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Author: NinePiece2

Solutions/2324-find-triangular-sum-of-an-array/README.md

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+<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, where <code>nums[i]</code> is a digit between <code>0</code> and <code>9</code> (<strong>inclusive</strong>).</p>
+
+<p>The <strong>triangular sum</strong> of <code>nums</code> is the value of the only element present in <code>nums</code> after the following process terminates:</p>
+
+<ol>
+	<li>Let <code>nums</code> comprise of <code>n</code> elements. If <code>n == 1</code>, <strong>end</strong> the process. Otherwise, <strong>create</strong> a new <strong>0-indexed</strong> integer array <code>newNums</code> of length <code>n - 1</code>.</li>
+	<li>For each index <code>i</code>, where <code>0 &lt;= i &lt;&nbsp;n - 1</code>, <strong>assign</strong> the value of <code>newNums[i]</code> as <code>(nums[i] + nums[i+1]) % 10</code>, where <code>%</code> denotes modulo operator.</li>
+	<li><strong>Replace</strong> the array <code>nums</code> with <code>newNums</code>.</li>
+	<li><strong>Repeat</strong> the entire process starting from step 1.</li>
+</ol>
+
+<p>Return <em>the triangular sum of</em> <code>nums</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png" style="width: 250px; height: 250px;" />
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5]
+<strong>Output:</strong> 8
+<strong>Explanation:</strong>
+The above diagram depicts the process from which we obtain the triangular sum of the array.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5]
+<strong>Output:</strong> 5
+<strong>Explanation:</strong>
+Since there is only one element in nums, the triangular sum is the value of that element itself.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 9</code></li>
+</ul>

Solutions/2324-find-triangular-sum-of-an-array/solution.py

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+class Solution:
+    def triangularSum(self, nums: List[int]) -> int:
+        # n = len(nums)
+        # sums = [[0] * (n - i) for i in range(n)]
+
+        # for j in range(n):
+        #     sums[0][j] = nums[j]
+
+        # for i in range(1, n):
+        #     for j in range(n - i):
+        #         sums[i][j] = sums[i - 1][j] + sums[i - 1][j + 1]
+        
+        # return sums[-1][-1] % 10
+
+        result = 0
+        n = len(nums)
+        for i, num in enumerate(nums):
+            result += num * comb(n - 1, i)
+
+        return result % 10