find-subarrays-with-equal-sum

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Algorithms

Given a 0-indexed integer array <code>nums</code>, determine whether there exist two subarrays of length <code>2</code> with equal sum. Note that the two subarrays must begin at different indices.

Return <code>true</code> if these subarrays exist, and <code>false</code> otherwise.

A subarray is a contiguous non-empty sequence of elements within an array.

&nbsp;
Example 1:

<pre>
Input: nums = [4,2,4]
Output: true
Explanation: The subarrays with elements [4,2] and [2,4] have the same sum of 6.
</pre>

Example 2:

<pre>
Input: nums = [1,2,3,4,5]
Output: false
Explanation: No two subarrays of size 2 have the same sum.
</pre>

Example 3:

<pre>
Input: nums = [0,0,0]
Output: true
Explanation: The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0. 
Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-109 &lt;= nums[i] &lt;= 109</code></li>
</ul>

findSubarrays

Find Subarrays With Equal Sum

Two Sum

two-sum

两数之和

Partition Equal Subset Sum

partition-equal-subset-sum

分割等和子集

Find Two Non-overlapping Sub-arrays Each With Target Sum

find-two-non-overlapping-sub-arrays-each-with-target-sum

找两个和为目标值且不重叠的子数组

Array

Hash Table

给你一个下标从 0&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，判断是否存在&nbsp;两个&nbsp;长度为&nbsp;<code>2</code>&nbsp;的子数组且它们的&nbsp;和&nbsp;相等。注意，这两个子数组起始位置的下标必须&nbsp;不相同&nbsp;。

如果这样的子数组存在，请返回&nbsp;<code>true</code>，否则返回&nbsp;<code>false</code>&nbsp;。

子数组 是一个数组中一段连续非空的元素组成的序列。

&nbsp;

示例 1：

<pre>输入：nums = [4,2,4]
输出：true
解释：元素为 [4,2] 和 [2,4] 的子数组有相同的和 6 。
</pre>

示例 2：

<pre>输入：nums = [1,2,3,4,5]
输出：false
解释：没有长度为 2 的两个子数组和相等。
</pre>

示例 3：

<pre>输入：nums = [0,0,0]
输出：true
解释：子数组 [nums[0],nums[1]] 和 [nums[1],nums[2]] 的和相等，都为 0 。
注意即使子数组的元素相同，这两个子数组也视为不相同的子数组，因为它们在原数组中的起始位置不同。
</pre>

&nbsp;

提示：

<ul>
	<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-109 &lt;= nums[i] &lt;= 109</code></li>
</ul>