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Commit d09f4b9

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feat: add solutions to lc problem: No.3284 (doocs#3515)
No.3284.Sum of Consecutive Subarrays
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‎lcof2/剑指 Offer II 098. 路径的数目/README.md‎

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@@ -254,7 +254,7 @@ class Solution {
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func uniquePaths(_ m: Int, _ n: Int) -> Int {
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var dp = Array(repeating: Array(repeating: 0, count: n), count: m)
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dp[0][0] = 1
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for i in 0..<m {
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for j in 0..<n {
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if i > 0 {
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}
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}
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}
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return dp[m - 1][n - 1]
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}
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}

‎lcof2/剑指 Offer II 099. 最小路径之和/README.md‎

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for i in 1..<m {
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dp[i][0] += dp[i-1][0]
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}
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for j in 1..<n {
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dp[0][j] += dp[0][j-1]
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}
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for i in 1..<m {
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for j in 1..<n {
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dp[i][j] += min(dp[i-1][j], dp[i][j-1])
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}
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}
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return dp[m-1][n-1]
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}
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}

‎solution/2100-2199/2146.K Highest Ranked Items Within a Price Range/README.md‎

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<b>输出:</b>[[2,1],[1,2]]
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<b>解释:</b>起点为 (2,3) 。
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价格范围为 [2,3] ,我们可以选择的物品坐标为 (0,1),(1,1),(1,2) 和 (2,1) 。
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这些物品的排名为:
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这些物品的排名为:
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- (2,1) 距离为 2 ,价格为 2
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- (1,2) 距离为 2 ,价格为 3
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- (1,1) 距离为 3

‎solution/2100-2199/2146.K Highest Ranked Items Within a Price Range/README_EN.md‎

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@@ -82,11 +82,11 @@ Thus, the 2 highest ranked items in the price range are (2,1) and (1,2).
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<strong>Input:</strong> grid = [[1,1,1],[0,0,1],[2,3,4]], pricing = [2,3], start = [0,0], k = 3
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<strong>Output:</strong> [[2,1],[2,0]]
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<strong>Explanation:</strong> You start at (0,0).
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With a price range of [2,3], we can take items from (2,0) and (2,1).
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The ranks of these items are:
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With a price range of [2,3], we can take items from (2,0) and (2,1).
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The ranks of these items are:
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- (2,1) with distance 5
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- (2,0) with distance 6
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Thus, the 2 highest ranked items in the price range are (2,1) and (2,0).
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Thus, the 2 highest ranked items in the price range are (2,1) and (2,0).
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Note that k = 3 but there are only 2 reachable items within the price range.
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</pre>
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3284.Sum%20of%20Consecutive%20Subarrays/README.md
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---
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<!-- problem:start -->
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# [3284. Sum of Consecutive Subarrays 🔒](https://leetcode.cn/problems/sum-of-consecutive-subarrays)
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[English Version](/solution/3200-3299/3284.Sum%20of%20Consecutive%20Subarrays/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>We call an array <code>arr</code> of length <code>n</code> <strong>consecutive</strong> if one of the following holds:</p>
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<ul>
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<li><code>arr[i] - arr[i - 1] == 1</code> for <em>all</em> <code>1 &lt;= i &lt; n</code>.</li>
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<li><code>arr[i] - arr[i - 1] == -1</code> for <em>all</em> <code>1 &lt;= i &lt; n</code>.</li>
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</ul>
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<p>The <strong>value</strong> of an array is the sum of its elements.</p>
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<p>For example, <code>[3, 4, 5]</code> is a consecutive array of value 12 and <code>[9, 8]</code> is another of value 17. While <code>[3, 4, 3]</code> and <code>[8, 6]</code> are not consecutive.</p>
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<p>Given an array of integers <code>nums</code>, return the <em>sum</em> of the <strong>values</strong> of all <strong>consecutive </strong><span data-keyword="subarray-nonempty">subarrays</span>.</p>
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<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9 </sup>+ 7.</code></p>
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<p><strong>Note</strong> that an array of length 1 is also considered consecutive.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3]</span></p>
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<p><strong>Output:</strong> <span class="example-io">20</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The consecutive subarrays are: <code>[1]</code>, <code>[2]</code>, <code>[3]</code>, <code>[1, 2]</code>, <code>[2, 3]</code>, <code>[1, 2, 3]</code>.<br />
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Sum of their values would be: <code>1 +たす 2 +たす 3 +たす 3 +たす 5 +たす 6 =わ 20</code>.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,3,5,7]</span></p>
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<p><strong>Output:</strong> <span class="example-io">16</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The consecutive subarrays are: <code>[1]</code>, <code>[3]</code>, <code>[5]</code>, <code>[7]</code>.<br />
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Sum of their values would be: <code>1 +たす 3 +たす 5 +たす 7 =わ 16</code>.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [7,6,1,2]</span></p>
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<p><strong>Output:</strong> <span class="example-io">32</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The consecutive subarrays are: <code>[7]</code>, <code>[6]</code>, <code>[1]</code>, <code>[2]</code>, <code>[7, 6]</code>, <code>[1, 2]</code>.<br />
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Sum of their values would be: <code>7 +たす 6 +たす 1 +たす 2 +たす 13 +たす 3 =わ 32</code>.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:递推
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我们定义两个变量 $f$ 和 $g,ドル分别表示以当前元素结尾的递增子数组的长度和以当前元素结尾的递减子数组的长度,用另外两个变量 $s$ 和 $t$ 分别表示以当前元素结尾的递增子数组的和和以当前元素结尾的递减子数组的和。初始时 $f = g = 1,ドル而 $s = t = \textit{nums}[0]$。
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接下来,我们从第二个元素开始遍历数组,对于当前元素 $\textit{nums}[i],ドル我们分别考虑以 $\textit{nums}[i]$ 结尾的递增子数组和递减子数组。
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如果 $\textit{nums}[i] - \textit{nums}[i - 1] = 1,ドル那么 $\textit{nums}[i]$ 可以加入到以 $\textit{nums}[i - 1]$ 结尾的递增子数组中,此时我们更新 $f$ 和 $s,ドル并将 $s$ 加到答案中;
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如果 $\textit{nums}[i] - \textit{nums}[i - 1] = -1,ドル那么 $\textit{nums}[i]$ 可以加入到以 $\textit{nums}[i - 1]$ 结尾的递减子数组中,此时我们更新 $g$ 和 $t,ドル并将 $t$ 加到答案中。
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否则,$\textit{nums}[i]$ 无法加入到以 $\textit{nums}[i - 1]$ 结尾的递增子数组或递减子数组中,我们将 $\textit{nums}[i]$ 加到答案中。
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遍历结束后,返回答案即可。
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时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def getSum(self, nums: List[int]) -> int:
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mod = 10**9 + 7
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f = g = 1
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s = t = nums[0]
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ans = nums[0]
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for x, y in pairwise(nums):
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if y - x == 1:
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f += 1
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s += f * y
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ans = (ans + s) % mod
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else:
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f = 1
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s = y
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if y - x == -1:
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g += 1
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t += g * y
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ans = (ans + t) % mod
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else:
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g = 1
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t = y
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if abs(y - x) != 1:
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ans = (ans + y) % mod
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return ans
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```
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#### Java
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```java
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class Solution {
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public int getSum(int[] nums) {
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final int mod = (int) 1e9 + 7;
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long s = nums[0], t = nums[0], ans = nums[0];
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int f = 1, g = 1;
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for (int i = 1; i < nums.length; ++i) {
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int x = nums[i - 1], y = nums[i];
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if (y - x == 1) {
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++f;
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s += 1L * f * y;
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ans = (ans + s) % mod;
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} else {
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f = 1;
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s = y;
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}
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if (y - x == -1) {
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++g;
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t += 1L * g * y;
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ans = (ans + t) % mod;
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} else {
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g = 1;
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t = y;
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}
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if (Math.abs(y - x) != 1) {
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ans = (ans + y) % mod;
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}
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}
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return (int) ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int getSum(vector<int>& nums) {
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const int mod = 1e9 + 7;
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long long s = nums[0], t = nums[0], ans = nums[0];
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int f = 1, g = 1;
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for (int i = 1; i < nums.size(); ++i) {
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int x = nums[i - 1], y = nums[i];
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if (y - x == 1) {
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++f;
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s += 1LL * f * y;
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ans = (ans + s) % mod;
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} else {
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f = 1;
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s = y;
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}
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if (y - x == -1) {
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++g;
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t += 1LL * g * y;
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ans = (ans + t) % mod;
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} else {
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g = 1;
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t = y;
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}
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if (abs(y - x) != 1) {
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ans = (ans + y) % mod;
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}
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func getSum(nums []int) int {
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const mod int = 1e9 + 7
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f, g := 1, 1
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s, t := nums[0], nums[0]
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ans := nums[0]
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for i := 1; i < len(nums); i++ {
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x, y := nums[i-1], nums[i]
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if y-x == 1 {
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f++
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s += f * y
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ans = (ans + s) % mod
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} else {
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f = 1
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s = y
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}
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if y-x == -1 {
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g++
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t += g * y
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ans = (ans + t) % mod
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} else {
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g = 1
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t = y
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}
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if abs(y-x) != 1 {
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ans = (ans + y) % mod
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}
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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#### TypeScript
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```ts
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function getSum(nums: number[]): number {
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const mod = 10 ** 9 + 7;
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let f = 1,
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g = 1;
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let s = nums[0],
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t = nums[0];
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let ans = nums[0];
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for (let i = 1; i < nums.length; i++) {
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const x = nums[i - 1];
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const y = nums[i];
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if (y - x === 1) {
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f++;
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s += f * y;
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ans = (ans + s) % mod;
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} else {
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f = 1;
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s = y;
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}
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if (y - x === -1) {
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g++;
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t += g * y;
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ans = (ans + t) % mod;
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} else {
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g = 1;
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t = y;
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}
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if (Math.abs(y - x) !== 1) {
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ans = (ans + y) % mod;
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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