You can pick any two different foods to make a good meal.

Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i​​​​​​th​​​​​​​​ item of food, return the number of different good meals you can make from this list modulo 109 + 7.

Note that items with different indices are considered different even if they have the same deliciousness value.

Example 1:

Input: deliciousness = [1,3,5,7,9]
Output: 4
Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9).
Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.

Example 2:

Input: deliciousness = [1,1,1,3,3,3,7]
Output: 15
Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.

Constraints:

• 1 <= deliciousness.length <= 105
• 0 <= deliciousness[i] <= 220

Solution: Hashtable

Same idea as LeetCode 1: Two Sum

Use a hashtable to store the occurrences of all the numbers added so far. For a new number x, check all possible 2^i – x. ans += freq[2^i – x] 0 <= i <= 21

Time complexity: O(22n)
Space complexity: O(n)

class Solution {
public:
  int countPairs(vector<int>& A) {
    constexpr int kMod = 1e9 + 7;
    unordered_map<int, int> m;    
    long ans = 0;
    for (int x : A) {
      for (int t = 1; t <= 1 << 21; t *= 2) {
        auto it = m.find(t - x);
        if (it != end(m)) ans += it->second;
      }
      ++m[x];
    }
    return ans % kMod;
  }
};

class Solution:
  def countPairs(self, deliciousness: List[int]) -> int:
    sums = [1<<i for i in range(22)]
    m = defaultdict(int)
    ans = 0
    for x in deliciousness:
      for t in sums:
        if t - x in m: ans += m[t - x]
      m[x] += 1
    return ans % (10**9 + 7)

The post 花花酱 LeetCode 1711. Count Good Meals appeared first on Huahua's Tech Road.

Given an array of integers arr and an integer target.

You have to find two non-overlapping sub-arrays of arr each with sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum.

Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.

Example 1:

Input: arr = [3,2,2,4,3], target = 3
Output: 2
Explanation: Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.

Example 2:

Input: arr = [7,3,4,7], target = 7
Output: 2
Explanation: Although we have three non-overlapping sub-arrays of sum = 7 ([7], [3,4] and [7]), but we will choose the first and third sub-arrays as the sum of their lengths is 2.

Example 3:

Input: arr = [4,3,2,6,2,3,4], target = 6
Output: -1
Explanation: We have only one sub-array of sum = 6.

Example 4:

Input: arr = [5,5,4,4,5], target = 3
Output: -1
Explanation: We cannot find a sub-array of sum = 3.

Example 5:

Input: arr = [3,1,1,1,5,1,2,1], target = 3
Output: 3
Explanation: Note that sub-arrays [1,2] and [2,1] cannot be an answer because they overlap.

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 1000
• 1 <= target <= 10^8

Solution: Sliding Window + Best so far

1. Use a sliding window to maintain a subarray whose sum is <= target
2. When the sum of the sliding window equals to target, we found a subarray [s, e]
3. Update ans with it’s length + shortest subarray which ends before s.
4. We can use an array to store the shortest subarray which ends before s.

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  int minSumOfLengths(vector<int>& arr, int target) {
    constexpr int kInf = 1e9;
    const int n = arr.size();
    // min_lens[i] := min length of a valid subarray ends or before i.
    vector<int> min_lens(n, kInf);
    int ans = kInf;
    int sum = 0;
    int s = 0;
    int min_len = kInf;
    for (int e = 0; e < n; ++e) {
      sum += arr[e];
      while (sum > target) sum -= arr[s++];
      if (sum == target) {       
        const int cur_len = e - s + 1;
        if (s > 0 && min_lens[s - 1] != kInf)
          ans = min(ans, cur_len + min_lens[s - 1]);
        min_len = min(min_len, cur_len);
      }
      min_lens[e] = min_len;
    }    
    return ans >= kInf ? -1 : ans;
  }
};

The post 花花酱 LeetCode 1477. Find Two Non-overlapping Sub-arrays Each With Target Sum appeared first on Huahua's Tech Road.

After doing so, return the head of the final linked list.  You may return any such answer.

(Note that in the examples below, all sequences are serializations of ListNode objects.)

Example 1:

Input: head = [1,2,-3,3,1]
Output: [3,1]
Note: The answer [1,2,1] would also be accepted.

Example 2:

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

Example 3:

Input: head = [1,2,3,-3,-2]
Output: [1]

Constraints:

• The given linked list will contain between 1 and 1000 nodes.
• Each node in the linked list has -1000 <= node.val <= 1000.

Solution: HashTable

Similar to target sum = 0, use a hashtable to store the first ListNode* that has a given prefix sum. Whenever the same prefix sum occurs, skip all the elements between the first occurrence and current one, e.g. first_sum_x.next = curr_sum_x.next

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  ListNode* removeZeroSumSublists(ListNode* head) {
    bool done = false;
    function<ListNode*(ListNode*)> helper = [&](ListNode* head) {
      ListNode dummy(0);
      dummy.next = head;    
      ListNode* prev = &dummy;
      ListNode* curr = prev->next;
      unordered_map<int, ListNode*> m{{0, prev}};    
      int s = 0;
      done = true;
      while (curr) {
        s += curr->val;      
        if (m.count(s)) {
          m[s]->next = curr->next;
          done = false;
        } else {
          m[s] = curr;
        }
        prev = curr;
        curr = curr->next;
      }
      return dummy.next;
    };
    while (!done) head = helper(head);
    return head;
  }
};

# Author: Huahua
class Solution:
  def removeZeroSumSublists(self, head: ListNode) -> ListNode:
    def helper(head: ListNode):      
      dummy = ListNode(0)
      dummy.next = head
      prev, curr = dummy, dummy.next    
      s = 0
      m = {s: prev}
      done = True
      while curr:
        s += curr.val
        if s in m:
          m[s].next = curr.next
          done = False
        else:
          m[s] = curr
        prev, curr = curr, curr.next
      return dummy.next, done
    while True:
      head, done = helper(head)
      if done: return head

The post 花花酱 LeetCode 1171. Remove Zero Sum Consecutive Nodes from Linked List appeared first on Huahua's Tech Road.