Posts tagged as “sliding window”

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 1456. Maximum Number of Vowels in a Substring of Given Length

By zxi on May 23, 2020

Given a string s and an integer k.

Return the maximum number of vowel letters in any substring of s with length k.

Vowel letters in English are (a, e, i, o, u).

Example 1:

Input: s = "abciiidef", k = 3
Output: 3
Explanation: The substring "iii" contains 3 vowel letters.

Example 2:

Input: s = "aeiou", k = 2
Output: 2
Explanation: Any substring of length 2 contains 2 vowels.

Example 3:

Input: s = "leetcode", k = 3
Output: 2
Explanation: "lee", "eet" and "ode" contain 2 vowels.

Example 4:

Input: s = "rhythms", k = 4
Output: 0
Explanation: We can see that s doesn't have any vowel letters.

Example 5:

Input: s = "tryhard", k = 4
Output: 1

Constraints:

1 <= s.length <= 10^5
s consists of lowercase English letters.
1 <= k <= s.length

Solution: Sliding Window

Keep tracking the number of vows in a window of size k.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxVowels(string s, int k) {

vector<int> v(128);

v['a'] = v['e'] = v['i'] = v['o'] = v['u'] = 1;

int total = 0;

int ans = 0;

for (int i = 1; i <= s.length(); ++i) {

total += v[s[i - 1]];

if (i >= k) {

ans = max(ans, total);

total -= v[s[i - k]];

}

return ans;

}

};

花花酱 LeetCode 1425. Constrained Subset Sum

By zxi on April 26, 2020

Given an integer array nums and an integer k, return the maximum sum of a non-empty subset of that array such that for every two consecutive integers in the subset, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.

A subset of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.

Example 1:

Input: nums = [10,2,-10,5,20], k = 2
Output: 37
Explanation: The subset is [10, 2, 5, 20].

Example 2:

Input: nums = [-1,-2,-3], k = 1
Output: -1
Explanation: The subset must be non-empty, so we choose the largest number.

Example 3:

Input: nums = [10,-2,-10,-5,20], k = 2
Output: 23
Explanation: The subset is [10, -2, -5, 20].

Constraints:

1 <= k <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4

Solution: DP / Sliding window / monotonic queue

dp[i] := max sum of a subset that include nums[i]
dp[i] := max(dp[i-1], dp[i-2], …, dp[i-k-1], 0) + nums[i]

C++

// Author: Huahua

class Solution {

public:

int constrainedSubsetSum(vector<int>& nums, int k) {

const int n = nums.size();

vector<int> dp(n);

multiset<int> s{INT_MIN};

int ans = INT_MIN;

for (int i = 0; i < nums.size(); ++i) {

if (i > k) s.erase(s.equal_range(dp[i - k - 1]).first);

dp[i] = max(*rbegin(s), 0) + nums[i];

s.insert(dp[i]);

ans = max(ans, dp[i]);

}

return ans;

}

};

Use a monotonic queue to track the maximum of a sliding window dp[i-k-1] ~ dp[i-1].

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

C++

// Author: Huahua 184 ms

class Solution {

public:

int constrainedSubsetSum(vector<int>& nums, int k) {

const int n = nums.size();

vector<int> dp(n);

deque<int> q;

int ans = INT_MIN;

for (int i = 0; i < nums.size(); ++i) {

if (i > k && q.front() == i - k - 1)

q.pop_front();

dp[i] = (q.empty() ? 0 : max(dp[q.front()], 0))

+ nums[i];

while (!q.empty() && dp[i] >= dp[q.back()])

q.pop_back();

q.push_back(i);

ans = max(ans, dp[i]);

}

return ans;

}

};

花花酱 LeetCode 1423. Maximum Points You Can Obtain from Cards

By zxi on April 25, 2020

There are several cards arranged in a row, and each card has an associated number of points The points are given in the integer array cardPoints.

In one step, you can take one card from the beginning or from the end of the row. You have to take exactly k cards.

Your score is the sum of the points of the cards you have taken.

Given the integer array cardPoints and the integer k, return the maximum score you can obtain.

Example 1:

Input: cardPoints = [1,2,3,4,5,6,1], k = 3
Output: 12
Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.

Example 2:

Input: cardPoints = [2,2,2], k = 2
Output: 4
Explanation: Regardless of which two cards you take, your score will always be 4.

Example 3:

Input: cardPoints = [9,7,7,9,7,7,9], k = 7
Output: 55
Explanation: You have to take all the cards. Your score is the sum of points of all cards.

Example 4:

Input: cardPoints = [1,1000,1], k = 1
Output: 1
Explanation: You cannot take the card in the middle. Your best score is 1.

Example 5:

Input: cardPoints = [1,79,80,1,1,1,200,1], k = 3
Output: 202

Constraints:

1 <= cardPoints.length <= 10^5
1 <= cardPoints[i] <= 10^4
1 <= k <= cardPoints.length

Solution: Sliding Window

Time complexity: O(k)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxScore(vector<int>& p, int k) {

const int n = p.size();

int cur = accumulate(begin(p), begin(p) + k, 0);

int ans = cur;

for (int i = 0; i < k; ++i) {

cur = cur - p[k - i - 1] + p[n - i - 1];

ans = max(ans, cur);

}

return ans;

}

};

花花酱 LeetCode 1383. Maximum Performance of a Team

By zxi on March 21, 2020

There are n engineers numbered from 1 to n and two arrays: speed and efficiency, where speed[i] and efficiency[i] represent the speed and efficiency for the i-th engineer respectively. Return the maximum performance of a team composed of at most k engineers, since the answer can be a huge number, return this modulo 10^9 + 7.

The performance of a team is the sum of their engineers’ speeds multiplied by the minimum efficiency among their engineers.

Example 1:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation: 
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.

Example 2:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.

Example 3:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
Output: 72

Constraints:

1 <= n <= 10^5
speed.length == n
efficiency.length == n
1 <= speed[i] <= 10^5
1 <= efficiency[i] <= 10^8
1 <= k <= n

Solution: Greedy + Sliding Window

Sort engineers by their efficiency in descending order.
For each window of K engineers (we can have less than K people in the first k-1 windows), ans is sum(speed) * min(efficiency).

Time complexity: O(nlogn) + O(nlogk)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {

vector<pair<int, int>> es;

for (int i = 0; i < n; ++i)

es.push_back({efficiency[i], speed[i]});

sort(rbegin(es), rend(es));

priority_queue<int, vector<int>, greater<int>> q;

long sum = 0;

long ans = 0;

for (int i = 0; i < n; ++i) {

if (i >= k) {

sum -= q.top();

q.pop();

}

sum += es[i].second;

q.push(es[i].second);

ans = max(ans, sum * es[i].first);

}

return ans % static_cast<int>(1e9 + 7);

}

};

Python3

# Author: Huahua

class Solution:

def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:

speeds = 0

ans = 0

q = []

for e, s in sorted(zip(efficiency, speed), reverse=True):

if len(q) >= k:

speeds -= heapq.heappop(q)

speeds += s

heapq.heappush(q, s)

ans = max(ans, speeds * e)

return ans % (10**9 + 7)