Lexicographically Smallest String After Applying Operations
Problem Statement - link #
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, if s ="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 ons
.
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'
.
Examples #
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
Solutions #
class Solution {
public:
string op_a(string s, int a){
for(int i=1; i<s.size(); i+=2)
s[i] = (s[i] - '0' + a) % 10 + '0';
return s;
}
string findLexSmallestString(string s, int a, int b) {
set<string> mp;
queue<string> q;
q.push(s);mp.insert(s);
string res = s;
while(q.empty()==false){
string temp = q.front();
q.pop();
res = min(res, temp);
string one = op_a(temp, a);
if(mp.find(one)==mp.end()) mp.insert(one), q.push(one);
rotate(temp.begin(), temp.begin()+b, temp.end());
if(mp.find(temp)==mp.end()) mp.insert(temp), q.push(temp);
}
return res;
}
};