Description#
Given two string arrays word1
and word2
, return true
if the two arrays represent the same string, and false
otherwise.
A string is represented by an array if the array elements concatenated in order forms the string.
Example 1:
Input: word1 = ["ab", "c"], word2 = ["a", "bc"]
Output: true
Explanation:
word1 represents string "ab" + "c" -> "abc"
word2 represents string "a" + "bc" -> "abc"
The strings are the same, so return true.
Example 2:
Input: word1 = ["a", "cb"], word2 = ["ab", "c"]
Output: false
Example 3:
Input: word1 = ["abc", "d", "defg"], word2 = ["abcddefg"]
Output: true
Constraints:
1 <= word1.length, word2.length <= 103
1 <= word1[i].length, word2[i].length <= 103
1 <= sum(word1[i].length), sum(word2[i].length) <= 103
word1[i]
and word2[i]
consist of lowercase letters.
Solutions#
Solution 1#
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| class Solution:
def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
return ''.join(word1) == ''.join(word2)
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| class Solution {
public boolean arrayStringsAreEqual(String[] word1, String[] word2) {
return String.join("", word1).equals(String.join("", word2));
}
}
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| class Solution {
public:
bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
return reduce(word1.cbegin(), word1.cend()) == reduce(word2.cbegin(), word2.cend());
}
};
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| func arrayStringsAreEqual(word1 []string, word2 []string) bool {
return strings.Join(word1, "") == strings.Join(word2, "")
}
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| function arrayStringsAreEqual(word1: string[], word2: string[]): boolean {
return word1.join('') === word2.join('');
}
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| impl Solution {
pub fn array_strings_are_equal(word1: Vec<String>, word2: Vec<String>) -> bool {
word1.join("") == word2.join("")
}
}
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| bool arrayStringsAreEqual(char** word1, int word1Size, char** word2, int word2Size) {
int i = 0;
int j = 0;
int x = 0;
int y = 0;
while (i < word1Size && j < word2Size) {
if (word1[i][x++] != word2[j][y++]) {
return 0;
}
if (word1[i][x] == '\0') {
x = 0;
i++;
}
if (word2[j][y] == '\0') {
y = 0;
j++;
}
}
return i == word1Size && j == word2Size;
}
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Solution 2#
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| class Solution:
def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
i = j = x = y = 0
while i < len(word1) and j < len(word2):
if word1[i][x] != word2[j][y]:
return False
x, y = x + 1, y + 1
if x == len(word1[i]):
x, i = 0, i + 1
if y == len(word2[j]):
y, j = 0, j + 1
return i == len(word1) and j == len(word2)
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| class Solution {
public boolean arrayStringsAreEqual(String[] word1, String[] word2) {
int i = 0, j = 0;
int x = 0, y = 0;
while (i < word1.length && j < word2.length) {
if (word1[i].charAt(x++) != word2[j].charAt(y++)) {
return false;
}
if (x == word1[i].length()) {
x = 0;
++i;
}
if (y == word2[j].length()) {
y = 0;
++j;
}
}
return i == word1.length && j == word2.length;
}
}
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| class Solution {
public:
bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
int i = 0, j = 0, x = 0, y = 0;
while (i < word1.size() && j < word2.size()) {
if (word1[i][x++] != word2[j][y++]) return false;
if (x == word1[i].size()) x = 0, i++;
if (y == word2[j].size()) y = 0, j++;
}
return i == word1.size() && j == word2.size();
}
};
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| func arrayStringsAreEqual(word1 []string, word2 []string) bool {
var i, j, x, y int
for i < len(word1) && j < len(word2) {
if word1[i][x] != word2[j][y] {
return false
}
x, y = x+1, y+1
if x == len(word1[i]) {
x, i = 0, i+1
}
if y == len(word2[j]) {
y, j = 0, j+1
}
}
return i == len(word1) && j == len(word2)
}
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| function arrayStringsAreEqual(word1: string[], word2: string[]): boolean {
let [i, j, x, y] = [0, 0, 0, 0];
while (i < word1.length && j < word2.length) {
if (word1[i][x++] !== word2[j][y++]) {
return false;
}
if (x === word1[i].length) {
x = 0;
++i;
}
if (y === word2[j].length) {
y = 0;
++j;
}
}
return i === word1.length && j === word2.length;
}
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| impl Solution {
pub fn array_strings_are_equal(word1: Vec<String>, word2: Vec<String>) -> bool {
let (n, m) = (word1.len(), word2.len());
let (mut i, mut j, mut x, mut y) = (0, 0, 0, 0);
while i < n && j < m {
if word1[i].as_bytes()[x] != word2[j].as_bytes()[y] {
return false;
}
x += 1;
y += 1;
if x == word1[i].len() {
x = 0;
i += 1;
}
if y == word2[j].len() {
y = 0;
j += 1;
}
}
i == n && j == m
}
}
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