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More LEETCODE
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‎Interview/Algorithm/Array.md‎

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4. Friends Of Appropriate Ages
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5. Trap Rain Water
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6. House Robber
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7. Degree of an array
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8. Decode Ways
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## Implementation
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return dp[nums.size()-1];
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}
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};
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```
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### **Degree of an array**
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***Big O:*** O(n) speed, O(n) space
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```
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Tips:
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Hash map to record the number frequency and another map to record the postions of each number. when encounter a number with the same degree, calculate the length and take the minium value of them.
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```
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```c++
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class Solution {
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public:
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int findShortestSubArray(vector<int>& nums) {
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map<int, int> freq;
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map<int, vector<int>> pos;
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int mx = INT_MIN;
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/*
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get frequency of each number in array
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get highest degree
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note the positions of frequencies
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*/
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for(int i = 0; i < nums.size(); i++)
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{
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mx = max(mx, ++freq[nums[i]]);
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pos[nums[i]].push_back(i);
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}
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//get shortest distance
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int dist = INT_MAX;
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for(auto num : nums)
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{
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if(freq[num] == mx)
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dist = min(dist, pos[num].back() - pos[num].front());
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}
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return dist + 1;
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}
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};
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```
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### **Decode ways**
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***Big O:*** O(n) speed, O(n) space
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```
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Tips:
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Dynamic programming
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```
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```c++
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class Solution {
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public:
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int numDecodings(string s) {
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// edge cases out - leading zero and single character string
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if (s[0] == '0') return 0;
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if (s.size() == 1) return 1;
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// support variables
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int len = s.size(), dp[len];
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// preparing dp
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dp[0] = 1;
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dp[1] = (s[0] == '1' || s[0] == '2' && s[1] < '7' ? 1 : 0) + (s[1] != '0');
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for (int i = 2; i < len; i++) {
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// edge case: we quit for 2 consecutive zeros
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if (s[i] == '0' && (s[i - 1] > '2' || s[i - 1] == '0')) return 0;
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// base case: we always keep the previous set of combinations, unless we met a 0
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dp[i] = s[i] != '0' ? dp[i - 1] : 0;
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// we go and look 2 positions behind if we can make a digit in the 10-26 range
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if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] < '7') dp[i] += dp[i - 2];
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}
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return dp[len - 1];
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}
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};
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```

‎Interview/Algorithm/LeetCode_for_embedded_advanced.md‎

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21. Friends Of Appropriate Ages v
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22. Trap Rain water v
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23. House Robber v
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24. Degree of an array v
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25. Decode Ways v
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***Math:***
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1. Add Binary v
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2. Plus One v
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3. Add strings v
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4. Fibonacci Number v
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5. Pow(x, n)
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5. Pow(x, n) v
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6. Divide Two Integers v
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7. Reverse Integer without long long v
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8. Palindrome Number v
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1. Insert Delete GetRandom O(1) v
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2. LRU Cache
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3. Design Add and Search Words Data Structure
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4. Find Median from Data Stream v
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***String:***
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1. Valid Palindrome II v
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2. Valid Palindrome v
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3. Decode String v
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### Impplementations
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Valid Palindrome II
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```c++
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class Solution {
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public:
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// Checks if string is a palindrome
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bool isPalin(string &s, int start, int end) {
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while(start < end) {
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if(s[start] != s[end])
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return false;
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++start, --end;
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}
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return true;
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}
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// TC: O(N)
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// SC: O(1)
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bool validPalindrome(string s) {
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for(int i = 0, j = s.size()-1; i < j; ++i, --j) {
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// mismatch found, only if it is the first time delete
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// a char and move on, else not possible
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if(s[i] != s[j]) {
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// s[0:i-1] and s[j+1, n-1] matched,
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// now we check if atleast s[i:j-1] or s[i+1:j] is a palindrome
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return (isPalin(s, i, j-1) || isPalin(s, i+1, j));
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}
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}
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return true;
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}
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};
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```
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Verifying an Alien Dictionary
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```c++
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class Solution {

‎Interview/Algorithm/dataStructure.md‎

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## Problems
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1. Insert Delete GetRandom O(1)
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2. LRU Cache
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3. Design Add and Search Words Data Structure
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4. Find Median from Data Stream
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## Implementation
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### **Find Median from Data Stream**
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***Big O:*** O(log(n)) speed, O(n) space
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```
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Tips:
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Two heaps.
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```
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```c++
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class MedianFinder {
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priority_queue<int> lo;
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priority_queue<int> hi;
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public:
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// Adds a number into the data structure.
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void addNum(int num)
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{
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lo.push(num); // Add to max heap
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hi.push(-lo.top()); // balancing step
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lo.pop();
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if (lo.size() < hi.size()) { // maintain size property
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lo.push(-hi.top());
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hi.pop();
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}
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}
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// Returns the median of current data stream
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double findMedian()
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{
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return lo.size() > hi.size() ? lo.top() : ((double) lo.top() - hi.top()) * 0.5;
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}
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};
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```

‎Interview/Algorithm/math.md‎

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1. Reverse Integer without using long long
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2. Palindrome Number
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3. Dot Product of Two Sparse Vectors
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4. Pow(x, n)
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## Implementation
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return (long long) x == dummy;
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}
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};
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```
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## **Pow(x, n)**
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***Big O:*** O(log(n)) speed, O(1) space
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```
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Tips:
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Fast Power Algorithm Iterative
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```
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```c++
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class Solution {
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public:
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double myPow(double x, int n) {
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long long N = n;
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if (N < 0) {
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x = 1 / x;
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N = -N;
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}
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double ans = 1;
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double current_product = x;
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for (long long i = N; i ; i /= 2) {
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if ((i % 2) == 1) {
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ans = ans * current_product;
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}
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current_product = current_product * current_product;
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}
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return ans;
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}
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};
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```

‎Interview/Algorithm/string.md‎

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## Problems
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1. Valid Palindrome II
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2. Valid Palindrome
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3. Decode String
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## Implementation
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### **Valid Palindrome**
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***Big O:*** O(n), O(1) space
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```
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Tips:
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Two pointers.
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```
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```c++
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class Solution {
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public:
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bool isPalindrome(string s) {
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int st = 0;
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int end = s.size()-1;
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while (st < end) {
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if (!isalnum(s[st]) || !isalnum(s[end])) {
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while(st < end && !isalnum(s[st])) {
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st ++;
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}
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while(st < end && !isalnum(s[end])) {
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end --;
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}
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} else {
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if ((isalpha(s[st]) && isalpha(s[end]) && tolower(s[st]) == tolower(s[end])) || (s[st] - '0' == s[end] - '0')) {
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st ++;
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end --;
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} else
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return false;
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}
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}
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return true;
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}
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};
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```
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### **Valid Palindrome II **
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***Big O:*** O(n), O(1) space
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```
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Tips:
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Implement is_palindrome first. return the answer of two is_palin or together.
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```
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```c++
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class Solution {
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public:
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// Checks if string is a palindrome
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bool isPalin(string &s, int start, int end) {
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while(start < end) {
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if(s[start] != s[end])
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return false;
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++start, --end;
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}
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return true;
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}
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// TC: O(N)
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// SC: O(1)
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bool validPalindrome(string s) {
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for(int i = 0, j = s.size()-1; i < j; ++i, --j) {
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// mismatch found, only if it is the first time delete
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// a char and move on, else not possible
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if(s[i] != s[j]) {
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// s[0:i-1] and s[j+1, n-1] matched,
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// now we check if atleast s[i:j-1] or s[i+1:j] is a palindrome
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return (isPalin(s, i, j-1) || isPalin(s, i+1, j));
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}
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}
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return true;
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}
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};
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```
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### **Decode String**
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***Big O:***
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Time Complexity: O(maxK⋅n), where maxK is the maximum value of kk and nn is the length of a given string ss. We traverse a string of size nn and iterate kk times to decode each pattern of form k[string]. This gives us worst case time complexity as O(maxK⋅n).
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Space Complexity: O(m+n), where mm is the number of letters(a-z) and nn is the number of digits(0-9) in string ss. In worst case, the maximum size of stringStack and countStack could be mm and nn respectively.
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```
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Tips:
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Use stack to store pair<int, int>(frequency, string). (Equivalent of using two stacks).
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```
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```c++
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class Solution {
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public:
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string decodeString(string s) {
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stack<pair<int, string>> decode_stk;
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string ret = "";
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int number = 0;
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for (int i = 0; i < s.size(); i++) {
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if (s[i] == '[') {
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string encode_str = "";
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decode_stk.push(make_pair(number, encode_str));
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number = 0;
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} else if (s[i] == ']') {
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pair<int, string> tmp = decode_stk.top();
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decode_stk.pop();
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string copy = tmp.second;
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while(--tmp.first) {
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copy += tmp.second;
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}
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if (decode_stk.empty()) {
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ret += copy;
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} else {
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tmp = decode_stk.top();
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decode_stk.pop();
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tmp.second += copy;
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decode_stk.push(tmp);
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}
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} else if (isalpha(s[i])) {
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if (decode_stk.empty()) {
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ret += s[i];
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} else {
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pair<int, string> tmp = decode_stk.top();
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decode_stk.pop();
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tmp.second += s[i];
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decode_stk.push(tmp);
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}
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} else if (isdigit(s[i])) {
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number = number*10 + s[i] - '0';
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}
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}
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return ret;
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}
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};
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```

‎README.md‎

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6. [Binary Search Tree](Data_Struct_Implementation/BST/README.md)
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7. Red Black Tree
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8. Minium Spanning Tree (MST)
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9. Bitwise tries
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10. Math
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1. Rolling average
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2. Taylor Series

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