Leetcode 2562: Find the Array Concatenation Value
You are given an array of integers, nums, and you need to calculate the total concatenation value. To calculate this, you repeatedly perform the following operations until the array is empty:
- If the array has more than one element, take the first and last elements, concatenate them, and add the result to the total concatenation value. Then remove the first and last elements from the array.
- If the array has only one element, add it to the total concatenation value and remove the element.
The goal is to return the total concatenation value after performing all the operations.
Problem
Approach
Steps
Complexity
Input: The input consists of a single integer array `nums` of length `n` (1 <= n <= 1000), where each element of the array is a positive integer (1 <= nums[i] <= 104).
Example: Input: nums = [8, 32, 5, 1]
Constraints:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 104
Output: Return the total concatenation value after performing the above operations on the array.
Example: Output: 398
Constraints:
• The concatenation value is guaranteed to fit within the range of a 32-bit integer.
Goal: The goal is to repeatedly concatenate the first and last elements of the array until it becomes empty and calculate the total concatenation value.
Steps:
• Initialize a variable to store the concatenation value.
• Repeat the operation of concatenating the first and last elements, removing them from the array, and adding the concatenated result to the value.
• If only one element is left, simply add it to the value.
Goal: The constraints are based on the array length and element values, ensuring that the solution can handle arrays of length up to 1000 and elements within the range [1, 104].
Steps:
• The length of the input array will always be at least 1 and at most 1000.
• Each element in the array will be a positive integer between 1 and 104.
Assumptions:
• The array will always have at least one element.
• The array elements are positive integers.
• Input: Example 1: Input: nums = [8, 32, 5, 1], Output: 398
• Explanation: In this example, the array starts as [8, 32, 5, 1]. The first operation concatenates 8 and 1 to form 81. Then the array becomes [32, 5]. The second operation concatenates 32 and 5 to form 325. Finally, the total value is 398.
• Input: Example 2: Input: nums = [1, 12, 8, 7, 2], Output: 160
• Explanation: In this example, the first concatenation adds 12, then 127, and finally adds 8, resulting in a total value of 160.
Approach: The approach to solving this problem is to simulate the process of repeatedly concatenating the first and last elements, updating the total concatenation value at each step.
Observations:
• The problem is straightforward and involves array manipulation.
• The challenge is ensuring the proper concatenation of the first and last elements at each step.
• We can perform this task in a loop, ensuring we handle the case of one remaining element properly.
Steps:
• Initialize a variable to store the result.
• Use a loop to repeatedly process the first and last elements of the array.
• Concatenate these elements, add the result to the total value, and remove them from the array.
• Handle the case where only one element remains.
Empty Inputs:
• If the array is empty (though not applicable in this case due to constraints), the result would be 0.
Large Inputs:
• Ensure the solution handles arrays up to the size limit of 1000 elements.
Special Values:
• If the array contains only one element, return the value of that element.
Constraints:
• The solution must ensure that all concatenations fit within the 32-bit integer limit.
long long findTheArrayConcVal(vector<int>& nums) {
long long res = 0, sz = nums.size();
for (int i = 0, j = sz - 1; i <= j; ++i, --j)
if (i < j)
res += nums[i] * pow(10, (int)log10(nums[j]) + 1) + nums[j];
else
res += nums[i];
return res;
}
1 : Function Declaration
long long findTheArrayConcVal(vector<int>& nums) {
This is the function definition where `findTheArrayConcVal` accepts a vector of integers and returns a long long integer representing the concatenation value.
2 : Variable Initialization
long long res = 0, sz = nums.size();
The variables `res` and `sz` are initialized. `res` will hold the final result, and `sz` stores the size of the input array `nums`.
3 : Loop Initialization
for (int i = 0, j = sz - 1; i <= j; ++i, --j)
A loop is set up where two pointers, `i` and `j`, start from the beginning and end of the array, respectively, and move towards the center. The loop runs as long as `i <= j`.
4 : Conditional Branch
if (i < j)
Inside the loop, there is a condition checking if `i` is less than `j`, which means there are still two elements to pair.
5 : Concatenation Logic
res += nums[i] * pow(10, (int)log10(nums[j]) + 1) + nums[j];
If `i < j`, the elements `nums[i]` and `nums[j]` are concatenated by multiplying `nums[i]` by 10 raised to the power of the number of digits in `nums[j]`, then adding `nums[j]`.
6 : Else Condition
else
If `i` is equal to `j`, there is only one element left, and no concatenation is needed.
7 : Single Element Addition
res += nums[i];
In this case, simply add the element `nums[i]` to the result as no concatenation is required.
8 : Return Statement
return res;
After the loop completes, return the accumulated value `res` which represents the concatenated result of the array.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: Since we are processing each pair of elements in the array once, the time complexity is linear, O(n).
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, O(1), as we are not using any additional space beyond a few variables.
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