Posts tagged as “medium”

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 542. 01 Matrix

By zxi on July 14, 2018

Problem

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:
Input:

0 0 0
0 1 0
0 0 0

Output:

0 0 0
0 1 0
0 0 0

Example 2:
Input:

0 0 0
0 1 0
1 1 1

Output:

0 0 0
0 1 0
1 2 1

Note:

The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.

Solution 1: DP

Two passes:

down, right
up, left

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 132 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

int m = matrix.size();

int n = matrix[0].size();

vector<vector<int>> ans(m, vector<int>(n, INT_MAX - m * n));

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

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

if (matrix[i][j]) {

if (i > 0) ans[i][j] = min(ans[i][j], ans[i - 1][j] + 1);

if (j > 0) ans[i][j] = min(ans[i][j], ans[i][j - 1] + 1);

} else {

ans[i][j] = 0;

}

for (int i = m - 1; i >= 0; --i)

for (int j = n - 1; j >= 0; --j) {

if (i < m - 1) ans[i][j] = min(ans[i][j], ans[i + 1][j] + 1);

if (j < n - 1) ans[i][j] = min(ans[i][j], ans[i][j + 1] + 1);

}

return ans;

}

};

Solution 2: BFS

Start from all 0 cells and find shortest paths to rest of the cells.

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 136 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

queue<pair<int, int>> q;

vector<vector<int>> seen(m, vector<int>(n, 0));

vector<vector<int>> ans(m, vector<int>(n, INT_MAX));

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

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

if (matrix[i][j] == 0) {

q.push({i, j});

seen[i][j] = 1;

}

vector<int> dirs{0, -1, 0, 1, 0};

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

auto pair = q.front(); q.pop();

int i = pair.first;

int j = pair.second;

ans[i][j] = steps;

for (int k = 0; k < 4; ++k) {

int x = j + dirs[k];

int y = i + dirs[k + 1];

if (x < 0 || x >= n || y < 0 || y >= m || seen[y][x]) continue;

seen[y][x] = 1;

q.push({y, x});

}

++steps;

}

return ans;

}

};

花花酱 LeetCode 523. Continuous Subarray Sum

By zxi on July 14, 2018

Problem

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

The length of the array won’t exceed 10,000.
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

Special case:

nums = [0,0], k = 0, return = True

Solution: Prefix Sum Reminder

Time complexity: O(n)

Space complexity: O(min(n, k))

// Author: Huahua

// Running time: 16 ms (<99.62%)

class Solution {

public:

bool checkSubarraySum(vector<int>& nums, int k) {

unordered_map<int, int> m; // sum % k -> first index

m[0] = -1;

int sum = 0;

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

sum += nums[i];

if (k != 0) sum %= k;

if (m.count(sum)) {

if (i - m.at(sum) > 1) return true;

} else {

m[sum] = i;

}

return false;

}

};

Posts tagged as “medium”

花花酱 LeetCode 869. Reordered Power of 2

Problem

Solution: HashTable

花花酱 LeetCode 287. Find the Duplicate Number

Problem

Solution1: Binary Search

Solution: Linked list cycle

花花酱 LeetCode 445. Add Two Numbers II

Problem

Solution: Simulation

Related Problems

花花酱 LeetCode 542. 01 Matrix

Problem

Solution 1: DP

Solution 2: BFS

花花酱 LeetCode 523. Continuous Subarray Sum

Problem

Special case:

Solution: Prefix Sum Reminder

Related Problems