# 2937. Make Three Strings Equal

## Description

You are given three strings s1, s2, and s3. You have to perform the following operation on these three strings as many times as you want.

In one operation you can choose one of these three strings such that its length is at least 2 and delete the rightmost character of it.

Return the minimum number of operations you need to perform to make the three strings equal if there is a way to make them equal, otherwise, return -1.

Example 1:

Input: s1 = "abc", s2 = "abb", s3 = "ab"
Output: 2
Explanation: Performing operations on s1 and s2 once will lead to three equal strings.
It can be shown that there is no way to make them equal with less than two operations.

Example 2:

Input: s1 = "dac", s2 = "bac", s3 = "cac"
Output: -1
Explanation: Because the leftmost letters of s1 and s2 are not equal, they could not be equal after any number of operations. So the answer is -1.


Constraints:

• 1 <= s1.length, s2.length, s3.length <= 100
• s1, s2 and s3 consist only of lowercase English letters.

## Solutions

Solution 1: Enumeration

According to the problem description, we know that if the three strings are equal after deleting characters, then they have a common prefix of length greater than $1$. Therefore, we can enumerate the position $i$ of the common prefix. If the three characters at the current index $i$ are not all equal, then the length of the common prefix is $i$. At this point, we check if $i$ is $0$. If it is, return $-1$. Otherwise, return $s - 3 \times i$, where $s$ is the sum of the lengths of the three strings.

The time complexity is $O(n)$, where $n$ is the minimum length of the three strings. The space complexity is $O(1)$.

• class Solution {
public int findMinimumOperations(String s1, String s2, String s3) {
int s = s1.length() + s2.length() + s3.length();
int n = Math.min(Math.min(s1.length(), s2.length()), s3.length());
for (int i = 0; i < n; ++i) {
if (!(s1.charAt(i) == s2.charAt(i) && s2.charAt(i) == s3.charAt(i))) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
}

• class Solution {
public:
int findMinimumOperations(string s1, string s2, string s3) {
int s = s1.size() + s2.size() + s3.size();
int n = min({s1.size(), s2.size(), s3.size()});
for (int i = 0; i < n; ++i) {
if (!(s1[i] == s2[i] && s2[i] == s3[i])) {
return i == 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}
};

• class Solution:
def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
s = len(s1) + len(s2) + len(s3)
n = min(len(s1), len(s2), len(s3))
for i in range(n):
if not s1[i] == s2[i] == s3[i]:
return -1 if i == 0 else s - 3 * i
return s - 3 * n


• func findMinimumOperations(s1 string, s2 string, s3 string) int {
s := len(s1) + len(s2) + len(s3)
n := min(len(s1), len(s2), len(s3))
for i := range s1[:n] {
if !(s1[i] == s2[i] && s2[i] == s3[i]) {
if i == 0 {
return -1
}
return s - 3*i
}
}
return s - 3*n
}

• function findMinimumOperations(s1: string, s2: string, s3: string): number {
const s = s1.length + s2.length + s3.length;
const n = Math.min(s1.length, s2.length, s3.length);
for (let i = 0; i < n; ++i) {
if (!(s1[i] === s2[i] && s2[i] === s3[i])) {
return i === 0 ? -1 : s - 3 * i;
}
}
return s - 3 * n;
}