花花酱 LeetCode 869. Reordered Power of 2

By zxi on July 14, 2018

Problem

Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2.

Example 1:

Input: 1
Output: true

Example 2:

Input: 10
Output: false

Example 3:

Input: 16
Output: true

Example 4:

Input: 24
Output: false

Example 5:

1 2	<strong>Input: </strong><span id="example-input-5-1">46</span> <strong>Output: </strong><span id="example-output-5">true</span>

Note:

1 <= N <= 10^9

Solution: HashTable

Compare the counter of digit string with that of all power of 2s.

e.g. 64 -> {4: 1, 6: 1} == 46 {4:1, 6: 1}

Time complexity: O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool reorderedPowerOf2(int N) {

auto m = countMap(N);

for (int i = 0; i < 31; ++i)

if (m == countMap(1 << i)) return true;

return false;

}

private:

map<int, int> countMap(int n) {

map<int, int> m;

while (n) {

++m[n % 10];

n /= 10;

}

return m;

}

};

花花酱 LeetCode 287. Find the Duplicate Number

By zxi on July 14, 2018

Problem

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

Note:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n²).
There is only one duplicate number in the array, but it could be repeated more than once.

Solution1: Binary Search

Time complexity: O(nlogn)

Space complexity: O(1)

Find the smallest m such that len(nums <= m) > m, which means m is the duplicate number.

In the sorted form [1, 2, …, m1, m2, m + 1, …, n]

There are m+1 numbers <= m

// Author: Huahua

// Running time: 8 ms

class Solution {

public:

int findDuplicate(vector<int>& nums) {

int l = 1;

int r = nums.size();

while (l < r) {

int m = (r - l) / 2 + l;

int count = 0; // len(nums <= m)

for (int num : nums)

if (num <= m) ++count;

if (count <= m)

l = m + 1;

else

r = m;

}

return l;

}

};

Solution: Linked list cycle

Convert the problem to find the entry point of the cycle in a linked list.

Take the number in the array as the index of next node.

[1,3,4,2,2]

0->1->3->2->4->2 cycle: 2->4->2

[3,1,3,4,2]

0->3->4->2->3->4->2 cycle 3->4->2->3

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findDuplicate(vector<int>& nums) {

int slow = 0;

int fast = 0;

while (true) {

slow = nums[slow];

fast = nums[nums[fast]];

if (slow == fast) break;

}

fast = 0;

while (fast != slow) {

slow = nums[slow];

fast = nums[fast];

}

return slow;

}

};

花花酱 LeetCode 641. Design Circular Deque

By zxi on July 14, 2018

Problem

Design your implementation of the circular double-ended queue (deque).
Your implementation should support following operations:

MyCircularDeque(k): Constructor, set the size of the deque to be k.
insertFront(): Adds an item at the front of Deque. Return true if the operation is successful.
insertLast(): Adds an item at the rear of Deque. Return true if the operation is successful.
deleteFront(): Deletes an item from the front of Deque. Return true if the operation is successful.
deleteLast(): Deletes an item from the rear of Deque. Return true if the operation is successful.
getFront(): Gets the front item from the Deque. If the deque is empty, return -1.
getRear(): Gets the last item from Deque. If the deque is empty, return -1.
isEmpty(): Checks whether Deque is empty or not.
isFull(): Checks whether Deque is full or not.

Example:

MyCircularDeque circularDeque = new MycircularDeque(3); // set the size to be 3
circularDeque.insertLast(1);			// return true
circularDeque.insertLast(2);			// return true
circularDeque.insertFront(3);			// return true
circularDeque.insertFront(4);			// return false, the queue is full
circularDeque.getRear();  				// return 32
circularDeque.isFull();				// return true
circularDeque.deleteLast();			// return true
circularDeque.insertFront(4);			// return true
circularDeque.getFront();				// return 4

Note:

All values will be in the range of [1, 1000].
The number of operations will be in the range of [1, 1000].
Please do not use the built-in Deque library.

Solution

Using head and tail to pointer to the head and the tail in the circular buffer.

// Author: Huahua

// Running time: 20 ms

class MyCircularDeque {

public:

/** Initialize your data structure here. Set the size of the deque to be k. */

MyCircularDeque(int k): k_(k), q_(k), head_(1), tail_(-1), size_(0) {}

/** Adds an item at the front of Deque. Return true if the operation is successful. */

bool insertFront(int value) {

if (isFull()) return false;

head_ = (head_ - 1 + k_) % k_;

q_[head_] = value;

if (size_ == 0) tail_ = head_;

++size_;

return true;

}

/** Adds an item at the rear of Deque. Return true if the operation is successful. */

bool insertLast(int value) {

if (isFull()) return false;

tail_ = (tail_ + 1 + k_) % k_;

q_[tail_] = value;

if (size_ == 0) head_ = tail_;

++size_;

return true;

}

/** Deletes an item from the front of Deque. Return true if the operation is successful. */

bool deleteFront() {

if (isEmpty()) return false;

head_ = (head_ + 1 + k_) % k_;

--size_;

return true;

}

/** Deletes an item from the rear of Deque. Return true if the operation is successful. */

bool deleteLast() {

if (isEmpty()) return false;

tail_ = (tail_ - 1 + k_) % k_;

--size_;

return true;

}

/** Get the front item from the deque. */

int getFront() {

if (isEmpty()) return -1;

return q_[head_];

}

/** Get the last item from the deque. */

int getRear() {

if (isEmpty()) return -1;

return q_[tail_];

}

/** Checks whether the circular deque is empty or not. */

bool isEmpty() {

return size_ == 0;

}

/** Checks whether the circular deque is full or not. */

bool isFull() {

return size_ == k_;

}

private:

const int k_;

int head_;

int tail_;

int size_;

vector<int> q_;

};

Problem

Given a positive integer N, find and return the longest distance between two consecutive 1’s in the binary representation of N.

If there aren’t two consecutive 1’s, return 0.

Example 1:

Input: 22
Output: 2
Explanation: 
22 in binary is 0b10110.
In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
The first consecutive pair of 1's have distance 2.
The second consecutive pair of 1's have distance 1.
The answer is the largest of these two distances, which is 2.

Example 2:

Input: 5
Output: 2
Explanation: 
5 in binary is 0b101.

Example 3:

Input: 6
Output: 1
Explanation: 
6 in binary is 0b110.

Example 4:

Input: 8
Output: 0
Explanation: 
8 in binary is 0b1000.
There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.\

Note:

1 <= N <= 10^9

Solution: Bit

Time complexity: O(logN)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int binaryGap(int N) {

int l = -1;

int ans = 0;

for (int i = 0; i < 31; ++i)

if (N & (1 << i)) {

if (l >= 0)

ans = max(ans, i - l);

l = i;

}

return ans;

}

};

花花酱 LeetCode 869. Reordered Power of 2

Problem

Solution: HashTable

花花酱 LeetCode 287. Find the Duplicate Number

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Solution1: Binary Search

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花花酱 LeetCode 641. Design Circular Deque

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