Today I have done Minimum Subsequence in Non-Increasing Order and leetcode's July LeetCoding Challenge with cpp.

Minimum Subsequence in Non-Increasing Order

Description

Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence.

If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements. A subsequence of an array can be obtained by erasing some (possibly zero) elements from the array.

Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order.

Example 1:

Input: nums = [4,3,10,9,8]
Output: [10,9] 
Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements is strictly greater than the sum of elements not included, however, the subsequence [10,9] has the maximum total sum of its elements. 

Example 2:

Input: nums = [4,4,7,6,7]
Output: [7,7,6] 
Explanation: The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to returned in non-decreasing order.  

Example 3:

Input: nums = [6]
Output: [6]

Constraints:

• 1 <= nums.length <= 500
• 1 <= nums[i] <= 100

Solution

little data...

class Solution {
public:
  vector<int> minSubsequence(vector<int>& nums) {
    int len = nums.size();
    priority_queue<int> pq;
    int sum = 0;
    for(int i = 0; i < nums.size(); ++i) {
      pq.push(nums[i]);
      sum += nums[i];
    }
    int cur = 0;
    vector<int> answer;
    while(cur <= sum) {
      answer.push_back(pq.top());
      pq.pop();
      cur += answer.back();
      sum -= answer.back();
    }

    return answer;
  }
};

class Solution {
public:
  vector<int> minSubsequence(vector<int>& nums) {
    int sum = accumulate(nums.begin(), nums.end(), 0);
    sort(nums.begin(), nums.end(), greater<int>());
    int cur = 0;
    for(int i = 0; i < nums.size(); ++i) {
      if(cur * 2 > sum) {
        nums.resize(i);
        return nums;
      }
      cur += nums[i];
    }
    return nums;
  }
};

July LeetCoding Challenge 23

Description

Binary Tree Pruning

Given the root of a binary tree, return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

A subtree of a node node is node plus every node that is a descendant of node.

Example 1:

Input: root = [1,null,0,0,1]
Output: [1,null,0,null,1]
Explanation: 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.

Example 2:

Input: root = [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

Example 3:

Input: root = [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]

Constraints:

• The number of nodes in the tree is in the range [1, 200].
• Node.val is either 0 or 1.

Solution

class Solution {
public:
  TreeNode* pruneTree(TreeNode* root) {
    if(!root) return nullptr;
    root->left = pruneTree(root->left);
    root->right = pruneTree(root->right);
    if(!root->val && !root->left && !root->right) return nullptr;
    return root;
  }
};