Posts published in “Algorithms”

花花酱 LeetCode 2824. Count Pairs Whose Sum is Less than Target

By zxi on August 23, 2023

Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs(i, j)where0 <= i < j < nandnums[i] + nums[j] < target.

Example 1:

Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation: There are 3 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
- (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target 
- (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.

Example 2:

Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation: There are 10 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
- (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
- (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
- (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
- (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
- (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
- (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
- (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
- (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target

Constraints:

1 <= nums.length == n <= 50
-50 <= nums[i], target <= 50

Solution 1: Brute force

Enumerate all pairs.

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

C++

// Author: Huahua

class Solution {

public:

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

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] < target;

return ans;

}

};

Solution 2: Two Pointers

Sort the numbers.

Use two pointers i, and j.
Set i to 0 and j to n – 1.
while (nums[i] + nums[j] >= target) –j
then we have nums[i] + nums[k] < target (i < k <= j), in total (j – i) pairs.
++i, move to the next starting number.
Time complexity: O(nlogn + n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

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

const int n = nums.size();

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

int ans = 0;

for (int i = 0, j = n - 1; i < n; ++i) {

while (j >= i && nums[i] + nums[j] >= target) --j;

ans += max(0, j - i);

}

return ans;

}

};

花花酱 LeetCode 2815. Max Pair Sum in an Array

By zxi on August 13, 2023

You are given a 0-indexed integer array nums. You have to find the maximum sum of a pair of numbers from nums such that the maximum digit in both numbers are equal.

Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [51,71,17,24,42]
Output: 88
Explanation: 
For i = 1 and j = 2, nums[i] and nums[j] have equal maximum digits with a pair sum of 71 + 17 = 88. 
For i = 3 and j = 4, nums[i] and nums[j] have equal maximum digits with a pair sum of 24 + 42 = 66.
It can be shown that there are no other pairs with equal maximum digits, so the answer is 88.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: No pair exists in nums with equal maximum digits.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

Solution: Brute Force

Enumerate all pairs of nums and check their sum and max digit.

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

C++

// Author: Huahua

class Solution {

public:

int maxSum(vector<int>& nums) {

auto maxDigit = [](int x) {

int ans = 0;

while (x) {

ans = max(ans, x % 10);

x /= 10;

}

return ans;

};

int ans = -1;

const int n = nums.size();

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

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

if (nums[i] + nums[j] > ans

&& maxDigit(nums[i]) == maxDigit(nums[j]))

ans = nums[i] + nums[j];

return ans;

}

};

花花酱 LeetCode 2657. Find the Prefix Common Array of Two Arrays

By zxi on April 30, 2023

You are given two 0-indexed integer permutations A and B of length n.

A prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B.

Return the prefix common array of A and B.

A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Example 1:

Input: A = [1,3,2,4], B = [3,1,2,4]
Output: [0,2,3,4]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: 1 and 3 are common in A and B, so C[1] = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.

Example 2:

Input: A = [2,3,1], B = [3,1,2]
Output: [0,1,3]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: only 3 is common in A and B, so C[1] = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.

Constraints:

1 <= A.length == B.length == n <= 50
1 <= A[i], B[i] <= n
It is guaranteed that A and B are both a permutation of n integers.

Solution 1: Bitset

Use bitsets to store the numbers seen so far for each array, and use sA & sB to count the common elements.

Time complexity: O(n*50)
Space complexity: O(50)

C++

// Author: Huahua

class Solution {

public:

vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {

bitset<51> sA;

bitset<51> sB;

vector<int> ans;

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

sA.set(A[i]);

sB.set(B[i]);

ans.push_back((sA & sB).count());

}

return ans;

}

};

Solution 2: Counter

Use a counter to track the frequency of each element, when the counter[x] == 2, we found a pair.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {

vector<int> count(A.size() + 1);

vector<int> ans;

for (int i = 0, s = 0; i < A.size(); ++i) {

if (++count[A[i]] == 2) ++s;

if (++count[B[i]] == 2) ++s;

ans.push_back(s);

}

return ans;

}

};

花花酱 LeetCode 2640. Find the Score of All Prefixes of an Array

By zxi on April 29, 2023

We define the conversion array conver of an array arr as follows:

conver[i] = arr[i] + max(arr[0..i]) where max(arr[0..i]) is the maximum value of arr[j] over 0 <= j <= i.

We also define the score of an array arr as the sum of the values of the conversion array of arr.

Given a 0-indexed integer array nums of length n, return an array ans of length n where ans[i] is the score of the prefix nums[0..i].

Example 1:

Input: nums = [2,3,7,5,10]
Output: [4,10,24,36,56]
Explanation: 
For the prefix [2], the conversion array is [4] hence the score is 4
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56

Example 2:

Input: nums = [1,1,2,4,8,16]
Output: [2,4,8,16,32,64]
Explanation: 
For the prefix [1], the conversion array is [2] hence the score is 2
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Solution: Prefix Sum

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

C++

// Author: Huahua

class Solution {

public:

vector<long long> findPrefixScore(vector<int>& nums) {

vector<long long> ans(nums.size());

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

m = max(m, nums[i]);

ans[i] = (i ? ans[i - 1] : 0) + nums[i] + m;

}

return ans;

}

};

花花酱 LeetCode 2644. Find the Maximum Divisibility Score

By zxi on April 28, 2023

You are given two 0-indexed integer arrays nums and divisors.

The divisibility score of divisors[i] is the number of indices j such that nums[j] is divisible by divisors[i].

Return the integer divisors[i] with the maximum divisibility score. If there is more than one integer with the maximum score, return the minimum of them.

Example 1:

Input: nums = [4,7,9,3,9], divisors = [5,2,3]
Output: 3
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors[0] is 0 since no number in nums is divisible by 5.
The divisibility score of divisors[1] is 1 since nums[0] is divisible by 2.
The divisibility score of divisors[2] is 3 since nums[2], nums[3], and nums[4] are divisible by 3.
Since divisors[2] has the maximum divisibility score, we return it.

Example 2:

Input: nums = [20,14,21,10], divisors = [5,7,5]
Output: 5
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors[0] is 2 since nums[0] and nums[3] are divisible by 5.
The divisibility score of divisors[1] is 2 since nums[1] and nums[2] are divisible by 7.
The divisibility score of divisors[2] is 2 since nums[0] and nums[3] are divisible by 5.
Since divisors[0], divisors[1], and divisors[2] all have the maximum divisibility score, we return the minimum of them (i.e., divisors[2]).

Example 3:

Input: nums = [12], divisors = [10,16]
Output: 10
Explanation: The divisibility score for every element in divisors is:
The divisibility score of divisors[0] is 0 since no number in nums is divisible by 10.
The divisibility score of divisors[1] is 0 since no number in nums is divisible by 16.
Since divisors[0] and divisors[1] both have the maximum divisibility score, we return the minimum of them (i.e., divisors[0]).

Constraints:

1 <= nums.length, divisors.length <= 1000
1 <= nums[i], divisors[i] <= 10⁹

Solution: Brute Force

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

C++

// Author: Huahua

class Solution {

public:

int maxDivScore(vector<int>& nums, vector<int>& divisors) {

int maxScore = 0;

int ans = INT_MAX;

for (int d : divisors) {

int score = 0;

for (int x : nums) {

score += x % d == 0;

}

if (score > maxScore) {

maxScore = score;

ans = d;

}

if (score == maxScore)

ans = min(ans, d);

}

return ans;

}

};