target sum Archives - Huahua's Tech Road

花花酱 LeetCode 1711. Count Good Meals

By zxi on January 2, 2021

A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two.

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)

C++

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;

}

};

Python3

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)

花花酱 LeetCode 1477. Find Two Non-overlapping Sub-arrays Each With Target Sum

By zxi on June 13, 2020

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

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

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

C++

// 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;

}

};

花花酱 LeetCode 1171. Remove Zero Sum Consecutive Nodes from Linked List

By zxi on August 24, 2019

Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0until there are no such sequences.

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)

C++

// 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;

}

};

Python3

# 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

Posts tagged as “target sum”

花花酱 LeetCode 1711. Count Good Meals

Solution: Hashtable

C++

Python3

花花酱 LeetCode 1477. Find Two Non-overlapping Sub-arrays Each With Target Sum

Solution: Sliding Window + Best so far

C++

花花酱 LeetCode 1171. Remove Zero Sum Consecutive Nodes from Linked List

Solution: HashTable

C++

Python3