Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
You are given two strings start
and target
, both of length n
. Each string consists only of the characters 'L'
, 'R'
, and '_'
where:
'L'
and 'R'
represent pieces, where a piece 'L'
can move to the left only if there is a blank space directly to its left, and a piece 'R'
can move to the right only if there is a blank space directly to its right.'_'
represents a blank space that can be occupied by any of the 'L'
or 'R'
pieces.Return true
if it is possible to obtain the string target
by moving the pieces of the string start
any number of times. Otherwise, return false
.
Example 1:
Input: start = "_L__R__R_", target = "L______RR"
Output: true
Explanation: We can obtain the string target from start by doing the following moves:
- Move the first piece one step to the left, start becomes equal to "L___R__R_".
- Move the last piece one step to the right, start becomes equal to "L___R___R".
- Move the second piece three steps to the right, start becomes equal to "L______RR".
Since it is possible to get the string target from start, we return true.
Example 2:
Input: start = "R_L_", target = "__LR"
Output: false
Explanation: The 'R' piece in the string start can move one step to the right to obtain "_RL_".
After that, no pieces can move anymore, so it is impossible to obtain the string target from start.
Example 3:
Input: start = "_R", target = "R_"
Output: false
Explanation: The piece in the string start can move only to the right, so it is impossible to obtain the string target from start.
Constraints:
n == start.length == target.length
1 <= n <= 105
start
and target
consist of the characters 'L'
, 'R'
, and '_'
. public boolean canChange(String start, String target) {
if (!(start.replaceAll("_", "")).equals(target.replaceAll("_", ""))) {
return false;
}
for (int i = 0, j = 0, n = start.length(); i < n && j < n; ++i, ++j) {
while (i < n && start.charAt(i) == '_') {
++i;
}
while (j < n && target.charAt(j) == '_') {
++j;
}
if (i < n && j < n && (start.charAt(i) == 'L' && i < j || target.charAt(j) == 'R' && i > j)) {
return false;
}
}
return true;
}
class Solution {
public:
bool canChange(string s, string t) {
queue<pair<char, int>> ss, ts;
// Fill queue with start
for(int i=0; i<s.size(); i++){
if(s[i] != '_')ss.push({s[i], i});
}
// Fill queue with target
for(int i=0; i<t.size(); i++){
if(t[i] != '_')ts.push({t[i], i});
}
if(ss.size() != ts.size())return false;
while(ss.size()){
pair<char, int> sp, tp;
sp = ss.front();
tp = ts.front();
ss.pop();
ts.pop();
// If both the letters don;t match return false, or check for the index to move Left and Right
if(sp.first != tp.first) return false;
if(sp.first == 'L' && tp.second>sp.second){
return false;
} else if(sp.first == 'R' && tp.second<sp.second){
return false;
}
}
return true;
}
};
class Solution:
def canChange(self, A: str, B: str) -> bool:
P = lambda c : c != '_'
I = lambda s,x : [i for i,c in enumerate(s) if c==x]
G = lambda d,p : all( p(x,y) for x,y in zip( I(A,d), I(B,d) ) )
S = lambda : [*filter(P,A)] == [*filter(P,B)]
return S() and G('L', ge) and G('R', le)
# 1. 2. 3.
In our experience, we suggest you solve this Move Pieces to Obtain a String LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.
If you are stuck anywhere between any coding problem, just visit Queslers to get the Move Pieces to Obtain a String LeetCode Solution
I hope this Move Pieces to Obtain a String LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.
This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.
Keep Learning!
More Coding Solutions >>