Add solutions and explanations for problems 3583, 3772, 3771, 3770, 3766, 3768, 3767, 3765, 3761, 3760

Author: vikahaze

books/All.md

@@ -8922,6 +8922,68 @@ class Solution:
         return res
 ```
 
+## 1431. Kids With the Greatest Number of Candies [Easy]
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
+
+### Explanation
+
+## 1431. Kids With the Greatest Number of Candies [Easy]
+
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies
+
+## Description
+There are `n` kids with candies. You are given an integer array `candies`, where each `candies[i]` represents the number of candies the `i`th kid has, and an integer `extraCandies`, denoting the number of extra candies that you have.
+
+Return a boolean array `result` of length `n`, where `result[i]` is `true` if, after giving the `i`th kid all the `extraCandies`, they will have the greatest number of candies among all the kids, or false otherwise.
+
+Note that multiple kids can have the greatest number of candies.
+
+**Examples**
+
+```text
+Input: candies = [2,3,5,1,3], extraCandies = 3
+Output: [true,true,true,false,true]
+Explanation: If you give all extraCandies to:
+- Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
+- Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+- Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
+- Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
+- Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+
+Input: candies = [4,2,1,1,2], extraCandies = 1
+Output: [true,false,false,false,false]
+
+Input: candies = [12,1,12], extraCandies = 10
+Output: [true,false,true]
+```
+
+**Constraints**
+
+```text
+- n == candies.length
+- 2 <= n <= 100
+- 1 <= candies[i] <= 100
+- 1 <= extraCandies <= 50
+```
+
+## Hint
+For each kid, check if their candies plus `extraCandies` is at least as much as the current maximum.
+
+## Explanation
+First, you want to know the highest number of candies any kid currently has. This is important because you need a reference point to see if giving extra candies to a kid will make them "the greatest."
+
+For each kid, you add the `extraCandies` to their current amount. You do this because you want to see if, after the bonus, they can reach or beat the current maximum. If they do, you mark them as `True` in our answer list; otherwise, False.
+
+You only need to find the maximum once, and then just compare each kid's total to it. Don't need to recalculate the maximum for every kid.
+
+### Solution
+
+```python
+def kidsWithCandies(candies, extraCandies):
+    max_candies = max(candies)  # Find the current maximum
+    return [(c + extraCandies) >= max_candies for c in candies]
+```
+
 ## 1448. Count Good Nodes in Binary Tree [Medium]
 https://leetcode.com/problems/count-good-nodes-in-binary-tree/
 

books/LeetCode_75.md

@@ -517,6 +517,68 @@ def uniqueOccurrences(arr):
     return len(set(count.values())) == len(count)
 ```
 
+## 1431. Kids With the Greatest Number of Candies [Easy]
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
+
+### Explanation
+
+## 1431. Kids With the Greatest Number of Candies [Easy]
+
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies
+
+## Description
+There are `n` kids with candies. You are given an integer array `candies`, where each `candies[i]` represents the number of candies the `i`th kid has, and an integer `extraCandies`, denoting the number of extra candies that you have.
+
+Return a boolean array `result` of length `n`, where `result[i]` is `true` if, after giving the `i`th kid all the `extraCandies`, they will have the greatest number of candies among all the kids, or false otherwise.
+
+Note that multiple kids can have the greatest number of candies.
+
+**Examples**
+
+```text
+Input: candies = [2,3,5,1,3], extraCandies = 3
+Output: [true,true,true,false,true]
+Explanation: If you give all extraCandies to:
+- Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
+- Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+- Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
+- Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
+- Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+
+Input: candies = [4,2,1,1,2], extraCandies = 1
+Output: [true,false,false,false,false]
+
+Input: candies = [12,1,12], extraCandies = 10
+Output: [true,false,true]
+```
+
+**Constraints**
+
+```text
+- n == candies.length
+- 2 <= n <= 100
+- 1 <= candies[i] <= 100
+- 1 <= extraCandies <= 50
+```
+
+## Hint
+For each kid, check if their candies plus `extraCandies` is at least as much as the current maximum.
+
+## Explanation
+First, you want to know the highest number of candies any kid currently has. This is important because you need a reference point to see if giving extra candies to a kid will make them "the greatest."
+
+For each kid, you add the `extraCandies` to their current amount. You do this because you want to see if, after the bonus, they can reach or beat the current maximum. If they do, you mark them as `True` in our answer list; otherwise, False.
+
+You only need to find the maximum once, and then just compare each kid's total to it. Don't need to recalculate the maximum for every kid.
+
+### Solution
+
+```python
+def kidsWithCandies(candies, extraCandies):
+    max_candies = max(candies)  # Find the current maximum
+    return [(c + extraCandies) >= max_candies for c in candies]
+```
+
 ## 2215. Find the Difference of Two Arrays [Easy]
 https://leetcode.com/problems/find-the-difference-of-two-arrays/
 

books/Visualization.md

@@ -86,6 +86,68 @@ def __init__(self, val=0, next=None):
         self.next = next
 ```
 
+## 1431. Kids With the Greatest Number of Candies [Easy]
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
+
+### Explanation
+
+## 1431. Kids With the Greatest Number of Candies [Easy]
+
+https://leetcode.com/problems/kids-with-the-greatest-number-of-candies
+
+## Description
+There are `n` kids with candies. You are given an integer array `candies`, where each `candies[i]` represents the number of candies the `i`th kid has, and an integer `extraCandies`, denoting the number of extra candies that you have.
+
+Return a boolean array `result` of length `n`, where `result[i]` is `true` if, after giving the `i`th kid all the `extraCandies`, they will have the greatest number of candies among all the kids, or false otherwise.
+
+Note that multiple kids can have the greatest number of candies.
+
+**Examples**
+
+```text
+Input: candies = [2,3,5,1,3], extraCandies = 3
+Output: [true,true,true,false,true]
+Explanation: If you give all extraCandies to:
+- Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
+- Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+- Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
+- Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
+- Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
+
+Input: candies = [4,2,1,1,2], extraCandies = 1
+Output: [true,false,false,false,false]
+
+Input: candies = [12,1,12], extraCandies = 10
+Output: [true,false,true]
+```
+
+**Constraints**
+
+```text
+- n == candies.length
+- 2 <= n <= 100
+- 1 <= candies[i] <= 100
+- 1 <= extraCandies <= 50
+```
+
+## Hint
+For each kid, check if their candies plus `extraCandies` is at least as much as the current maximum.
+
+## Explanation
+First, you want to know the highest number of candies any kid currently has. This is important because you need a reference point to see if giving extra candies to a kid will make them "the greatest."
+
+For each kid, you add the `extraCandies` to their current amount. You do this because you want to see if, after the bonus, they can reach or beat the current maximum. If they do, you mark them as `True` in our answer list; otherwise, False.
+
+You only need to find the maximum once, and then just compare each kid's total to it. Don't need to recalculate the maximum for every kid.
+
+### Solution
+
+```python
+def kidsWithCandies(candies, extraCandies):
+    max_candies = max(candies)  # Find the current maximum
+    return [(c + extraCandies) >= max_candies for c in candies]
+```
+
 ## 1798. Maximum Number of Consecutive Values You Can Make [Medium]
 https://leetcode.com/problems/maximum-number-of-consecutive-values-you-can-make/
 

data/book-sets.json

@@ -57,13 +57,13 @@
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-      3766, 3767, 3768, 3769, 3770, 3771, 3772
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+      3761, 3762, 3764, 3765, 3766, 3767, 3768, 3769, 3770, 3771, 3772
     ]
   },
   {"title": "Visualization", "description": "", "tags": [], "problems": [1, 2, 11, 1431, 1679, 1768, 1798, 2215, 3603, 3622, 3623]},
@@ -176,13 +176,13 @@
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explanations/3583/en.md

@@ -0,0 +1,56 @@
+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We need to count special triplets (i, j, k) where i < j < k, nums[i] == nums[j] * 2, and nums[k] == nums[j] * 2. In other words, for each middle element j, we need to count how many i < j have nums[i] == target and how many k > j have nums[k] == target, where target = nums[j] * 2.
+
+**1.1 Constraints & Complexity:**
+- Input size: `3 <= n <= 10^5`, `0 <= nums[i] <= 10^5`
+- **Time Complexity:** O(n) - single pass through array with hash map operations
+- **Space Complexity:** O(n) for hash maps storing frequencies
+- **Edge Case:** If no triplets exist, return 0
+
+**1.2 High-level approach:**
+We process each position j as the middle element. We maintain two frequency maps: one for elements before j (freq_before) and one for elements after j (freq_after). For each j, we count how many i < j have nums[i] == target and how many k > j have nums[k] == target, then multiply these counts.
+
+![Frequency counting for triplets visualization](https://assets.leetcode.com/static_assets/others/frequency-counting-triplets.png)
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Check all triplets (i, j, k) where i < j < k, which is O(n³)
+- **Optimized Strategy:** For each j, use hash maps to count valid i and k positions in O(1), achieving O(n) time
+- **Emphasize the optimization:** Hash maps allow us to count frequencies before and after each position efficiently without nested loops
+
+**1.4 Decomposition:**
+1. Initialize freq_before (empty) and freq_after (contains all elements initially)
+2. Process each position j from left to right
+3. Remove nums[j] from freq_after (since j is no longer "after")
+4. Calculate target = nums[j] * 2
+5. Count valid i positions (freq_before[target]) and valid k positions (freq_after[target])
+6. Multiply counts and add to result (modulo 10^9 + 7)
+7. Add nums[j] to freq_before for future j positions
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use the example: `nums = [6, 3, 6]`
+- Initialize: `freq_before = {}`, `freq_after = {6: 2, 3: 1}`, `res = 0`, `MOD = 10^9 + 7`
+
+**2.2 Start Processing:**
+We iterate through each position j from 0 to n-1.
+
+**2.3 Trace Walkthrough:**
+
+| j | nums[j] | Remove from freq_after | target | freq_before[target] | freq_after[target] | Count | res |
+|---|---------|------------------------|--------|---------------------|-------------------|-------|-----|
+| 0 | 6 | freq_after = {6: 1, 3: 1} | 12 | 0 | 0 | 0 * 0 = 0 | 0 |
+| 0 | - | Add to freq_before | - | freq_before = {6: 1} | - | - | 0 |
+| 1 | 3 | freq_after = {6: 1} | 6 | 1 (from freq_before) | 1 (from freq_after) | 1 * 1 = 1 | 1 |
+| 1 | - | Add to freq_before | - | freq_before = {6: 1, 3: 1} | - | - | 1 |
+| 2 | 6 | freq_after = {} | 12 | 0 | 0 | 0 * 0 = 0 | 1 |
+
+**2.4 Increment and Loop:**
+For each j, we update frequency maps and count triplets where j is the middle element.
+
+**2.5 Return Result:**
+After processing all positions, `res = 1`, which is the number of special triplets. The triplet is (0, 1, 2) where nums[0] = 6 = 3 * 2 and nums[2] = 6 = 3 * 2.
+