The concatenation of two numbers is the number formed by concatenating their numerals.

• For example, the concatenation of 15, 49 is 1549.

The concatenation value of nums is initially equal to 0. Perform this operation until nums becomes empty:

• If there exists more than one number in nums, pick the first element and last element in nums respectively and add the value of their concatenation to the concatenation value of nums, then delete the first and last element from nums.
• If one element exists, add its value to the concatenation value of nums, then delete it.

Return the concatenation value of the nums.

Example 1:

Input: nums = [7,52,2,4]
Output: 596
Explanation: Before performing any operation, nums is [7,52,2,4] and concatenation value is 0.
 - In the first operation:
We pick the first element, 7, and the last element, 4.
Their concatenation is 74, and we add it to the concatenation value, so it becomes equal to 74.
Then we delete them from nums, so nums becomes equal to [52,2].
 - In the second operation:
We pick the first element, 52, and the last element, 2.
Their concatenation is 522, and we add it to the concatenation value, so it becomes equal to 596.
Then we delete them from the nums, so nums becomes empty.
Since the concatenation value is 596 so the answer is 596.

Example 2:

Input: nums = [5,14,13,8,12]
Output: 673
Explanation: Before performing any operation, nums is [5,14,13,8,12] and concatenation value is 0.
 - In the first operation:
We pick the first element, 5, and the last element, 12.
Their concatenation is 512, and we add it to the concatenation value, so it becomes equal to 512.
Then we delete them from the nums, so nums becomes equal to [14,13,8].
 - In the second operation:
We pick the first element, 14, and the last element, 8.
Their concatenation is 148, and we add it to the concatenation value, so it becomes equal to 660.
Then we delete them from the nums, so nums becomes equal to [13].
 - In the third operation:
nums has only one element, so we pick 13 and add it to the concatenation value, so it becomes equal to 673.
Then we delete it from nums, so nums become empty.
Since the concatenation value is 673 so the answer is 673.

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 104

Solution: Follow the rules

Time complexity: O(sum(log(nums[i]))
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  long long findTheArrayConcVal(vector<int>& nums) {
    long long ans = 0;
    const int n = nums.size();
    for (int i = 0; i < n / 2; ++i) {
      int cur  = nums[i];
      for (int t = nums[n - i - 1]; t > 0; t /= 10)
        cur *= 10;    
      cur += nums[n - i - 1];
      ans += cur;
    }          
    if (n & 1) ans += nums[n / 2];
    return ans;
  }
};

The post 花花酱 LeetCode 2562. Find the Array Concatenation Value appeared first on Huahua's Tech Road.

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.

Solution1: Construct the string

Time complexity: O(l1 + l2)
Space complexity: O(l1 + l2)

class Solution {
public:
  bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
    string s1, s2;
    for (const string& w1 : word1) s1 += w1;
    for (const string& w2 : word2) s2 += w2;
    return s1 == s2;
  }
};

Solution 2: Pointers

Time complexity: O(l1 + l2)
Space complexity: O(1)

class Solution {
public:
  bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
    int i1 = 0, j1 = 0;
    int i2 = 0, j2 = 0;
    while (i1 < word1.size() || i2 < word2.size()) {
      char c1 = i1 < word1.size() ? word1[i1][j1++] : '\0';
      char c2 = i2 < word2.size() ? word2[i2][j2++] : '\0';
      if (c1 != c2) return false;
      if (i1 < word1.size() && j1 == word1[i1].length())
        ++i1, j1 = 0;
      if (i2 < word2.size() && j2 == word2[i2].length())
        ++i2, j2 = 0;
    }
    return true;
  }
};

The post 花花酱 LeetCode 1662. Check If Two String Arrays are Equivalent appeared first on Huahua's Tech Road.