Posts tagged as “easy”

花花酱 LeetCode 922. Sort Array By Parity II

By zxi on October 13, 2018

Problem

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Note:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

Solution 1: Brute Force

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<int> sortArrayByParityII(vector<int>& A) {

vector<int> evens;

vector<int> odds;

for (int a : A)

if (a % 2 == 0)

evens.push_back(a);

else

odds.push_back(a);

auto it1 = begin(evens);

auto it2 = begin(odds);

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

A[i] = *it1++;

swap(it1, it2);

}

return A;

}

};

花花酱 LeetCode 917. Reverse Only Letters

By zxi on October 6, 2018

Problem

Given a string S, return the “reversed” string where all characters that are not a letter stay in the same place, and all letters reverse their positions.

Example 1:

Input: "ab-cd"
Output: "dc-ba"

Example 2:

Input: "a-bC-dEf-ghIj"
Output: "j-Ih-gfE-dCba"

Example 3:

Input: "Test1ng-Leet=code-Q!"
Output: "Qedo1ct-eeLg=ntse-T!"

Note:

S.length <= 100
33 <= S[i].ASCIIcode <= 122
S doesn’t contain \ or "

Solution: Two Pointers

Time complexity: O(n)

Space complexity: O(1) – in place

C++/index

// Author: Huahua

class Solution {

public:

string reverseOnlyLetters(string S) {

int i = 0;

int j = S.length() - 1;

while (i < j) {

if (isalpha(S[i]) && isalpha(S[j])) {

swap(S[i++], S[j--]);

} else {

if (!isalpha(S[i])) ++i;

if (!isalpha(S[j])) --j;

}

return S;

}

};

C++/iter

// Author: Huahua

class Solution {

public:

string reverseOnlyLetters(string S) {

auto it1 = begin(S);

auto it2 = prev(end(S));

while (it1 < it2) {

if (isalpha(*it1) && isalpha(*it2)) {

swap(*it1++, *it2--);

} else {

if (!isalpha(*it1)) ++it1;

if (!isalpha(*it2)) --it2;

}

return S;

}

};

花花酱 LeetCode 28. Implement strStr()

By zxi on October 2, 2018

Problem

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf().

Solution 1: Brute Force

Time complexity: O(mn)

Space complexity: O(1)

C++

// Author: Huahua, 4 ms

class Solution {

public:

int strStr(string haystack, string needle) {

const int l1 = haystack.length();

const int l2 = needle.length();

for (int i = 0; i <= l1 - l2; ++i) {

int j = 0;

while (j < l2 && haystack[i + j] == needle[j]) ++j;

if (j == l2) return i;

}

return -1;

}

};

Python3

# Author: Huahua

class Solution:

def strStr(self, haystack, needle):

l1 = len(haystack)

l2 = len(needle)

for i in range(l1 - l2 + 1):

if haystack[i:i+l2] == needle: return i

return -1

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;

}

};

Posts tagged as “easy”

花花酱 LeetCode 922. Sort Array By Parity II

Problem

Solution 1: Brute Force

C++

花花酱 LeetCode 917. Reverse Only Letters

Problem

Solution: Two Pointers

C++/index

C++/iter

花花酱 LeetCode 28. Implement strStr()

Problem

Solution 1: Brute Force

C++

Python3

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++