# 164. Maximum Gap

Difficulty: Hard

Frequency: N/A

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

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```class Solution {
public:
vector countSort(vector&amp; nums, int w) {
vector counts(11, 0);
for (int n : nums) {
int v = n / w % 10;
counts[v + 1]++;
}
for (int i = 1; i &lt;= 10; i++) counts[i] += counts[i - 1];
vector results(nums.size(), 0);
for (int n : nums) {
int v = n / w % 10;
results[counts[v]++] = n;
}
return results;
}
int max_n = 0;
for (int n : nums) max_n = max(max_n, n);
int w = 1;
while (w &lt;= max_n) {
nums = countSort(nums, w);
w *= 10;
}
}
int maximumGap(vector&amp; nums) {
int gap = 0, n = nums.size();
for (int i = 1; i &lt; n; i++) {
gap = max(gap, nums[i] - nums[i - 1]);
}
return gap;
}
};
```

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# 057. Insert Interval

Difficulty: Hard

Frequency: N/A

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals `[1,3],[6,9]`, insert and merge `[2,5]` in as `[1,5],[6,9]`.

Example 2:
Given `[1,2],[3,5],[6,7],[8,10],[12,16]`, insert and merge `[4,9]` in as `[1,2],[3,10],[12,16]`.

This is because the new interval `[4,9]` overlaps with `[3,5],[6,7],[8,10]`.

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```/**
* Definition for an interval.
* struct Interval {
*     int start;
*     int end;
*     Interval() : start(0), end(0) {}
*     Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> results;
intervals.push_back(newInterval);
sort(intervals.begin(), intervals.end(), compare);
int i = 0, n = intervals.size();
while (i < n) {
int j = i + 1;
while (j < n && intervals[i].end >= intervals[j].start) {
intervals[i].end = max(intervals[i].end, intervals[j].end);
j++;
}
results.push_back(intervals[i]);
i = j;
}
return results;
}
static bool compare(Interval& i1, Interval& i2) {
return i1.start < i2.start;
}
};
```

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# 056. Merge Intervals

Difficulty: Hard

Frequency: N/A

Given a collection of intervals, merge all overlapping intervals.

For example,
Given `[1,3],[2,6],[8,10],[15,18]`,
return `[1,6],[8,10],[15,18]`.

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```/**
* Definition for an interval.
* struct Interval {
*     int start;
*     int end;
*     Interval() : start(0), end(0) {}
*     Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), compare);
vector<Interval> results;
int i = 0, n = intervals.size();
while (i < n) {
int j = i + 1;
while (j < n && intervals[i].end >= intervals[j].start) {
intervals[i].end = max(intervals[i].end, intervals[j].end);
j++;
}
results.push_back(intervals[i]);
i = j;
}
return results;
}
static bool compare(Interval& i1, Interval& i2) {
return i1.start < i2.start;
}
};
```

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# 324. Wiggle Sort II

Difficulty: Medium

Frequency: N/A

Given an unsorted array `nums`, reorder it such that `nums[0] < nums[1] > nums[2] < nums[3]...`.

Example:
(1) Given `nums = [1, 5, 1, 1, 6, 4]`, one possible answer is `[1, 4, 1, 5, 1, 6]`.
(2) Given `nums = [1, 3, 2, 2, 3, 1]`, one possible answer is `[2, 3, 1, 3, 1, 2]`.

Note:
You may assume all input has valid answer.

Can you do it in O(n) time and/or in-place with O(1) extra space?

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```class Solution {
public:
void wiggleSort(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size(), m = (n + 1) / 2;
vector<int> small(nums.begin(), nums.begin() + m);
vector<int> large(nums.begin() + m, nums.end());
reverse(small.begin(), small.end());
reverse(large.begin(), large.end());

vector<int> result;
for (int i = 0; i < m; i++) {
result.push_back(small[i]);
if (i < large.size()) result.push_back(large[i]);
}
nums = result;
}
};
```

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# 280. (Locked)Wiggle Sort

Difficulty: Medium

Frequency: N/A

Given an unsorted array `nums`, reorder it in-place such that `nums[0] <= nums[1] >= nums[2] <= nums[3]...`.

For example, given `nums = [3, 5, 2, 1, 6, 4]`, one possible answer is `[1, 6, 2, 5, 3, 4]`.

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```class Solution {
public:
void wiggleSort(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
for (int i = 1; i < n - 1; i += 2) swap(nums[i], nums[i + 1]);
}
};
```

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# 179. Largest Number

Difficulty: Medium

Frequency: N/A

Given a list of non negative integers, arrange them such that they form the largest number.

For example, given `[3, 30, 34, 5, 9]`, the largest formed number is `9534330`.

Note: The result may be very large, so you need to return a string instead of an integer.

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```class Solution {
public:
string largestNumber(vector<int>& nums) {
sort(nums.begin(), nums.end(), compare);
string result;
for (int i = nums.size() - 1; i >= 0; i--) {
result += to_string(nums[i]);
}
if (result.find_first_not_of('0') == string::npos) return "0";
else return result = result.substr(result.find_first_not_of('0'));
}
private:
static bool compare(int& num1, int& num2) {
string s1 = to_string(num1);
string s2 = to_string(num2);
if ((s1 + s2) < (s2 + s2)) return true;
else return false;
}
};
```

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# 148. Sort List

Difficulty: Medium

Frequency: N/A

Sort a linked list in O(n log n) time using constant space complexity.

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```/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int n = 0;
while (p) {
n++;
p = p->next;
}
}
ListNode* divide(ListNode *&head, int n) {
if (n == 0) return NULL;
if (n == 1) {
}
}
ListNode *dummy = new ListNode(0);
ListNode *p = dummy;
} else {
}
p = p->next;
}
return dummy->next;
}
};
```

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# 252. (Locked) Meeting rooms

Difficulty: Easy

Frequency: N/A

Given an array of meeting time intervals consisting of start and end times `[[s1,e1],[s2,e2],...]` (si < ei), determine if a person could attend all meetings.

For example,
Given `[[0, 30],[5, 10],[15, 20]]`,
return `false`.

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```/**
* Definition for an interval.
* struct Interval {
*     int start;
*     int end;
*     Interval() : start(0), end(0) {}
*     Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
bool canAttendMeetings(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), compare);
for (int i = 1; i < intervals.size(); i++) {
if (intervals[i].start < intervals[i - 1].end) return false;
}
return true;
}
private:
static bool compare(Interval& interval1, Interval& interval2) {
return interval1.start < interval2.start;
}
};
```

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# 075. Sort Colors

Difficulty: Medium

Frequency: N/A

Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note:
You are not suppose to use the library’s sort function for this problem.

A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0’s, 1’s, and 2’s, then overwrite array with total number of 0’s, then 1’s and followed by 2’s.

Could you come up with an one-pass algorithm using only constant space?

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```class Solution {
public:
void sortColors(vector<int>& nums) {
int sec = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] != 0) {
for (int j = i + 1; j < nums.size(); j++) {
if (nums[j] == 0) {
nums[j] = nums[i];
nums[i] = 0;
sec = i + 1;
}
}
}
}
for (int i = sec; i < nums.size(); i++) {
if (nums[i] == 2) {
for (int j = i + 1; j < nums.size(); j++) {
if (nums[j] == 1) {
nums[j] = nums[i];
nums[i] = 1;
}
}
}
}
}
};
```

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# 147. Insertion Sort List

Difficulty: Medium

Frequency: N/A

Sort a linked list using insertion sort.

My solution:
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```/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *dummy = new ListNode(INT_MIN);
while (cur) {
if (!curNext) {
break;
} else if (head->val < curNext->val) {
break;
} else {
cur = cur->next;
curNext = curNext->next;
}
}
}
delete dummy;