Posts tagged as “medium”

花花酱 LeetCode 954. Array of Doubled Pairs

By zxi on December 9, 2018

Problem

Given an array of integers A with even length, return true if and only if it is possible to reorder it such that A[2 * i + 1] = 2 * A[2 * i] for every 0 <= i < len(A) / 2.

Example 1:

Input: [3,1,3,6]
Output: false

Example 2:

Input: [2,1,2,6]
Output: false

Example 3:

Input: [4,-2,2,-4]
Output: true
Explanation: We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4].

Example 4:

Input: [1,2,4,16,8,4]
Output: false

Note:

0 <= A.length <= 30000
A.length is even
-100000 <= A[i] <= 100000

Solution 1:

Time complexity: O(N + 100000 * 2)

Space complexity: O(100000 * 2)

C++

// Author: Huahua, 108 ms

class Solution {

public:

bool canReorderDoubled(vector<int>& A) {

const int kMaxN = 100000;

vector<int> p(kMaxN + 1);

vector<int> n(kMaxN + 1);

for (int a : A) {

if (a >= 0) ++p[a];

else ++n[-a];

}

for (int i = 0; i <= kMaxN / 2; ++i) {

if (p[i] && (p[i * 2] -= p[i]) < 0) return false;

if (n[i] && (n[i * 2] -= n[i]) < 0) return false;

}

return true;

}

};

Solution 2:

Time complexity: O(NlogN)

Space complexity: O(N)

C++

// Author: Huahua, 108 ms

class Solution {

public:

bool canReorderDoubled(vector<int>& A) {

unordered_map<int, int> c;

for (int a : A) ++c[a];

vector<int> keys;

for (const auto &kv : c) keys.push_back(kv.first);

sort(begin(keys), end(keys), [](int a, int b) { return abs(a) < abs(b); });

for (int key : keys)

if (c[key] && (c[key * 2] -= c[key]) < 0) return false;

return true;

}

};

花花酱 LeetCode 948. Bag of Tokens

By zxi on November 29, 2018

Problem

You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1 point.
If we have at least 1 point, we may play the token face down, gaining token[i]power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

tokens.length <= 1000
0 <= tokens[i] < 10000
0 <= P < 10000

Solution: Greedy + Two Pointers

Sort the tokens, gain points from the low end gain power from the high end.

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, Running time: 8 ms

class Solution {

public:

int bagOfTokensScore(vector<int>& tokens, int P) {

sort(begin(tokens), end(tokens));

int points = 0;

int ans = 0;

int i = 0;

int j = tokens.size() - 1;

while (i <= j)

if (P >= tokens[i]) {

P -= tokens[i++];

ans = max(ans, ++points);

} else if (points > 0) {

P += tokens[j--];

--points;

} else {

break;

}

return ans;

}

};

花花酱 LeetCode 946. Validate Stack Sequences

By zxi on November 29, 2018

Problem

Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.

Example 1:

Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
Output: true
Explanation: We might do the following sequence:
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1

Example 2:

Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
Output: false
Explanation: 1 cannot be popped before 2.

Note:

0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed is a permutation of popped.
pushed and popped have distinct values.

Solution: Simulation

Simulate the push/pop operation.

Push element from |pushed sequence| onto stack s one by one and pop when top of the stack s is equal the current element in the |popped sequence|.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua, 8 ms

class Solution {

public:

bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {

stack<int> s;

auto it = begin(popped);

for (int e : pushed) {

s.push(e);

while (!s.empty() && s.top() == *it) {

s.pop();

++it;

}

return it == end(popped);

}

};

Python3

# Author: Huahua, Running time: 40 ms

class Solution:

def validateStackSequences(self, pushed, popped):

s = []

i = 0

for e in pushed:

s.append(e)

while s and s[-1] == popped[i]:

s.pop()

i += 1

return i == len(popped)

花花酱 LeetCode 945. Minimum Increment to Make Array Unique

By zxi on November 29, 2018

Problem

Given an array of integers A, a move consists of choosing any A[i], and incrementing it by 1.

Return the least number of moves to make every value in A unique.

Example 1:

Input: [1,2,2]
Output: 1
Explanation:  After 1 move, the array could be [1, 2, 3].

Example 2:

Input: [3,2,1,2,1,7]
Output: 6
Explanation:  After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown with 5 or less moves that it is impossible for the array to have all unique values.

Note:

0 <= A.length <= 40000
0 <= A[i] < 40000

Solution: Greedy

Sort the elements, make sure A[i] >= A[i-1] + 1, if not increase A[i] to A[i – 1] + 1

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, 56 ms

class Solution {

public:

int minIncrementForUnique(vector<int>& A) {

int ans = 0;

sort(begin(A), end(A));

for (int i = 1; i < A.size(); ++i) {

if (A[i] > A[i - 1]) continue;

ans += (A[i - 1] - A[i]) + 1;

A[i] = A[i - 1] + 1;

}

return ans;

}

};

Python3

# Author: Huahua, Running time: 196 ms

class Solution(object):

def minIncrementForUnique(self, A):

A.sort()

ans = 0

for i in range(1, len(A)):

if A[i] > A[i - 1]: continue

ans += A[i - 1] - A[i] + 1

A[i] = A[i - 1] + 1

return ans

花花酱 LeetCode 944. Delete Columns to Make Sorted

By zxi on November 18, 2018

Problem

We are given an array A of N lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have a string "abcdef" and deletion indices {0, 2, 3}, then the final string after deletion is "bef".

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.

Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]]

Return the minimum possible value of D.length.

Example 1:

Input: ["cba","daf","ghi"]
Output: 1

Example 2:

Input: ["a","b"]
Output: 0

Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3

Note:

1 <= A.length <= 100
1 <= A[i].length <= 1000

Solution: Simulation

Check descending case column by column.

Time complexity: O(NL)

Space complexity: O(1)

C++

// Author: Huahua, 64 ms

class Solution {

public:

int minDeletionSize(vector<string>& A) {

int ans = 0;

for (int c = 0; c < A[0].size(); ++c)

for (int r = 1; r < A.size(); ++r) {

if (A[r][c] < A[r - 1][c]) {

++ans;

break;

}

return ans;

}

};