2023-10-26 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 26

Description

Binary Trees With Factors

Given an array of unique integers, arr, where each integer arr[i] is strictly greater than 1.

We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node's value should be equal to the product of the values of its children.

Return the number of binary trees we can make. The answer may be too large so return the answer modulo 109 + 7.

 

Example 1:

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

Example 2:

Input: arr = [2,4,5,10]
Output: 7
Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].

 

Constraints:

  • 1 <= arr.length <= 1000
  • 2 <= arr[i] <= 109
  • All the values of arr are unique.

Solution

const int MOD = 1e9 + 7;
class Solution {
public:
  int numFactoredBinaryTrees(vector<int>& arr) {
    sort(arr.begin(), arr.end());
    int len = arr.size();
    vector<int> count(len, 1);
    for(int i = 1; i < len; ++i) {
      for(int j = 0; arr[i] / arr[j] >= arr[j]; ++j) {
        if(arr[i] % arr[j]) continue;
        int pos = lower_bound(arr.begin(), arr.end(), arr[i] / arr[j]) - arr.begin();
        if(arr[i] / arr[j] != arr[pos]) continue;
        count[i] += (2LL - (arr[j] == arr[pos])) * count[j] * count[pos] % MOD;
        count[i] %= MOD;
      }
    }
    
    int answer = accumulate(count.begin(), count.end(), 0, [](int prev, int cur) {
      return (prev + cur) % MOD;
    });
    return answer;
  }
};

// Accepted
// 48/48 cases passed (28 ms)
// Your runtime beats 91.73 % of cpp submissions
// Your memory usage beats 92.91 % of cpp submissions (8.7 MB)
Algorithm