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OJ notes/pages/Leetcode Two-Sum-II-Input-Array-Is-Sorted.md

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+# Leetcode Two-Sum-II-Input-Array-Is-Sorted
+
+#### 2022-07-11 14:54
+
+> ##### Algorithms:
+> #algorithm #two_pointers 
+> ##### Data structures:
+> #DS #array 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy 
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#two_pointers
+```
+ 
+- [[Leetcode Intersection-of-Two-Arrays-II]]
+- [[Leetcode Merge-Sorted-Array]]
+- [[Leetcode Merge-Two-Sorted-Lists]]
+- [[Leetcode Move-Zeroes]]
+- [[Leetcode Squares-of-a-Sorted-Array]]
+- [[Two pointers approach]]
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/)
+___
+### Problem
+
+Given a **1-indexed** array of integers `numbers` that is already **_sorted in non-decreasing order_**, find two numbers such that they add up to a specific `target` number. Let these two numbers be `numbers[index1]` and `numbers[index2]` where `1 <= index1 < index2 <= numbers.length`.
+
+Return _the indices of the two numbers,_ `index1` _and_ `index2`_, **added by one** as an integer array_ `[index1, index2]` _of length 2._
+
+The tests are generated such that there is **exactly one solution**. You **may not** use the same element twice.
+
+Your solution must use only constant extra space.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** numbers = [2,7,11,15], target = 9
+**Output:** [1,2]
+**Explanation:** The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
+
+**Example 2:**
+
+**Input:** numbers = [2,3,4], target = 6
+**Output:** [1,3]
+**Explanation:** The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
+
+**Example 3:**
+
+**Input:** numbers = [-1,0], target = -1
+**Output:** [1,2]
+**Explanation:** The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
+
+#### Constraints
+
+-   `2 <= numbers.length <= 3 * 104`
+-   `-1000 <= numbers[i] <= 1000`
+-   `numbers` is sorted in **non-decreasing order**.
+-   `-1000 <= target <= 1000`
+-   The tests are generated such that there is **exactly one solution**.
+
+### Thoughts
+
+> [!summary]
+> This is a #two_pointers problem.
+
+decrement the right pointer makes the sum smaller, increment the left pointer makes the sum bigger, and thud we get the answer.
+
+### Solution
+
+```cpp
+class Solution {
+public:
+  vector<int> twoSum(vector<int> &numbers, int target) {
+    // Sorted array, use two pointers
+    // remember to subtract 1
+    int r = numbers.size() - 1;
+    int l = 0;
+    int sum;
+
+    while (l < r) {
+      sum = numbers[l] + numbers[r];
+      if (sum < target) {
+        l++;
+      } else if (sum > target) {
+        r--;
+      } else {
+        return {l + 1, r + 1};
+      }
+    }
+
+    // Shouldn't return this, since there is a solution
+    return {0, 0};
+  }
+};
+```