You are given a 0-indexed string s
that you must perform k
replacement operations on. The replacement operations are given as three 0-indexed parallel arrays, indices
, sources
, and targets
, all of length k
.
To complete the ith
replacement operation:
- Check if the substring
sources[i]
occurs at indexindices[i]
in the original strings
. - If it does not occur, do nothing.
- Otherwise if it does occur, replace that substring with
targets[i]
.
For example, if s = "abcd"
, indices[i] = 0
, sources[i] = "ab"
, and targets[i] = "eee"
, then the result of this replacement will be "eeecd"
.
All replacement operations must occur simultaneously, meaning the replacement operations should not affect the indexing of each other. The testcases will be generated such that the replacements will not overlap.
- For example, a testcase with
s = "abc"
,indices = [0, 1]
, andsources = ["ab","bc"]
will not be generated because the"ab"
and"bc"
replacements overlap.
Return the resulting string after performing all replacement operations on s
.
A substring is a contiguous sequence of characters in a string.
Example 1:
Input: s = "abcd", indices = [0, 2], sources = ["a", "cd"], targets = ["eee", "ffff"]
Output: "eeebffff"
Explanation:
"a" occurs at index 0 in s, so we replace it with "eee".
"cd" occurs at index 2 in s, so we replace it with "ffff".
Example 2:
Input: s = "abcd", indices = [0, 2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation:
"ab" occurs at index 0 in s, so we replace it with "eee".
"ec" does not occur at index 2 in s, so we do nothing.
Constraints:
-
1 <= s.length <= 1000
-
k == indices.length == sources.length == targets.length
-
1 <= k <= 100
-
0 <= indexes[i] < s.length
-
1 <= sources[i].length, targets[i].length <= 50
-
s
consists of only lowercase English letters. -
sources[i]
andtargets[i]
consist of only lowercase English letters.
SOLUTION:
class Solution:
def findReplaceString(self, s: str, indices: List[int], sources: List[str], targets: List[str]) -> str:
n = len(indices)
op = list(s)
for i in range(n):
if s[indices[i]:indices[i]+len(sources[i])] == sources[i]:
op[indices[i]] = targets[i]
for j in range(indices[i] + 1, indices[i] + len(sources[i])):
op[j] = ""
return "".join(op)
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