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" Add: "5121" Rotate: "2151" Add: "2050" 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" Add: "42" Rotate: "24" 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
Related Topics:
Depth-first Search, Breadth-first Search
// OJ: https://leetcode.com/problems/lexicographically-smallest-string-after-applying-operations/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
string first(string &s, int a) {
string ans = s;
for (int i = 1; i < s.size(); i += 2) ans[i] = '0' + (ans[i] - '0' + a) % 10;
return ans;
}
string second(string &s, int b) {
string ans = s;
for (int i = 0, j = s.size() - b; i < s.size(); ++i, ++j) ans[i] = s[j % s.size()];
return ans;
}
public:
string findLexSmallestString(string s, int a, int b) {
unordered_set<string> seen;
queue<string> q;
q.push(s);
string ans = s;
while (q.size()) {
auto s = q.front();
q.pop();
if (s < ans) ans = s;
string x = first(s, a), y = second(s, b);
if (seen.count(x) == 0) {
q.push(x);
seen.insert(x);
}
if (seen.count(y) == 0) {
q.push(y);
seen.insert(y);
}
}
return ans;
}
};