Posts published in “Array”

花花酱 LeetCode 2256. Minimum Average Difference

By zxi on May 1, 2022

You are given a 0-indexed integer array nums of length n.

The average difference of the index i is the absolute difference between the average of the first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages should be rounded down to the nearest integer.

Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.

Note:

The absolute difference of two numbers is the absolute value of their difference.
The average of n elements is the sum of the n elements divided (integer division) by n.
The average of 0 elements is considered to be 0.

Example 1:

Input: nums = [2,5,3,9,5,3]
Output: 3
Explanation:
- The average difference of index 0 is: |2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3.
- The average difference of index 1 is: |(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2.
- The average difference of index 2 is: |(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2.
- The average difference of index 3 is: |(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0.
- The average difference of index 4 is: |(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1.
- The average difference of index 5 is: |(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4.
The average difference of index 3 is the minimum average difference so return 3.

Example 2:

Input: nums = [0]
Output: 0
Explanation:
The only index is 0 so return 0.
The average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵

Solution: Prefix / Suffix Sum

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

C++

// Author: Huahua

class Solution {

public:

int minimumAverageDifference(vector<int>& nums) {

const int n = nums.size();

long long suffix = accumulate(begin(nums), end(nums), 0LL);

long long prefix = 0;

int best = INT_MAX;

int ans = 0;

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

prefix += nums[i];

suffix -= nums[i];

int cur = abs(prefix / (i + 1) - (i == n - 1 ? 0 : suffix / (n - i - 1)));

if (cur < best) {

best = cur;

ans = i;

}

return ans;

}

};

花花酱 LeetCode 2239. Find Closest Number to Zero

By zxi on April 16, 2022

Given an integer array nums of size n, return the number with the value closest to 0 in nums. If there are multiple answers, return the number with the largest value.

Example 1:

Input: nums = [-4,-2,1,4,8]
Output: 1
Explanation:
The distance from -4 to 0 is |-4| = 4.
The distance from -2 to 0 is |-2| = 2.
The distance from 1 to 0 is |1| = 1.
The distance from 4 to 0 is |4| = 4.
The distance from 8 to 0 is |8| = 8.
Thus, the closest number to 0 in the array is 1.

Example 2:

Input: nums = [2,-1,1]
Output: 1
Explanation: 1 and -1 are both the closest numbers to 0, so 1 being larger is returned.

Constraints:

1 <= n <= 1000
-10⁵ <= nums[i] <= 10⁵

Solution: ABS

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

C++

// Author: Huahua

class Solution {

public:

int findClosestNumber(vector<int>& nums) {

return *min_element(begin(nums), end(nums), [](int a, int b){

return abs(a) < abs(b) || (abs(a) == abs(b) && a > b);

});

}

};

花花酱 LeetCode 2231. Largest Number After Digit Swaps by Parity

By zxi on April 9, 2022

You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).

Return the largest possible value of num after any number of swaps.

Example 1:

Input: num = 1234
Output: 3412
Explanation: Swap the digit 3 with the digit 1, this results in the number 3214.
Swap the digit 2 with the digit 4, this results in the number 3412.
Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.

Example 2:

Input: num = 65875
Output: 87655
Explanation: Swap the digit 8 with the digit 6, this results in the number 85675.
Swap the first digit 5 with the digit 7, this results in the number 87655.
Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.

Constraints:

1 <= num <= 10⁹

Solution:

Put all even digits into one array, all odd digits into another one, all digits into the third. Sort two arrays, and generate a new number from sorted arrays.

Time complexity: O(logn*loglogn)
Space complexity: O(logn)

C++

// Author: Huahua

class Solution {

public:

int largestInteger(int num) {

vector<int> odd, even, org;

for (int cur = num; cur; cur /= 10) {

((cur & 1) ? odd : even).push_back(cur % 10);

org.push_back(cur % 10);

}

sort(begin(odd), end(odd));

sort(begin(even), end(even));

int ans = 0;

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

if (i) ans *= 10;

auto& cur = (org[org.size() - i - 1] & 1) ? odd : even;

ans += cur.back();

cur.pop_back();

}

return ans;

}

};

花花酱 LeetCode 2176. Count Equal and Divisible Pairs in an Array

By zxi on March 4, 2022

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs(i, j)where0 <= i < j < n, such thatnums[i] == nums[j]and(i * j)is divisible byk.

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

Constraints:

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

Solution: Brute Force

Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

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

const int n = nums.size();

int ans = 0;

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

for (int j = i + 1; j < n; ++j)

ans += (nums[i] == nums[j]) && (i * j % k == 0);

return ans;

}

};

花花酱 LeetCode 2155. All Divisions With the Highest Score of a Binary Array

By zxi on February 4, 2022

You are given a 0-indexed binary array nums of length n. nums can be divided at index i (where 0 <= i <= n) into two arrays (possibly empty) nums_left and nums_right:

nums_left has all the elements of nums between index 0 and i - 1 (inclusive), while nums_right has all the elements of nums between index i and n - 1 (inclusive).
If i == 0, nums_left is empty, while nums_right has all the elements of nums.
If i == n, nums_left has all the elements of nums, while nums_right is empty.

The division score of an index i is the sum of the number of 0‘s in nums_left and the number of 1‘s in nums_right.

Return all distinct indices that have the highest possible division score. You may return the answer in any order.

Example 1:

Input: nums = [0,0,1,0]
Output: [2,4]
Explanation: Division at index
- 0: nums_left is []. nums_right is [0,0,1,0]. The score is 0 + 1 = 1.
- 1: nums_left is [0]. nums_right is [0,1,0]. The score is 1 + 1 = 2.
- 2: nums_left is [0,0]. nums_right is [1,0]. The score is 2 + 1 = 3.
- 3: nums_left is [0,0,1]. nums_right is [0]. The score is 2 + 0 = 2.
- 4: nums_left is [0,0,1,0]. nums_right is []. The score is 3 + 0 = 3.
Indices 2 and 4 both have the highest possible division score 3.
Note the answer [4,2] would also be accepted.

Example 2:

Input: nums = [0,0,0]
Output: [3]
Explanation: Division at index
- 0: nums_left is []. nums_right is [0,0,0]. The score is 0 + 0 = 0.
- 1: nums_left is [0]. nums_right is [0,0]. The score is 1 + 0 = 1.
- 2: nums_left is [0,0]. nums_right is [0]. The score is 2 + 0 = 2.
- 3: nums_left is [0,0,0]. nums_right is []. The score is 3 + 0 = 3.
Only index 3 has the highest possible division score 3.

Example 3:

Input: nums = [1,1]
Output: [0]
Explanation: Division at index
- 0: nums_left is []. nums_right is [1,1]. The score is 0 + 2 = 2.
- 1: nums_left is [1]. nums_right is [1]. The score is 0 + 1 = 1.
- 2: nums_left is [1,1]. nums_right is []. The score is 0 + 0 = 0.
Only index 0 has the highest possible division score 2.

Constraints:

n == nums.length
1 <= n <= 10⁵
nums[i] is either 0 or 1.

Solution: Precompute + Prefix Sum

Count how many ones in the array, track the prefix sum to compute score for each index in O(1).

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

C++

// Author: Huahua

class Solution {

public:

vector<int> maxScoreIndices(vector<int>& nums) {

const int n = nums.size();

const int t = accumulate(begin(nums), end(nums), 0);

vector<int> ans;

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

const int s = (i - p) + (t - p);

if (s > b) {

b = s; ans.clear();

}

if (s == b) ans.push_back(i);

if (i != n) p += nums[i];

}

return ans;

}

};