Posts tagged as “math”

花花酱 LeetCode 1250. Check If It Is a Good Array

By zxi on November 8, 2019

Given an array nums of positive integers. Your task is to select some subset of nums, multiply each element by an integer and add all these numbers. The array is said to be good if you can obtain a sum of 1 from the array by any possible subset and multiplicand.

Return True if the array is good otherwise return False.

Example 1:

Input: nums = [12,5,7,23]
Output: true
Explanation: Pick numbers 5 and 7.
5*3 + 7*(-2) = 1

Example 2:

Input: nums = [29,6,10]
Output: true
Explanation: Pick numbers 29, 6 and 10.
29*1 + 6*(-3) + 10*(-1) = 1

Example 3:

Input: nums = [3,6]
Output: false

Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9

Solution: Math

Bézout’s lemma: b[1] * A[1] + b[2] * A[2] + …. + b[n] * A[n] = 1 <==> gcd(A[1], A[2], …, A[n]) = 1 <=> at least two numbers are co-prime

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

C++

// Author: Huahua

class Solution {

public:

bool isGoodArray(vector<int>& nums) {

int g = nums[0];

for (int x : nums)

g = gcd(g, x);

return g == 1;

}

};

花花酱 LeetCode 1247. Minimum Swaps to Make Strings Equal

By zxi on November 7, 2019

You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j].

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: 
Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: 
Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you can't swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

Example 4:

Input: s1 = "xxyyxyxyxx", s2 = "xyyxyxxxyx"
Output: 4

Constraints:

1 <= s1.length, s2.length <= 1000
s1, s2 only contain 'x' or 'y'.

Solution: Math

if s1[i] == s2[i] than no need to swap, so we can only look for
case1. s1[i] = x, s2[i] = y, xy
case2. s1[i] = y, s2[i] = x, yx

If case1 + case2 is odd, then there’s no solution.

Otherwise we can use one swap to fix two xys (or two yxs)
xx, yy => xy, yx

One special case is there an extra xy and and extra yx, which takes two swaps
xy, yx => yy, xx => xy, xy

Finally,
ans = (case1 + 1) / 2 + (case2 + 1) / 2

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

C++

// Author: Huahua

class Solution {

public:

int minimumSwap(string s1, string s2) {

int xy = 0;

int yx = 0;

for (int i = 0; i < s1.length(); ++i) {

if (s1[i] == 'x' && s2[i] == 'y') ++xy;

if (s1[i] == 'y' && s2[i] == 'x') ++yx;

}

if ((xy + yx) % 2) return -1;

return (xy + 1) / 2 + (yx + 1) / 2;

}

};

花花酱 LeetCode 1217. Play with Chips

By zxi on October 5, 2019

There are some chips, and the i-th chip is at position chips[i].

You can perform any of the two following types of moves any number of times (possibly zero) on any chip:

Move the i-th chip by 2 units to the left or to the right with a cost of 0.
Move the i-th chip by 1 unit to the left or to the right with a cost of 1.

There can be two or more chips at the same position initially.

Return the minimum cost needed to move all the chips to the same position (any position).

Example 1:

Input: chips = [1,2,3]
Output: 1
Explanation: Second chip will be moved to positon 3 with cost 1. First chip will be moved to position 3 with cost 0. Total cost is 1.

Example 2:

Input: chips = [2,2,2,3,3]
Output: 2
Explanation: Both fourth and fifth chip will be moved to position two with cost 1. Total minimum cost will be 2.

Constraints:

1 <= chips.length <= 100
1 <= chips[i] <= 10^9

Solution: Math

We can choose either:
1. move all odd positions to an arbitrary even position and move the rest to the same position
2. move all even positions to an arbitrary odd position and move the rest to the same position
ans = min(# of odd pos, # of even pos)

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

C++

// Author: Huahua

class Solution {

public:

int minCostToMoveChips(vector<int>& chips) {

int odd = 0;

int even = 0;

for (int p : chips) {

if (p % 2) {

++odd;

} else {

++even;

}

return min(odd, even);

}

};

花花酱 LeetCode 528. Random Pick with Weight

By zxi on September 29, 2019

Given an array w of positive integers, where w[i] describes the weight of index i, write a function pickIndex which randomly picks an index in proportion to its weight.

Note:

1 <= w.length <= 10000
1 <= w[i] <= 10^5
pickIndex will be called at most 10000 times.

Example 1:

Input: 
["Solution","pickIndex"]
[[[1]],[]]
Output: [null,0]

Example 2:

Input: 
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output: [null,0,1,1,1,0]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution‘s constructor has one argument, the array w. pickIndex has no arguments. Arguments are always wrapped with a list, even if there aren’t any.

Solution: Binary Search

Crate a cumulative weight array, random sample a “weight”, do a binary search to see which bucket that weight falls in.
e.g. w = [2, 3, 1, 4], sum = [2, 5, 6, 10]
sample 3 => index = 1
sample 7 => index = 3

Time complexity: Init: O(n) Pick: O(logn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

Solution(vector<int> w): sums_(std::move(w)) {

partial_sum(begin(sums_), end(sums_), begin(sums_));

}

int pickIndex() {

int s = rand() % sums_.back();

return upper_bound(begin(sums_), end(sums_), s) - begin(sums_);

}

private:

vector<int> sums_;

};

花花酱 LeetCode 60. Permutation Sequence

By zxi on September 29, 2019

The set [1,2,3,...,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order, we get the following sequence for n = 3:

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the k^th permutation sequence.

Note:

Given n will be between 1 and 9 inclusive.
Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

Solution: Math

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

C++

// Author: Huahua

class Solution {

public:

string getPermutation(int n, int k) {

vector<int> num;

vector<int> fact(10, 1);

for (int i = 1; i <= 9; i++) {

num.push_back(i);

fact[i] = fact[i - 1] * i;

}

string s;

k--;

while (n--) {

int d = k / fact[n];

k %= fact[n];

s += ('0' + num[d]);

for (int i = d + 1; i <= 9; i++)

num[i - 1] = num[i];

}

return s;

}

};