833. Find And Replace in String
To some string
S
, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).Each replacement operation has
3
parameters: a starting indexi
, a source wordx
and a target wordy
. The rule is that ifx
starts at positioni
in the original stringS
, then we will replace that occurrence ofx
withy
. If not, we do nothing.For example, if we have
S = "abcd"
and we have some replacement operationi = 2, x = "cd", y = "ffff"
, then because"cd"
starts at position2
in the original stringS
, we will replace it with"ffff"
.Using another example on
S = "abcd"
, if we have both the replacement operationi = 0, x = "ab", y = "eee"
, as well as another replacement operationi = 2, x = "ec", y = "ffff"
, this second operation does nothing because in the original stringS[2] = 'c'
, which doesn't matchx[0] = 'e'
.All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example,
S = "abc", indexes = [0, 1], sources = ["ab","bc"]
is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"] Output: "eeebffff" Explanation: "a" starts at index 0 in S, so it's replaced by "eee". "cd" starts at index 2 in S, so it's replaced by "ffff".Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"] Output: "eeecd" Explanation: "ab" starts at index 0 in S, so it's replaced by "eee". "ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
- All characters in given inputs are lowercase letters.
Approach #1: String. [Java]
class Solution { public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < indexes.length; ++i) { if (S.startsWith(sources[i], indexes[i])) map.put(indexes[i], i); } StringBuilder sb = new StringBuilder(); for (int i = 0; i < S.length(); ) { if (map.containsKey(i)) { sb.append(targets[map.get(i)]); i += sources[map.get(i)].length(); } else { sb.append(S.charAt(i)); i++; } } return sb.toString(); } }
Analysis:
Idea: Use a StringBuilder to build up the result.
1. Iterate through the indexes array and find out all indices that support replacement. Then, store mapping from those index values to their indices in the indexes array into a map named map.
2. Iterate through str, at each iteration i, check whether we can perform replacement, i.e., map.get(i) != null, if yes, append targets[i] to the StringBuilder and increase i by sources[i].length()-1. if no, append str.charAt(i) and increment i.
Reference:
https://leetcode.com/problems/find-and-replace-in-string/discuss/134758/Java-O(n)-solution