Input: s = "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]
Explanation: The large groups are "ddd", "eeee", and "bbb".

using System.Collections.Generic;

public class Solution {
    public IList<IList<int>> LargeGroupPositions(string S) {
        var ans = new List<IList<int>>();
        int i = 0, N = S.Length;

        for (int j = 0; j < N; ++j) {
            if (j == N - 1 || S[j] != S[j + 1]) {
                if (j - i + 1 >= 3) {
                    ans.Add(new List<int> { i, j });
                }
                i = j + 1;
            }
        }

        return ans;
    }
}

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

public class Solution {
    public ListNode DeleteDuplicates(ListNode head) {
        ListNode sentinel = new ListNode(0, head);
        ListNode pred = sentinel;
        while (head != null) {
            if (head.next != null && head.val == head.next.val) {
                while (head.next != null && head.val == head.next.val) {
                    head = head.next;
                }
                pred.next = head.next;
            } else {
                pred = pred.next;
            }
            head = head.next;
        }
        return sentinel.next;
    }
}

Input: arr = [2,4]
Output: 3
Explanation: We can make these trees: [2], [4], [4, 2, 2]

public class Solution {
    public int NumFactoredBinaryTrees(int[] A) {
        const int MOD = 1_000_000_007;
        Array.Sort(A);
        var dp = new long[A.Length];
        Array.Fill(dp, 1);
        var index = new Dictionary<int, int>();
        
        for (int i = 0; i < A.Length; i++) {
            index[A[i]] = i;
        }
        
        for (int i = 0; i < A.Length; i++) {
            for (int j = 0; j < i; j++) {
                if (A[i] % A[j] == 0) {
                    int right = A[i] / A[j];
                    if (index.ContainsKey(right)) {
                        dp[i] = (dp[i] + dp[j] * dp[index[right]]) % MOD;
                    }
                }
            }
        }
        
        long result = 0;
        foreach (long x in dp) {
            result = (result + x) % MOD;
        }
        return (int)result;
    }
}

Input: nums = [113,215,221]
Output: 12
Explanation: The tree that the list represents is shown.
The path sum is (3 + 5) + (3 + 1) = 12.

public class Solution {
    public int PathSum(int[] nums) {
        var tree = new Dictionary<int, int>();
        
        foreach (int num in nums) {
            int depth = num / 100;
            int pos = (num / 10) % 10;
            int val = num % 10;
            tree[depth * 10 + pos] = val;
        }
        
        return Dfs(tree, 1, 1, 0);
    }
    
    private int Dfs(Dictionary<int, int> tree, int depth, int pos, int currentSum) {
        int key = depth * 10 + pos;
        if (!tree.ContainsKey(key)) {
            return 0;
        }
        currentSum += tree[key];
        int leftChild = (depth + 1) * 10 + 2 * pos - 1;
        int rightChild = (depth + 1) * 10 + 2 * pos;
        if (!tree.ContainsKey(leftChild) && !tree.ContainsKey(rightChild)) {
            return currentSum;
        }
        return Dfs(tree, depth + 1, 2 * pos - 1, currentSum) + Dfs(tree, depth + 1, 2 * pos, currentSum);
    }
}

Input: arr = [1,2,3,4,5], k = 4, x = 3
Output: [1,2,3,4]

public class Solution {
    public IList<int> FindClosestElements(int[] arr, int k, int x) {
        int left = 0, right = arr.Length - k;

        while (left < right) {
            int mid = (left + right) / 2;
            if (x - arr[mid] > arr[mid + k] - x) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return arr.Skip(left).Take(k).ToList();
    }
}

Input: nums = [1,2,1,2,3], k = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2]

public class Solution {
    public int CountGoodSubarrays(int[] nums, int k) {
        int count = 0;
        int left = 0;
        int right = 0;
        int distinctCount = 0;
        Dictionary<int, int> freq = new Dictionary<int, int>();
        
        while (right < nums.Length) {
            if (!freq.ContainsKey(nums[right])) {
                freq[nums[right]] = 0;
            }
            freq[nums[right]]++;
            if (freq[nums[right]] == 1) {
                distinctCount++;
            }
            right++;
            
            while (distinctCount > k) {
                freq[nums[left]]--;
                if (freq[nums[left]] == 0) {
                    distinctCount--;
                }
                left++;
            }
            
            if (distinctCount == k) {
                count++;
            }
        }
        
        return count;
    }
}

Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2
Output: 45

public class Solution {
    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int areaOfA = (ay2 - ay1) * (ax2 - ax1);
        int areaOfB = (by2 - by1) * (bx2 - bx1);

        int left = Math.Max(ax1, bx1);
        int right = Math.Min(ax2, bx2);
        int xOverlap = right - left;

        int top = Math.Min(ay2, by2);
        int bottom = Math.Max(ay1, by1);
        int yOverlap = top - bottom;

        int areaOfOverlap = 0;
        if (xOverlap > 0 && yOverlap > 0) {
            areaOfOverlap = xOverlap * yOverlap;
        }

        int totalArea = areaOfA + areaOfB - areaOfOverlap;
        return totalArea;
    }
}

Input: nums = [1,-1,5,-2,3], k = 3
Output: 4
Explanation: The subarray [1, -1, 5, -2] sums to 3 and is the longest.

public class Solution {
    public int MaxSubArrayLen(int[] nums, int k) {
        int prefixSum = 0;
        int longestSubarray = 0;
        Dictionary<int, int> indices = new Dictionary<int, int>();
        
        for (int i = 0; i < nums.Length; i++) {
            prefixSum += nums[i];
            
            if (prefixSum == k) {
                longestSubarray = i + 1;
            }
            if (indices.ContainsKey(prefixSum - k)) {
                longestSubarray = Math.Max(longestSubarray, i - indices[prefixSum - k]);
            }
            if (!indices.ContainsKey(prefixSum)) {
                indices[prefixSum] = i;
            }
        }
        
        return longestSubarray;
    }
}

Input: intervals = [[0,30],[5,10],[15,20]]
Output: false

class Solution {
    public bool Overlap(int[] interval1, int[] interval2) {
        return (interval1[0] >= interval2[0] && interval1[0] < interval2[1]) ||
               (interval2[0] >= interval1[0] && interval2[0] < interval1[1]);
    }
    
    public bool CanAttendMeetings(int[][] intervals) {
        for (int i = 0; i < intervals.Length; i++) {
            for (int j = i + 1; j < intervals.Length; j++) {
                if (Overlap(intervals[i], intervals[j])) {
                    return false;
                }
            }
        }
        return true;
    }
}

Input: nums = [1,1,2]  
Output: 2, nums = [1,2,_]  

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        if (nums.Length == 0) return 0;
        
        int i = 0;
        for (int j = 1; j < nums.Length; j++) {
            if (nums[j] != nums[i]) {
                i++;
                nums[i] = nums[j];
            }
        }
        
        return i + 1;
    }
}

Input: nums = [3,2,1]
Output: 1

public class Solution {
    public int ThirdMax(int[] nums) {
        int? first = null;
        int? second = null;
        int? third = null;
        
        foreach (int num in nums) {
            if (num == first || num == second || num == third) {
                continue;
            }
            if (first == null || num > first) {
                third = second;
                second = first;
                first = num;
            } else if (second == null || num > second) {
                third = second;
                second = num;
            } else if (third == null || num > third) {
                third = num;
            }
        }
        
        return third ?? first.Value;
    }
}

Input: arr = [3,1,3,6]
Output: false

using System;
using System.Collections.Generic;

public class Solution {
    public bool CanReorderDoubled(int[] arr) {
        Dictionary<int, int> count = new Dictionary<int, int>();
        foreach (int num in arr) {
            if (count.ContainsKey(num)) {
                count[num]++;
            } else {
                count[num] = 1;
            }
        }
        Array.Sort(arr, (a, b) => Math.Abs(a).CompareTo(Math.Abs(b)));
        foreach (int num in arr) {
            if (count[num] == 0) continue;
            if (!count.ContainsKey(2 * num) || count[2 * num] == 0) return false;
            count[num]--;
            count[2 * num]--;
        }
        return true;
    }
}

Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; }
}

public class Solution {
    public TreeNode InsertIntoMaxTree(TreeNode root, int val) {
        if (root == null || val > root.val) {
            TreeNode newNode = new TreeNode(val);
            newNode.left = root;
            return newNode;
        }
        root.right = InsertIntoMaxTree(root.right, val);
        return root;
    }
}