Today I have done Minimize Maximum Pair Sum in Array and leetcode's June LeetCoding Challenge with cpp.

Minimize Maximum Pair Sum in Array

Description

The pair sum of a pair (a,b) is equal to a + b. The maximum pair sum is the largest pair sum in a list of pairs.

• For example, if we have pairs (1,5), (2,3), and (4,4), the maximum pair sum would be max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8.

Given an array nums of even length n, pair up the elements of nums into n / 2 pairs such that:

• Each element of nums is in exactly one pair, and
• The maximum pair sum is minimized.

Return the minimized maximum pair sum after optimally pairing up the elements.

Example 1:

Input: nums = [3,5,2,3]
Output: 7
Explanation: The elements can be paired up into pairs (3,3) and (5,2).
The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.

Example 2:

Input: nums = [3,5,4,2,4,6]
Output: 8
Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.

Constraints:

• n == nums.length
• 2 <= n <= 10^5
• n is even.
• 1 <= nums[i] <= 10^5

Solution

auto speedup = []() {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int minPairSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int len = nums.size();
    int answer = 0;
    for(int i = 0; i * 2 < len; ++i) {
      answer = max(answer, nums[i] + nums[len - i - 1]);
    }
    return answer;
  }
};

// Accepted
// 37/37 cases passed (140 ms)
// Your runtime beats 99.92 % of cpp submissions
// Your memory usage beats 17.97 % of cpp submissions (96.3 MB)

June LeetCoding Challenge 30

Description

Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

Solution

class Solution {
public:
  TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if(root == p || root == q) return root;
    if(!root) return nullptr;
    auto left = lowestCommonAncestor(root->left, p, q);
    auto right = lowestCommonAncestor(root->right, p, q);
    if(left && right) return root;
    if(!left && !right) return nullptr;
    if(!left) return right;
    return left;
  }
};