Leetcode 2221: Find Triangular Sum of an Array
You are given a 0-indexed integer array
nums, where each element is a digit between 0 and 9. The triangular sum of nums is computed by repeatedly reducing the array by replacing each element with the sum of two consecutive elements, modulo 10. This process continues until only one element remains, which is the triangular sum. Your task is to return the triangular sum of the array.Problem
Approach
Steps
Complexity
Input: The input consists of an array `nums`, where each element is a digit between 0 and 9.
Example: nums = [3, 7, 4, 9, 1]
Constraints:
• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 9
Output: Return the triangular sum of the array `nums`.
Example: Output: 5
Constraints:
• The array will always have at least one element.
Goal: The goal is to compute the triangular sum of the array `nums` by applying the given process until one element remains.
Steps:
• Start with the input array `nums`.
• While the array has more than one element, create a new array by replacing each element with the sum of two consecutive elements, modulo 10.
• Repeat the process until only one element remains, which will be the triangular sum.
Goal: The array `nums` can have up to 1000 elements, and each element is a digit between 0 and 9.
Steps:
• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 9
Assumptions:
• The array will always contain at least one element.
• Input: Input: nums = [3, 7, 4, 9, 1]
• Explanation: The triangular sum is computed as follows: First, compute the sums of consecutive elements modulo 10: [3+7, 7+4, 4+9, 9+1] = [0, 1, 3, 0]. Then compute the sums of consecutive elements in this new array: [0+1, 1+3, 3+0] = [1, 4, 3]. Finally, compute the sum of the remaining elements: [1+4] = [5]. Hence, the triangular sum is 5.
• Input: Input: nums = [5]
• Explanation: Since there is only one element in the array, the triangular sum is the value of that element itself, which is 5.
Approach: The approach involves repeatedly reducing the array `nums` by applying the modulo 10 sum of consecutive elements until only one element remains, which is the triangular sum.
Observations:
• Each reduction step generates a new array that is shorter by one element.
• Modulo 10 operation ensures that the values remain within the range of a single digit.
• This is a straightforward problem that can be solved with a loop that continuously reduces the array until only one element is left.
Steps:
• Initialize the array `nums`.
• While the length of the array is greater than 1, iterate through the array and compute the sum of consecutive elements modulo 10, storing the result in a new array.
• Repeat this process until only one element remains in the array, which will be the triangular sum.
Empty Inputs:
• There will always be at least one element in the input array, so there are no empty arrays to consider.
Large Inputs:
• The solution should efficiently handle arrays with up to 1000 elements.
Special Values:
• If the array has only one element, the triangular sum is the element itself.
Constraints:
• The problem must handle arrays with elements ranging from 0 to 9.
int triangularSum(vector<int>& nums) {
int n = nums.size();
vector<int> arr(n, 0);
for(int i = n; i > 1; i--) {
int k = i - 1;
for(int j = k - 1; j >= 0; j--)
arr[j] = (nums[j] % 10 + nums[j + 1] % 10) % 10;
nums = arr;
}
return nums[0];
}
1 : Function Declaration
int triangularSum(vector<int>& nums) {
This is the function declaration for `triangularSum`, which takes a vector of integers `nums` and returns an integer representing the final triangular sum.
2 : Variable Initialization
int n = nums.size();
The size of the input vector `nums` is stored in the variable `n`, which represents the number of elements in the array.
3 : Array Initialization
vector<int> arr(n, 0);
A new vector `arr` of size `n` is created and initialized with zeros. This vector will hold intermediate results during the iteration.
4 : Outer Loop
for(int i = n; i > 1; i--) {
An outer loop is used to iterate through the array, reducing the size of the array in each step. The loop runs as long as the size of the array is greater than 1.
5 : Index Calculation
int k = i - 1;
The variable `k` is used to store the current index for the inner loop, which will process elements up to the `i-1` position.
6 : Inner Loop
for(int j = k - 1; j >= 0; j--)
An inner loop is used to iterate through the array and calculate the new values for `arr[j]` by summing adjacent elements from the original `nums` array.
7 : Element Calculation
arr[j] = (nums[j] % 10 + nums[j + 1] % 10) % 10;
In this step, the adjacent elements `nums[j]` and `nums[j + 1]` are summed modulo 10 and stored in `arr[j]` to create the new triangular sum.
8 : Empty Line
An empty line used for separation of code logic.
9 : Update Vector
nums = arr;
The updated vector `arr` is assigned to `nums` so that it will be used in the next iteration of the outer loop.
10 : Return Final Result
return nums[0];
The function returns the final value from the `nums` array, which represents the triangular sum after all iterations.
Best Case: O(n), where n is the length of the input array, as each reduction step involves iterating through the array.
Average Case: O(n), as each reduction step takes linear time.
Worst Case: O(n^2), as the number of iterations grows with the length of the array.
Description: The time complexity is linear in the number of elements in the array, but the process repeats for multiple iterations.
Best Case: O(n), as we always need to store the array at each step.
Worst Case: O(n), as a new array is created in each iteration, requiring O(n) space.
Description: The space complexity is O(n), where n is the length of the input array, as we need to store intermediate arrays.
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