You are given a string s
of even length consisting of digits from 0
to 9
, and two integers a
and b
.
You can apply either of the following two operations any number of times and in any order on s
:
- Add
a
to all odd indices ofs
(0-indexed). Digits post9
are cycled back to0
. For example, ifs = "3456"
anda = 5
,s
becomes"3951"
. - Rotate
s
to the right byb
positions. For example, ifs = "3456"
andb = 1
,s
becomes"6345"
.
Return the lexicographically smallest string you can obtain by applying the above operations any number of times on s
.
A string a
is lexicographically smaller than a string b
(of the same length) if in the first position where a
and b
differ, string a
has a letter that appears earlier in the alphabet than the corresponding letter in b
. For example, "0158"
is lexicographically smaller than "0190"
because the first position they differ is at the third letter, and '5'
comes before '9'
.
Example 1:
Input: s = "5525", a = 9, b = 2 Output: "2050" Explanation: We can apply the following operations: Start: "5525" Rotate: "2555" Add: "2454" Add: "2353" Rotate: "5323" Add: "5222" โโโโโโโAdd: "5121" โโโโโโโRotate: "2151" โโโโโโโAdd: "2050"โโโโโโโโโโโโ There is no way to obtain a string that is lexicographically smaller then "2050".
Example 2:
Input: s = "74", a = 5, b = 1 Output: "24" Explanation: We can apply the following operations: Start: "74" Rotate: "47" โโโโโโโAdd: "42" โโโโโโโRotate: "24"โโโโโโโโโโโโ There is no way to obtain a string that is lexicographically smaller then "24".
Example 3:
Input: s = "0011", a = 4, b = 2 Output: "0011" Explanation: There are no sequence of operations that will give us a lexicographically smaller string than "0011".
Example 4:
Input: s = "43987654", a = 7, b = 3 Output: "00553311"
Constraints:
2 <= s.length <= 100
s.length
is even.s
consists of digits from0
to9
only.1 <= a <= 9
1 <= b <= s.length - 1
Solution: Search
C++
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// Author: Huahua class Solution { public: string findLexSmallestString(string s, int a, int b) { unordered_set<string> seen; string ans(s); function<void(string)> dfs = [&](const string& s) { if (!seen.insert(s).second) return; ans = min(ans, s); string t(s); for (int i = 1; i < s.length(); i += 2) t[i] = (t[i] - '0' + a) % 10 + '0'; dfs(t); dfs(s.substr(b) + s.substr(0, b)); }; dfs(s); return ans; } }; |