Formatted question description: https://leetcode.ca/all/1055.html
1055. Shortest Way to Form String
Level
Medium
Description
From any string, we can form a subsequence of that string by deleting some number of characters (possibly no deletions).
Given two strings source
and target
, return the minimum number of subsequences of source
such that their concatenation equals target
. If the task is impossible, return -1
.
Example 1:
Input: source = “abc”, target = “abcbc”
Output: 2
Explanation: The target “abcbc” can be formed by “abc” and “bc”, which are subsequences of source “abc”.
Example 2:
Input: source = “abc”, target = “acdbc”
Output: -1
Explanation: The target string cannot be constructed from the subsequences of source string due to the character “d” in target string.
Example 3:
Input: source = “xyz”, target = “xzyxz”
Output: 3
Explanation: The target string can be constructed as follows “xz” + “y” + “xz”.
Constraints:
- Both the
source
andtarget
strings consist of only lowercase English letters from “a”-“z”. - The lengths of
source
andtarget
string are between1
and1000
.
Solution
Use two pointers in source
and target
respectively. Both pointers are initialized to 0.
Each time starting from the pointer of target
, find the longest subsequence of source
in target
. A subsequence in target
ends when either the pointer in source
reaches the end or the pointer in target
reaches the end. If a longest subsequence has length 0, then the character in target
does not exist in source
, so return -1. Otherwise, increase the number of subsequences by 1, and find the next subsequence in target
starting from the updated pointer in target
. Finally, return the number of subsequences.
class Solution {
public int shortestWay(String source, String target) {
int count = 0;
int targetLength = target.length();
int startIndex = 0;
while (startIndex < targetLength) {
startIndex = longestSubsequence(source, target, startIndex);
if (startIndex < 0)
return -1;
else
count++;
}
return count;
}
public int longestSubsequence(String source, String target, int startIndex) {
int sourceLength = source.length(), targetLength = target.length();
int sourceIndex = 0, targetIndex = startIndex;
while (sourceIndex < sourceLength && targetIndex < targetLength) {
char c1 = source.charAt(sourceIndex), c2 = target.charAt(targetIndex);
if (c1 == c2)
targetIndex++;
sourceIndex++;
}
if (targetIndex == startIndex)
return -1;
else
return targetIndex;
}
}