Find Triangular Sum of an Array - Complete Solution Guide
Find Triangular Sum of an Array is LeetCode problem 2221, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.
Problem Statement
You are given a 0-indexed integer array nums , where nums[i] is a digit between 0 and 9 ( inclusive ). The triangular sum of nums is the value of the only element present in nums after the following process terminates: Let nums comprise of n elements. If n == 1 , end the process. Otherwise, create a new 0-indexed integer array newNums of length n - 1 . For each index i , where 0 <= i < n - 1 , assign the value of newNums[i] as (nums[i] + nums[i+1]) % 10 , where % denotes modulo operator. Replace
Detailed Explanation
The problem requires you to calculate the 'triangular sum' of an integer array. The array initially contains digits between 0 and 9. The triangular sum is computed iteratively. In each iteration, a new array is created where each element is the sum (modulo 10) of adjacent elements in the previous array. This process continues until only one element remains in the array, which is the triangular sum.
Solution Approach
The provided code directly implements the algorithm described in the problem statement. It uses a `while` loop to continue the process until the array size reduces to 1. Inside the `while` loop, a `for` loop iterates through the array and updates each element with the sum of it and the next element modulo 10. After each iteration of the `while` loop, the effective size of the array decreases by 1 because the last element is 'removed' in terms of the algorithm's process. This process continues until only one element remains, which is then returned.
Step-by-Step Algorithm
- Step 1: Determine the size of the input array `nums` and store it in the variable `n`.
- Step 2: Start a `while` loop that continues as long as `n` is greater than 1. This ensures the algorithm continues until the array size reduces to 1.
- Step 3: Inside the `while` loop, iterate through the array using a `for` loop from index `i = 0` to `n - 2`.
- Step 4: Within the `for` loop, update the element at index `i` with the sum of the element at `i` and the element at `i + 1`, taken modulo 10: `nums[i] = (nums[i] + nums[i + 1]) % 10`.
- Step 5: After the inner `for` loop finishes, decrement `n` by 1 to reflect the reduction in the effective array size.
- Step 6: After the `while` loop terminates (when `n` becomes 1), return the element at index 0 of the array `nums`, as this is the triangular sum.
Key Insights
- Insight 1: The problem can be solved by directly simulating the described process.
- Insight 2: The modulo operator (%) is crucial to keep the values within the range of 0-9.
- Insight 3: The in-place modification of the input array helps reduce space complexity.
Complexity Analysis
Time Complexity: O(n^2)
Space Complexity: O(1)
Topics
This problem involves: Array, Math, Simulation, Combinatorics.
Companies
Asked at: Zoho.