3135. Equalize Strings by Adding or Removing Characters at Ends π ο
Descriptionο
Given two strings initial and target, your task is to modify initial by performing a series of operations to make it equal to target.
In one operation, you can add or remove one character only at the beginning or the end of the string initial.
Return the minimum number of operations required to transform initial into target.
Example 1:
Input: initial = "abcde", target = "cdef"
Output: 3
Explanation:
Remove 'a' and 'b' from the beginning of initial, then add 'f' to the end.
Example 2:
Input: initial = "axxy", target = "yabx"
Output: 6
Explanation:
| Operation | Resulting String |
|---|---|
Add 'y' to the beginning |
"yaxxy" |
| Remove from end | "yaxx" |
| Remove from end | "yax" |
| Remove from end | "ya" |
Add 'b' to the end |
"yab" |
Add 'x' to the end |
"yabx" |
Example 3:
Input: initial = "xyz", target = "xyz"
Output: 0
Explanation:
No operations are needed as the strings are already equal.
Constraints:
1 <= initial.length, target.length <= 1000initialandtargetconsist only of lowercase English letters.
Solutionsο
Solution 1: Dynamic Programmingο
Let's assume that the lengths of the strings initial and target are $m$ and $n$, respectively.
According to the problem description, we only need to find the length $mx$ of the longest common substring of initial and target. Then, we can delete $m - mx$ characters from initial and add $n - mx$ characters to transform initial into target. Therefore, the answer is $m + n - 2 \times mx$.
We can use dynamic programming to find the length $mx$ of the longest common substring of initial and target. We define a two-dimensional array $f$, where $f[i][j]$ represents the length of the longest common substring ending with initial[i - 1] and target[j - 1]. Then, we can get the state transition equation:
$$ f[i][j] = \begin{cases} f[i - 1][j - 1] + 1, & \textit{if } \textit{initial}[i - 1] = \textit{target}[j - 1], \ 0, & \textit{otherwise}. \end{cases} $$
Then $mx = \max f[i][j]$, and the final answer is $m + n - 2 \times mx$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of the strings initial and target, respectively.
Python3ο
class Solution:
def minOperations(self, initial: str, target: str) -> int:
m, n = len(initial), len(target)
f = [[0] * (n + 1) for _ in range(m + 1)]
mx = 0
for i, a in enumerate(initial, 1):
for j, b in enumerate(target, 1):
if a == b:
f[i][j] = f[i - 1][j - 1] + 1
mx = max(mx, f[i][j])
return m + n - mx * 2
Javaο
class Solution {
public int minOperations(String initial, String target) {
int m = initial.length(), n = target.length();
int[][] f = new int[m + 1][n + 1];
int mx = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (initial.charAt(i - 1) == target.charAt(j - 1)) {
f[i][j] = f[i - 1][j - 1] + 1;
mx = Math.max(mx, f[i][j]);
}
}
}
return m + n - 2 * mx;
}
}
C++ο
class Solution {
public:
int minOperations(string initial, string target) {
int m = initial.size(), n = target.size();
int f[m + 1][n + 1];
memset(f, 0, sizeof(f));
int mx = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (initial[i - 1] == target[j - 1]) {
f[i][j] = f[i - 1][j - 1] + 1;
mx = max(mx, f[i][j]);
}
}
}
return m + n - 2 * mx;
}
};
Goο
func minOperations(initial string, target string) int {
m, n := len(initial), len(target)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
mx := 0
for i, a := range initial {
for j, b := range target {
if a == b {
f[i+1][j+1] = f[i][j] + 1
mx = max(mx, f[i+1][j+1])
}
}
}
return m + n - 2*mx
}
TypeScriptο
function minOperations(initial: string, target: string): number {
const m = initial.length;
const n = target.length;
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
let mx: number = 0;
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (initial[i - 1] === target[j - 1]) {
f[i][j] = f[i - 1][j - 1] + 1;
mx = Math.max(mx, f[i][j]);
}
}
}
return m + n - 2 * mx;
}