Posts tagged as “easy”

花花酱 LeetCode 27. Remove Element

By zxi on October 2, 2018

Problem

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Solution:

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int removeElement(vector<int>& nums, int val) {

int len = 0;

for (int num : nums)

if (num != val) nums[len++] = num;

return len;

}

};

Python3

class Solution:

def removeElement(self, nums, val):

l = 0

for num in nums:

if num != val:

nums[l], l = num, l + 1

return l

Problem

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Solution:

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int removeDuplicates(vector<int>& nums) {

const int n = nums.size();

if (n <= 1) return n;

int ans = 0;

int i = 0;

while (i < n) {

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

int j = i + 1;

while (j < n && nums[j] == nums[j - 1]) ++j;

i = j;

}

return ans;

}

};

花花酱 LeetCode 24. Swap Nodes in Pairs

By zxi on October 2, 2018

Problem

Given a linked list, swap every two adjacent nodes and return its head.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

Your algorithm should use only constant extra space.
You may not modify the values in the list’s nodes, only nodes itself may be changed.

Solution

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode *swapPairs(ListNode *head) {

if (!head || !head->next) return head;

ListNode d(0);

d.next = head;

head = &d;

while (head && head->next && head->next->next) {

auto n1 = head->next;

auto n2 = n1->next;

n1->next = n2->next;

n2->next = n1;

head->next = n2;

head = n1;

}

return d.next;

}

};

Python3

class Solution:

def swapPairs(self, head):

if not head or not head.next: return head

dummy = ListNode(0)

dummy.next = head

head = dummy

while head.next and head.next.next:

n1, n2 = head.next, head.next.next

n1.next = n2.next

n2.next = n1

head.next = n2

head = n1

return dummy.next

Problem

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

Solution: Stack

Using a stack to track the existing open parentheses, if the current one is a close parenthesis but does not match the top of the stack, return false, otherwise pop the stack. Check whether the stack is empty in the end.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua, 0 ms

class Solution {

public:

bool isValid(string s) {

map<char, char> p{{')', '('}, {']', '['}, {'}', '{'}};

stack<char> st;

for (char c : s) {

if (!p.count(c)) {

st.push(c);

} else {

if (st.empty() || p[c] != st.top()) return false;

st.pop();

}

return st.empty();

}

};

Python3

# Author: Huahua

class Solution:

def isValid(self, s):

p = {')' : '(', ']': '[', '}' : '{'}

stack = []

for c in s:

if c not in p:

stack.append(c)

else:

if not stack or stack[-1] != p[c]: return False

stack.pop()

return not stack

Problem

n a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

Each group has exactly X cards.
All the cards in each group have the same integer.

Example 1:

Input: [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

Input: [1,1,1,2,2,2,3,3]
Output: false
Explanation: No possible partition.

Example 3:

Input: [1]
Output: false
Explanation: No possible partition.

Example 4:

Input: [1,1]
Output: true
Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2]

Note:

1 <= deck.length <= 10000
0 <= deck[i] < 10000

Solution 1: HashTable + Brute Force

Try all possible Xs. 2 ~ deck.size()

Time complexity: ~O(n)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms

class Solution {

public:

bool hasGroupsSizeX(vector<int>& deck) {

unordered_map<int, int> counts;

for (int card : deck) ++counts[card];

for (int i = 2; i <= deck.size(); ++i) {

if (deck.size() % i) continue;

bool ok = true;

for (const auto& pair : counts)

if (pair.second % i) {

ok = false;

break;

}

if (ok) return true;

}

return false;

}

};

Solution 2: HashTable + GCD

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms

class Solution {

public:

bool hasGroupsSizeX(vector<int>& deck) {

unordered_map<int, int> counts;

for (int card : deck) ++counts[card];

int X = deck.size();

for (const auto& p : counts)

X = __gcd(X, p.second);

return X >= 2;

}

};

Java

// Author: Somebody

class Solution {

public boolean hasGroupsSizeX(int[] deck) {

HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

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

int curr = deck[i];

if (map.containsKey(curr)) {

map.put(curr, map.get(curr) + 1);

} else {

map.put(curr, 1);

}

// O(nLogn)

int gcd = 0;

for(Integer key: map.keySet()) {

gcd = findGCDOfTwoNumbers(map.get(key), gcd);

}

return gcd >= 2;

}

// O(Log n)

public static int findGCDOfTwoNumbers (int a, int b) {

if (b == 0) {

return a;

}

return findGCDOfTwoNumbers(b, a%b);

}

Posts tagged as “easy”

花花酱 LeetCode 27. Remove Element

Problem

Solution:

C++

Python3

Related Problems

花花酱 LeetCode 26. Remove Duplicates from Sorted Array

Problem

Solution:

C++

花花酱 LeetCode 20. Valid Parentheses

Problem

Solution: Stack

C++

Python3

Related Problems

花花酱 LeetCode 914. X of a Kind in a Deck of Cards

Problem

Solution 1: HashTable + Brute Force

C++

Solution 2: HashTable + GCD

C++

Java