = "00011", x = 4\nOutput: -1\nExplanation: It is not possible to make the two strings equal.\n\n\n
\n
Constraints:
\n\nn == s1.length == s2.length
1 <= n, x <= 500
s1
and s2
consist only of the characters '0'
and '1'
.给你两个下标从 0 开始的二进制字符串 s1
和 s2
,两个字符串的长度都是 n
,再给你一个正整数 x
。
你可以对字符串 s1
执行以下操作 任意次 :
i
和 j
,将 s1[i]
和 s1[j]
都反转,操作的代价为 x
。i < n - 1
的下标 i
,反转 s1[i]
和 s1[i + 1]
,操作的代价为 1
。请你返回使字符串 s1
和 s2
相等的 最小 操作代价之和,如果无法让二者相等,返回 -1
。
注意 ,反转字符的意思是将 0
变成 1
,或者 1
变成 0
。
\n\n
示例 1:
\n\n\n输入:s1 = \"1100011000\", s2 = \"0101001010\", x = 2\n输出:4\n解释:我们可以执行以下操作:\n- 选择 i = 3 执行第二个操作。结果字符串是 s1 = \"1101111000\" 。\n- 选择 i = 4 执行第二个操作。结果字符串是 s1 = \"1101001000\" 。\n- 选择 i = 0 和 j = 8 ,执行第一个操作。结果字符串是 s1 = \"0101001010\" = s2 。\n总代价是 1 + 1 + 2 = 4 。这是最小代价和。\n\n\n
示例 2:
\n\n\n输入:s1 = \"10110\", s2 = \"00011\", x = 4\n输出:-1\n解释:无法使两个字符串相等。\n\n\n
\n\n
提示:
\n\nn == s1.length == s2.length
1 <= n, x <= 500
s1
和 s2
只包含字符 '0'
和 '1'
。s1
and s2
into a list, and work only with this list.",
"Try to use dynamic programming on this list to solve the problem. What will be the states and transitions of this dp?"
],
"solution": null,
"status": null,
"sampleTestCase": "\"1100011000\"\n\"0101001010\"\n2",
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