# Zigzag Conversion

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R


And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);


Example 1:

Input:

 s = "PAYPALISHIRING", numRows = 3


Output:

 "PAHNAPLSIIGYIR"


Example 2:

Input:

 s = "PAYPALISHIRING", numRows = 4


Output:

 "PINALSIGYAHRPI"


Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I


Example 3:

Input:

 s = "A", numRows = 1


Output:

 "A"


This problem was taken from Leet code, ZigZag conversion

## Solution

It’s take the first example string: “PAYPALISHIRING

An array representation of the string would be done by pushing each character into the array.

Then if we want to iterate through this array in zig zag pattern with 3 rows, it will look like this:

Then, we have to create 3 arrays representing the 3 rows as follows:

Dots represent empty strings which we have to ignore before concatenating all 3 rows, convert them to string and this is the answer to this problem.

But how to create the loop to traverse through the zig zag pattern (figure 1) ?

• first we set up headingDown flag to determine if we are going down or up
• each time when the row reaches numRows we are flipping the value of headingDown
• when headingDown is false (meaning that we are heading up) we have to do it diagonally: row — , col ++
• when row become 0 we are flipping again headingDown and it becomes true again: row ++ and we keep storing values of every character we traversed in the result array.
 /**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function(s, numRows) {

// if we don't have any rows or there is an empty string just return back empty string
if(s.length === 0 || numRows === 0)
return '';

// if rows are numRows is just one, we return the same string
if(numRows === 1) {
return s;
}

var l = s.length;
// put the string into single dimension array to iterate through
var arr = s.split('');

var rowsArray = [];
var row = 0;
var col = 0;
var headingDown = true; // this determines if the cursor is heading down in the same column, or up leaping to the next column (dizgonal)

// instantiate numRows arrays to store values of each row
for(var i = 0; i < numRows; i ++) {
rowsArray[i] = [];
}

// loop through each element in arr and fill rowsArray ()
for(var i = 0; i < l; i ++) {
rowsArray[row][col] = arr[i];
row ++;
}
else {
row --;
col ++;
}

if(row == numRows -1) {
} else if(row == 0) {
}

}

// Read 2D array and assemble the string
var result = [];
for(var i = 0; i < numRows; i ++) {
for(var j = 0; j < rowsArray[i].length; j ++) {
if(typeof rowsArray[i][j] != 'undefined') {
result.push(rowsArray[i][j]);
}
}
}
return result.join('');
};

convert('PAYPALISHIRING', 3);
convert('AB', 1);
//convert('ABC', 1);

How can we optimize this ?

the above example is good to read but not a good practical example. First we don’t need two dimensional array to store characters for each row. We could replace this with string array and just keep adding characters till we reached the end of the string. This way we also don’t need to keep track of cols

/**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function(s, numRows) {

// if we don't have any rows or there is an empty string just return back empty string
if(s.length === 0 || numRows === 0)
return '';

// if rows are numRows is just one, we return the same string
if(numRows === 1) {
return s;
}

var l = s.length;
// put the string into single dimension array to iterate through
var arr = s.split('');

var left = 0;
var arrStrings = [];
var row = 0;
var col = 0;
var headingDown = true; // this determines if the cursor is heading down in the same column, or up leaping to the next column (dizgonal)

// instantiate numRows arrays to store values of each row
for(var i = 0; i < numRows; i ++) {
arrStrings[i] = '';
}

// loop through each element in arr and fill arrStrings ()
for(var i = 0; i < l; i ++) {
//arrStrings[row][col] = arr[i];
arrStrings[row] += arr[i];
row ++;
}
else {
arrStrings[row] += arr[i];
row --;
col ++;
}

if(row == numRows -1) {
} else if(row == 0) {
}

}

var result = '';
// combine all strings and return as one
for(var i = 0; i < numRows; i ++) {
result += arrStrings[i];
}

return result;
};

# Longest Common Substring

Given two strings text1 and text2, return the length of their longest common subsequenceIf there is no common subsequence, return 0.

subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

common subsequence of two strings is a subsequence that is common to both strings.

Example 1:

Input:

 text1 = "abcde", text2 = "ace"


Output:

 3


Explanation:

 The longest common subsequence is "ace" and its length is 3.


Example 2:

Input:

 text1 = "abc", text2 = "abc"


Output:

 3


Explanation:

 The longest common subsequence is "abc" and its length is 3.


Example 3:

Input:

 text1 = "abc", text2 = "def"


Output:

 0


Explanation:

 There is no such common subsequence, so the result is 0.


Constraints:

• 1 <= text1.length, text2.length <= 1000
• text1 and text2 consist of only lowercase English characters.

This problem was taken from Leetcode Longest Common Subsequence

## Solution

Dynamic programming with memoization.

We create a matrix to store (memoize) the response from previous calculations so we won’t re-calculate them again.

We are starting from the first row comparing every character from column 1 with the first character from row one.

There are two cases:

• either the character match, then we add 1 and add it with the value from the diagonal to the left of the cell where we are.

Longest Common Substring Step 1

• the characters don’t match, then we get the MAX of the value above current cell  and the value to the left of the current cell.

Longest Common Substring Step 2

Here again c int the column (first string) matches c in the row (the second string)

so we get the value of last comparison (to the left and up diagonal of the current cell)  which is 1 and add 1 again since characters match.

Longest Common Substring 3

and keep going till we reach the end of both strings which is the answer.

Longest Common Substring 4

/**
* @param {string} text1
* @param {string} text2
* @return {number}
*/
var longestCommonSubsequence = function(text1, text2) {
var txt1 = text1.split('');
var txt2 = text2.split('');

txt1.unshift('0');
txt2.unshift('1');

var l1 = txt1.length;
var l2 = txt2.length;

var matrix = [];
for(var i = 0; i < l2; i ++) {
matrix[i] = new Array(l1).fill(0);
}

var maxCommon = 0;

for(var row = 0; row < l2; row ++) {
for(var col = 0; col < l1; col ++) {
var last = 0;

if(txt1[col] == txt2[row]) {
var previousDiagonalRowVal = row == 0 || col == 0  ? 0 : matrix[row - 1][col - 1];
last =  1 + previousDiagonalRowVal;
}
else {
var prevUp = row == 0 ?  0 : matrix[row - 1][col];
var prevLeft = col == 0 ? 0 : matrix[row][col - 1];
last = Math.max(prevUp, prevLeft);
}
matrix[row][col] = last;
maxCommon = last > maxCommon ? last : maxCommon;

}
}
return maxCommon;
};

var text1 = "abcde", text2 = "ace";
var r = longestCommonSubsequence(text1, text2);
console.log(">>", r);

# Copy List with Random Pointer

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

• val: an integer representing Node.val
• random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

Example 1:

Input:

 head = [[7,null],[13,0],[11,4],[10,2],[1,0]]


Output:

 [[7,null],[13,0],[11,4],[10,2],[1,0]]


Example 2:

Input:

 head = [[1,1],[2,1]]


Output:

 [[1,1],[2,1]]


Example 3:

Input:

 head = [[3,null],[3,0],[3,null]]


Output:

 [[3,null],[3,0],[3,null]]


Constraints:

• 0 <= n <= 1000
• -104 <= Node.val <= 104
• Node.random is null or is pointing to some node in the linked list.

This problem was taken from https://leetcode.com/problems/copy-list-with-random-pointer/

# Brute force using tree traversal

/**
* // Definition for a Node.
* function Node(val, next, random) {
*    this.val = val;
*    this.next = next;
*    this.random = random;
* };
*/

/**
* @return {Node}
*/

function dfsTraverse(node, visited={}) {
if(!node) {
return null;
}

// if a new node is created, return it. Otherwise you will fall into circular loops
if(node?.clone) {
return node?.clone;
}

var newNode = new Node(node?.val, null, null);
node.clone = newNode;
var next = dfsTraverse(node?.next);
var random = dfsTraverse(node?.random);

newNode.next = next;
newNode.random = random;
return newNode;
}

return result;
};

function Node(val, next, random) {
this.val = val;
this.next = next;
this.random = random;
}
;// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];

//console.log("root");
//console.log(nodes['7']);
var result = copyRandomList(nodes['1']);
console.log(result);



# More elegant solution

/**
* // Definition for a Node.
* function Node(val, next, random) {
*    this.val = val;
*    this.next = next;
*    this.random = random;
* };
*/

/**
* @return {Node}
*/
const copy = new Map();

// add all new nodes and values for now
while (cur) {
copy.set(cur, new Node(cur.val));
cur = cur.next;
}

// iterate again and point curent node to the newly created nodes using the key
while (cur) {
copy.get(cur).next = copy.get(cur.next) || null;
copy.get(cur).random = copy.get(cur.random) || null;
cur = cur.next;
}

}

function Node(val, next, random) {
this.val = val;
this.next = next;
this.random = random;
};

// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];

var result = copyRandomList(nodes['1']);
console.log(result);



# Range Sum of BST

Given the root node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high].

Example 1:

Input:

 root = [10,5,15,3,7,null,18], low = 7, high = 15


Output:

 32


Explanation:

 Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.


Example 2:

Input:

 root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10


Output:

 23


Explanation:

 Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.


Constraints:

• The number of nodes in the tree is in the range [1, 2 * 104].
• 1 <= Node.val <= 105
• 1 <= low <= high <= 105
• All Node.val are unique.

This problem was taken from https://leetcode.com/problems/range-sum-of-bst/

## DFS Solution

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} low
* @param {number} high
* @return {number}
*/
var rangeSumBST = function(root, low, high) {
if(!root) {
return;
}
console.log(root.val);
var queue = [root];
var visited = {};
var result = 0.0;

function dfsTraverse(root, low, high) {

var childNodes = [root?.left, root?.right];

for(const childNode of childNodes) {
if(childNode) {
var val = childNode.val;
console.log(val);
dfsTraverse(childNode, low, high);
}
}

}

dfsTraverse(root, low, high);
return result;
};

function TreeNode(val, left, right) {
this.val = (val===undefined ? 0 : val)
this.left = (left===undefined ? null : left)
this.right = (right===undefined ? null : right)
}

const treeNode = new TreeNode(
10,
new TreeNode(
5,
new TreeNode(3),
new TreeNode(7)
),
new TreeNode(
15,
null,
new TreeNode(18)
)
)

rangeSumBST(treeNode, 7, 15);

## BFS solution

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} low
* @param {number} high
* @return {number}
*/
var rangeSumBST = function(root, low, high) {
if(!root) {
return 0;
}
var queue = [root];
var visited = {};
var result = 0.0;
if(root.val >= low && root.val <= high) {
result = root.val;
}

var leftPointer = 0;
while(leftPointer <= queue.length) {
var curentNode = queue[leftPointer];
leftPointer ++;
var childNodes = [curentNode?.left, curentNode?.right];

for(const childNode of childNodes) {
if(childNode) {
console.log(childNode.val)
if(childNode.val >= low && childNode.val <= high) {
result += childNode.val;
}
queue.push(childNode);
}
}

}
return result;

};

# Moving Average from Data Stream

Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

Implement the MovingAverage class:

• MovingAverage(int size) Initializes the object with the size of the window size.
• double next(int val) Returns the moving average of the last size values of the stream.

Example 1:

Input

["MovingAverage", "next", "next", "next", "next"]
[[3], [1], [10], [3], [5]]


Output

[null, 1.0, 5.5, 4.66667, 6.0]



Explanation

MovingAverage movingAverage = new MovingAverage(3);
movingAverage.next(1); // return 1.0 = 1 / 1
movingAverage.next(10); // return 5.5 = (1 + 10) / 2
movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3


Constraints:

• 1 <= size <= 1000
• -105 <= val <= 105
• At most 104 calls will be made to next.

## Solution

/**
* @param {number} size
*/
var MovingAverage = function(size) {
this.n = size;
this.queue = [];
this.average = 0.0;
};

/**
* @param {number} val
* @return {number}
*/
MovingAverage.prototype.next = function(val) {
var removedVal;
if(this.queue.length >= this.n) {
removedVal = this.queue.shift();
this.average = this.average - removedVal;
}

this.queue.push(val);

this.average += val;
console.log(this.average / this.queue.length);
return this.average / this.queue.length;
};

var movingAverage = new MovingAverage(3);
movingAverage.next(1); // return 1.0 = 1 / 1
movingAverage.next(10); // return 5.5 = (1 + 10) / 2
movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3

# Implement pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

Example 1:

Input:

 x = 2.00000, n = 10


Output:

 1024.00000


Example 2:

Input:

 x = 2.10000, n = 3


Output:

 9.26100


Example 3:

Input:

 x = 2.00000, n = -2


Output:

 0.25000


Explanation:

 2

-2

 = 1/2

2

 = 1/4 = 0.25


Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• -104 <= xn <= 104

This problem was taken from https://leetcode.com/problems/powx-n/

# Brute force solution:

Straight forward, but we have to consider negative ranges.

/**
* @param {number} x
* @param {number} n
* @return {number}
*/
var myPow = function(x, n) {

if(n === 0) {
return 1;
}

var N = n;
var X = x;
var result = 1;

if(n < 0) {
X = 1 / x;
N = -n;
}

for(var i = 0; i < N; i ++) {
result = result * X;
}
return result;
};

# Approach 2: Fast Power Algorithm Recursive

• divide n so you immediately cut the computation time in half to logarithmic one.

/**
* @param {number} x
* @param {number} n
* @return {number}
*/
var myPow = function(x, n) {

function fastPow(x, n) {
if(n < 1) {
return 1;
}
var half = fastPow(x, Math.floor(n / 2));

if(n % 2 == 0) {
return half * half;
}
else {
return half * half * x;
}
}

var X = x;
var N = n;
if(n < 0) {
X = 1 / x;
N = -n;
}
return fastPow(X, N);

};

# Longest Common Prefix

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input:

 strs = ["flower","flow","flight"]


Output:

 "fl"


Example 2:

Input:

 strs = ["dog","racecar","car"]


Output:

 ""


Explanation:

 There is no common prefix among the input strings.


Constraints:

• 1 <= strs.length <= 200
• 0 <= strs[i].length <= 200
• strs[i] consists of only lower-case English letters.

This problem was taken from Leetcode Longest Common Prefix

## Solution

Let’s examine this example:

 strs = ["flower","flow","flight"]


the longest common substring is:

 "fl"

Solution 1: Horizontal scanning

We could assume that the whole word could be the common one so we set prefix = ‘flower’
Then we would compare with the rest words and keep removing last character until prefix becomes empty (meaning no common substring was found) or until we have the common substring.

 prefix flower flow flight flower flower flower flower flowe flowe flowe flowe flow flow flow flow flo flo flo flo fl fl fl fl

/**
* @param {string[]} strs
* @return {string}
*/
var longestCommonPrefix = function(strs) {
var prefix = strs[0];
for(var i = 1; i < strs.length; i ++ ) {
while(strs[i].indexOf(prefix) != 0) {
prefix = prefix.substring(0, prefix.length - 1);
if(prefix == "")
return '';
}
}
return prefix;
};

what we just did:
– set up prefix to be the whole 1st word strs[0]
– compare prefix with the second word (strs[1]) and if there is no match, remove the last symbol and keep going until it finds match.

Complexity Analysis

• Time complexity : O(S) , where S is the sum of all characters in all strings.In the worst case all n strings are the same. The algorithm compares the string S1 with the other strings [S_2 \ldots S_n] There are S character comparisons, where S is the sum of all characters in the input array.
• Space complexity : O(1). We only used constant extra space.

Solution 2: Vertical scanning

Similar but optimized for cases like the one above where we have very short common substring, and we don’t want to scan the whole word.

 prefix flower flow flight f f f f fl fl fl fl flo flo flo flo

/**
* @param {string[]} strs
* @return {string}
*/
var longestCommonPrefix = function(strs) {
var prefix;
for(var i = 0; i < strs[0].length; i ++ ) {
var c = strs[0][i];
for(var j = 0; j < strs.length; j++) {
if(strs[j][i] != c) {
return strs[0].substring(0, i);
}
}
}
return strs[0];
};

what we just did:
– Iterate through the words like they are in column.
– compare each character (starting with the first one) between all words. When we find a mismatch, remove the last (mismatched) character and return truncates strs[0]

# Roman to Integer

Roman numerals are represented by seven different symbols: IVXLCD and M.

Symbol                Value

I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two one’s added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9.
• X can be placed before L (50) and C (100) to make 40 and 90.
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer.

Example 1:

Input:

 s = "III"


Output:

 3


Example 2:

Input:

 s = "IV"


Output:

 4


Example 3:

Input:

 s = "IX"


Output:

 9


Example 4:

Input:

 s = "LVIII"


Output:

 58


Explanation:

 L = 50, V= 5, III = 3.


Example 5:

Input:

 s = "MCMXCIV"


Output:

 1994


Explanation:

 M = 1000, CM = 900, XC = 90 and IV = 4.


Constraints:

• 1 <= s.length <= 15
• s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
• It is guaranteed that s is a valid roman numeral in the range [1, 3999].

This problem was taken from Leetcode Roman To Integer

## Solution

Solution 1: Left to right pass

/**
* @param {string} s
* @return {number}
*/
var romanToInt = function(s) {
var len = s.length;
var i = 0;
var map = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
}
var sum = 0;
while(i < len) {
var currentVal = map[ s[i] ];
var nextVal = map[ s[i + 1] ];
if( currentVal < nextVal) {
sum += nextVal - currentVal;
i ++;
}
else {
sum += currentVal;
}
i ++;
}
return sum;
};

Solution 2: Left to right (or right to left) pass improved

/**
* @param {string} s
* @return {number}
*/
var romanToInt = function(s) {
var len = s.length;
var i = 0;
var map = {
'I': 1,
'IV': 4,
'V': 5,
'IX': 9,
'X': 10,
'XL': 40,
'L': 50,
'XC': 90,
'C': 100,
'CD': 400,
'D': 500,
'CM': 900,
'M': 1000
}
var sum = 0;
while(i < len) {
var currentVal = map[ s[i] ];
var nextVal = map[ s[i + 1] ];
if( currentVal < nextVal) {
var sumbol = s[i] + s[i+1];
sum += map[sumbol];
i ++;
}
else {
sum += currentVal;
}
i ++;
}
return sum;
};

Solution3: Right to left pass

In the “subtraction” cases, such as XC, we’ve been updating our running sum as follows:

sum += value(C) - value(X)


However, notice that this is mathematically equivalent to the following:

sum += value(C)
sum -= value(X)


Utilizing this means that we can process one symbol each time we go around the main loop. We still need to determine whether or not our current symbol should be added or subtracted by looking at the neighbour though.

This way we could start from the most right symbol an initialize the sym with it, since every most right symbol will always be added to the sum.

/**
* @param {string} s
* @return {number}
*/
var romanToInt = function(s) {
var len = s.length;
var i = len - 1;
var map = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
}
var sum = map[ s[i] ];
i --;
while(i > -1) {
var currentVal = map[ s[i] ];
var prevVal = map[ s[i + 1] ];
if( currentVal < prevVal) {
sum -= currentVal;
}
else {
sum += currentVal;
}
i --;
}
return sum;
};

# Container With Most Water

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai)n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

Example 1:

Input:

 height = [1,8,6,2,5,4,8,3,7]


Output:

 49


Explanation:

 The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.


Example 2:

Input:

 height = [1,1]


Output:

 1


Example 3:

Input:

 height = [4,3,2,1,4]


Output:

 16


Example 4:

Input:

 height = [1,2,1]


Output:

 2

This problem was taken from Leetcode Container With Most Water

## Solution

A better than brute force solution is to use a variation of “sliding doors” algorithm.

Let’s consider this case: [1,3,4,3]. The area with most water will be the one with highest height and length.

To find it we set up two pointers: one at position 0, and one at the end of the array. The amount of water that could be collected here is min(leftPointerValue, rightPointerValue) * length,
where length is rightPointer – leftPointer. Which is 4.

Now it’s clear that if rightPointerValue > leftPointerValue there is no point of keep moving rightPointer because we won’t get any bigger amount of water since it will always be limited by the leftPointerValue (height) and the length will always be smaller than the previous length.

So in this case we will move the leftPointer forward to evaluate the next case.

Here the amount of the water collected is min(leftPointerValue, rightPointerValue) * length which is min(3, 3) * 3 = 9.

Nex we continue evaluating all cases till leftPointer = rightPointer (length = 0), and we didn’t find bigger amount of water collected, so the answer we found on the second evaluation is the right answer: 9.

/**
* @param {number[]} height
* @return {number}
*/
var maxArea = function(height) {

var maxArea = 0;
var pLeft = 0;
var pRight = height.length - 1;
var len = pRight - pLeft;

while(len > 0) {
var pLeftVal = height[pLeft];
var pRightVal = height[pRight];

if(pLeftVal > pRightVal) {
maxArea = Math.max( len * pRightVal, maxArea );
pRight --;
}
else {
maxArea = Math.max( len * pLeftVal, maxArea );
pLeft ++;
}
len --;
}
return maxArea;
};

var height = [1,8,6,2,5,4,8,3,7];
console.log( maxArea(height) );

# Unique-paths

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input:

 m = 3, n = 2


Output:

 3


Explanation:

From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right


Example 2:

Input:

 m = 7, n = 3


Output:

 28

This problem was taken from Leetcode unique paths and Leetcode_unique_paths_part_II

## Solution

Since we can move only right or down on every cell in the first row we will have only one place from where we can come and this is the cell before. And same for the first vertical row.

Then after we figured out that there is only one way to reach each cell in the first row and the first column (which is from the cell before) we could start calculating possible ways to go to the next cells.
Let’s look at the cell in the second row and second column.There are actually two possible ways to go there: from the cell above, and the cell before, so 2 possible ways. (figure below).
The cell in the third column on the second row: same 1 way from the cell above, and from the cell before. But since there are already 2 ways to reach the cell before the total ways to reach this cell is: 1 + 2 = 3.

The solution:

/**
* @param {number} m
* @param {number} n
* @return {number}
*/
var uniquePaths = function(m, n) {

var memo = [];

for(var i=0;i < n; i ++) {
for(var j = 0; j < m; j ++) {
var index = (i * m) + j;
if(i == 0) {
memo[index] = 1;
}
else if(j == 0) {
memo[index] = 1;
}
else {
var up = index - m;
var left = index - 1;
memo[index] = memo[up] + memo[left];
}
}
}
return memo[memo.length - 1];
}

console.log(uniquePaths(7,3));

## Unique paths with obstacles.

/**
* @param {number[][]} obstacleGrid
* @return {number}
*/
var uniquePathsWithObstacles = function(obstacleGrid) {
var m = obstacleGrid[0].length;
var n = obstacleGrid.length;
var row = 0;
if(obstacleGrid[0][0] == 1)
return 0;

var memo = [];

for(var i=0;i < n; i ++) {
for(var j = 0; j < m; j ++) {
var index = (i * m) + j;
if(i == 0) {
if(obstacleGrid[i][j] == 1 || (j > 0 && memo[index -1] == 0))
memo[index] = 0;
else
memo[index] = 1;
}
else if(j == 0) {
if(obstacleGrid[i][j] == 1 || (i > 0 && memo[index - m] == 0))
memo[index] = 0;
else
memo[index] = 1;
}
else {
var up = index - m;
var left = index - 1;
if(obstacleGrid[i][j] == 1)
memo[index] = 0;
else
memo[index] = memo[up] + memo[left];
}
row += memo[index] ? 0 : 1;
}
if(row == m)
return 0;
row = 0;
}
return memo[memo.length - 1];
};

var grid = [
[0,0,0],
[0,1,0],
[0,0,0]
];

console.log(uniquePathsWithObstacles(grid));