1055. Shortest Way to Form String π
Description
A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace"
is a subsequence of "abcde"
while "aec"
is not).
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:
1 <= source.length, target.length <= 1000
source
andtarget
consist of lowercase English letters.
Solutions
Solution 1: Two Pointers
We can use the two pointers method, where pointer $j$ points to the target string target
. Then we traverse the source string source
with pointer $i$ pointing to the source string source
. If $source[i] = target[j]$, then both $i$ and $j$ move one step forward, otherwise only pointer $i$ moves. When both pointers $i$ and $j$ reach the end of the string, if no equal character is found, return $-1$, otherwise the subsequence count increases by one, and then set pointer $i$ to $0$ and continue to traverse.
After the traversal ends, return the subsequence count.
The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of the strings source
and target
respectively. The space complexity is $O(1)$.
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