Posts published in “Array”

花花酱 LeetCode 215. Kth Largest Element in an Array

By zxi on December 5, 2021

Given an integer array nums and an integer k, return the k^th largest element in the array.

Note that it is the k^th largest element in the sorted order, not the k^th distinct element.

Example 1:

Input: nums = [3,2,1,5,6,4], k = 2
Output: 5

Example 2:

Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
Output: 4

Constraints:

1 <= k <= nums.length <= 10⁴
-10⁴ <= nums[i] <= 10⁴

Solution: Quick selection

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

C++

// Author: Huahua

class Solution {

public:

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

nth_element(nums.begin(), nums.begin() + k - 1, nums.end(), std::greater<int>());

return nums[k - 1];

}

};

花花酱 LeetCode 152. Maximum Product Subarray

By zxi on November 28, 2021

Given an integer array nums, find a contiguous non-empty subarray within the array that has the largest product, and return the product.

It is guaranteed that the answer will fit in a 32-bit integer.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: nums = [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.

Example 2:

Input: nums = [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

Constraints:

1 <= nums.length <= 2 * 10⁴
-10 <= nums[i] <= 10
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Solution: Track high and low

Compute the low / high of prefix product, reset if nums[i] is higher than high or lower than low.

Swap low and high if nums[i] is a negative number.

e.g. [2, 3, -1, 8, -2]
nums[i] = 2, low = 2, high = 2
nums[i] = 3, low = 3, high = 2 * 3 = 6
nums[i] = -1, low = 6 * -1 = -6, high = -1
nums[i] = 8, low = -6 * 8 = -48, high = 8
nums[i] = -2, low = 8*-2 = -16, high = -48 * -2 = 96

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

C++

// Author: Huahua

class Solution {

public:

int maxProduct(vector<int>& nums) {

int ans = nums[0];

for (int i = 1, h = ans, l = ans; i < nums.size(); ++i) {

if (nums[i] < 0) swap(l, h);

l = min(nums[i], nums[i] * l);

h = max(nums[i], nums[i] * h);

ans = max(ans, h);

}

return ans;

}

};

花花酱 LeetCode 2091. Removing Minimum and Maximum From Array

By zxi on November 28, 2021

You are given a 0-indexed array of distinct integers nums.

There is an element in nums that has the lowest value and an element that has the highest value. We call them the minimum and maximum respectively. Your goal is to remove both these elements from the array.

A deletion is defined as either removing an element from the front of the array or removing an element from the back of the array.

Return the minimum number of deletions it would take to remove both the minimum and maximum element from the array.

Example 1:

Input: nums = [2,10,7,5,4,1,8,6]
Output: 5
Explanation: 
The minimum element in the array is nums[5], which is 1.
The maximum element in the array is nums[1], which is 10.
We can remove both the minimum and maximum by removing 2 elements from the front and 3 elements from the back.
This results in 2 + 3 = 5 deletions, which is the minimum number possible.

Example 2:

Input: nums = [0,-4,19,1,8,-2,-3,5]
Output: 3
Explanation: 
The minimum element in the array is nums[1], which is -4.
The maximum element in the array is nums[2], which is 19.
We can remove both the minimum and maximum by removing 3 elements from the front.
This results in only 3 deletions, which is the minimum number possible.

Example 3:

Input: nums = [101]
Output: 1
Explanation:  
There is only one element in the array, which makes it both the minimum and maximum element.
We can remove it with 1 deletion.

Constraints:

1 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵
The integers in nums are distinct.

Solution: Three ways

There are only three ways to remove min/max elements.
1) Remove front elements
2) Remove back elements
3) Remove one with front elements, and another one with back elements.

Just find the best way to do it.

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

C++

// Author: Huahua

class Solution {

public:

int minimumDeletions(vector<int>& nums) {

const int n = nums.size();

auto [it1, it2] = minmax_element(begin(nums), end(nums));

int i = it1 - begin(nums);

int j = it2 - begin(nums);

return min({max(i, j) + 1, // remove front elements

n - min(i, j), // remove back elements

min(i, j) + 1 + n - max(i, j)} // front + back

);

}

};

花花酱 LeetCode 2089. Find Target Indices After Sorting Array

By zxi on November 28, 2021

You are given a 0-indexed integer array nums and a target element target.

A target index is an index i such that nums[i] == target.

Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

Example 1:

Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.

Example 2:

Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.

Example 3:

Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.

Example 4:

Input: nums = [1,2,5,2,3], target = 4
Output: []
Explanation: There are no elements in nums with value 4.

Constraints:

1 <= nums.length <= 100
1 <= nums[i], target <= 100

Solution: Sorting

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

C++

// Author: Huahua

class Solution {

public:

vector<int> targetIndices(vector<int>& nums, int target) {

sort(begin(nums), end(nums));

vector<int> ans;

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

if (nums[i] == target) ans.push_back(i);

return ans;

}

};

花花酱 LeetCode 41. First Missing Positive

By zxi on November 26, 2021

Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses constant extra space.

Example 1:

Input: nums = [1,2,0]
Output: 3

Example 2:

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

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1

Constraints:

1 <= nums.length <= 5 * 10⁵
-2³¹ <= nums[i] <= 2³¹ - 1

Solution: Marking

First pass, marking nums[i] to INT_MAX if nums[i] <= 0
Second pass, use a negative number to mark the presence of a number x at nums[x – 1]
Third pass, the first positive number is the missing index i, return i +1
If not found return n + 1.

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

C++

// Author: Huahua

class Solution {

public:

int firstMissingPositive(vector<int>& nums) {

const int n = nums.size();

for (int& x : nums)

if (x <= 0) x = INT_MAX;

for (int x : nums) {

x = abs(x);

if (x >= 1 && x <= n && nums[x - 1] > 0)

nums[x - 1] *= -1;

}

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

if (nums[i] > 0)

return i + 1;

return n + 1;

}

};