## 991. Broken Calculator

On a broken calculator that has a number showing on its display, we can perform two operations:

Double: Multiply the number on the display by 2, or;
Decrement: Subtract 1 from the number on the display.
Initially, the calculator is displaying the number X.

Return the minimum number of operations needed to display the number Y.

Example 1:

``````Input: X = 2, Y = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
``````

Example 2:

``````Input: X = 5, Y = 8
Output: 2
``````

Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

``````Input: X = 3, Y = 10
Output: 3
``````

Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Example 4:

``````Input: X = 1024, Y = 1
Output: 1023
``````

Explanation: Use decrement operations 1023 times.

Note:

1 <= X <= 109
1 <= Y <= 109

``````class Solution {

/**
* @param Integer \$x
* @param Integer \$y
* @return Integer
*/
function brokenCalc(\$x, \$y) {
\$c = 0;
while(\$x < \$y){
if(\$y % 2 == 0){
\$y /= 2;
} else {
\$y ++;
}
\$c ++;
}
return \$c + \$x - \$y;
}
}
``````

## 884. Uncommon Words from Two Sentences

We are given two sentences `A` and `B`.  (A sentence is a string of space separated words.  Each word consists only of lowercase letters.)

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Return a list of all uncommon words.

You may return the list in any order.

Example 1:

```Input: A = "this apple is sweet", B = "this apple is sour"
Output: ["sweet","sour"]
```

Example 2:

```Input: A = "apple apple", B = "banana"
Output: ["banana"]
```

Note:

1. 0 <= A.length <= 200
2. 0 <= B.length <= 200
3. A and B both contain only spaces and lowercase letters.
Runtime: 8 ms
Memory Usage: 14.9 MB
```class Solution {

/**
* @param String \$A
* @param String \$B
* @return String[]
*/
function uncommonFromSentences(\$A, \$B) {
\$x = []; \$y = [];
foreach(explode(' ',\$A) as \$aa) {
}
foreach(explode(' ',\$B) as \$bb) {
}
\$this->rmCommon(\$x, \$y);
\$x = array_keys(\$x);
\$y = array_keys(\$y);
return array_merge(array_diff(\$x,\$y), array_diff(\$y, \$x));
}

if(isset(\$arr[\$v])){
\$arr[\$v] ++;
} else {
\$arr[\$v] = 1;
}
}

function rmCommon(&\$a,&\$b){
foreach(\$a as \$k => \$v){
if (\$v > 1){
unset (\$a[\$k]);
unset (\$b[\$k]);
}
}
foreach(\$b as \$k => \$v){
if (\$v > 1){
unset (\$b[\$k]);
unset (\$a[\$k]);
}
}
}
}```

## 994. Rotting Oranges

In a given grid, each cell can have one of three values:

the value 0 representing an empty cell;
the value 1 representing a fresh orange;
the value 2 representing a rotten orange.
Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1 instead.

Example 1:

Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
Example 2:

Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
Example 3:

Input: [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.

Note:

1 <= grid.length <= 10
1 <= grid[0].length <= 10
grid[i][j] is only 0, 1, or 2.

24 ms 38.6 MB

```class Solution {
public int orangesRotting(int[][] grid) {
if(this.nothing(grid)){
return 0;
}
if(this.neverR(grid)){
return -1;
}

int count = 0;
while(!this.isAllR(grid)){
grid = this.infect(grid);

count ++;
if(count > grid.length * grid[0].length){
return -1;
}
}
return count;
}

public int[][] infect(int[][] grid){
int[][] next = new int[grid.length][grid[0].length];
for(int x  = 0; x < grid.length; x ++){
for(int y = 0; y < grid[0].length; y++){
next[x][y] = grid[x][y];
}
}
for(int x  = 0; x < grid.length; x ++){
for(int y = 0; y < grid[0].length; y++){
if(grid[x][y] == 2){
if(x>0 && next[x-1][y] > 0){
next[x-1][y] = 2;
}
if(y>0 && next[x][y-1] > 0){
next[x][y-1] = 2;
}
if(x< grid.length-1 && next[x+1][y] > 0){
next[x+1][y] = 2;
}
if(y < grid[0].length - 1 && next[x][y+1] > 0){
next[x][y+1] = 2;
}
}
}
}
return next;
}

public boolean isAllR(int[][] grid){
boolean allR = true;
for (int[] col : grid){
for (int c : col){
if(c == 1){
allR = false;
}
}
}
return allR;
}

public boolean neverR(int[][] grid){
boolean allFresh = true;
for(int x  = 0; x < grid.length; x ++){
for(int y = 0; y < grid[0].length; y++){

if(grid[x][y] == 2){
allFresh = false;
}

if(grid[x][y] == 1){

boolean alone = true;
if(x>0 && grid[x-1][y] != 0){
alone = false;
}
if(y>0 && grid[x][y-1] != 0){
alone = false;
}
if(x< grid.length-1 && grid[x+1][y] != 0){
alone = false;
}
if(y < grid[0].length - 1 && grid[x][y+1] != 0){
alone = false;
}
System.out.println(alone);
if(alone){
return true;
}
}
}
}
return allFresh;
}

public boolean nothing(int[][] grid){
boolean nothing = true;
for(int x  = 0; x < grid.length; x ++){
for(int y = 0; y < grid[0].length; y++){

if(grid[x][y] > 0){
nothing = false;
}

}
}
return nothing;
}
}```

## 996. Number of Squareful Arrays

Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square.

Return the number of permutations of A that are squareful. Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].

Example 1:

``````Output: 2
Explanation:
[1,8,17] and [17,8,1] are the valid permutations.
``````

Example 2:

``````Input: [2,2,2]
Output: 1
``````

Note:

``````1 <= A.length <= 12
0 <= A[i] <= 1e9
``````

Solved this with PHP.

1. Go through the input and construct array ‘peers’
1. peers holds an array of pointers pointing to all possible ‘partners’ in the input.
2. For example, `[1 => [8=>ref(8)], 8 => [1=>ref(1), 17=>ref(17)], 17 => [8=>ref(8)]]`
2. Meanwhile, another array stat collect all numbers and its occurrence.
3. Then peers is recursive. so it’s like a tree, with infinite depth.
4. Starting from the root of peers, if found any number in ‘stat’, update (-1) stat, and move on to peers’ children
1. if cannot find any match at some points, break;
2. since stat is finite, the recursive func will run forever
5. Calculate the sum.

Works well with even larger input length.

16ms runtime w/ 14.7M RAM

```Class Solution {

/**
* @param Integer[] \$ls
* @return Integer
*/
function numSquarefulPerms(\$ls) {

if(count(\$ls) <= 1){
return 0;
}

\$peers = [];
\$stat = [];

for(\$i = 0; \$i < count(\$ls); \$i++){
// Insert/Update stat
if(!isset(\$stat[\$ls[\$i]])){
\$stat[\$ls[\$i]] = 1;
} else {
\$stat[\$ls[\$i]] ++;
}

for(\$j = \$i + 1; \$j < count(\$ls); \$j++){

\$value_i = \$ls[\$i];
\$value_j = \$ls[\$j];

if(!isset(\$peers[\$value_i])){
\$peers[\$value_i] = [];
}
if(!isset(\$peers[\$value_j])){
\$peers[\$value_j] = [];
}
if(\$this -> isPerfectSquare(\$ls[\$i] + \$ls[\$j])){
\$peers[\$value_i][\$value_j] = &\$peers[\$value_j];
if(\$ls[\$j] !== \$ls[\$i]){
\$peers[\$value_j][\$value_i] = &\$peers[\$value_i];
}
}
}
}
// If any number has no partner
foreach(\$peers as \$p){
if(\$p === []){
return 0;
}
}
return \$this->traversal(\$stat, \$peers, count(\$ls));
}

// Check if number is perfect square w/ small optimization
function isPerfectSquare(\$num){
\$last_digit = \$num % 10;
if(\$last_digit == 2 ||
\$last_digit == 3 ||
\$last_digit == 7 ||
\$last_digit == 8
)
return false;

\$sqrt = sqrt(\$num);
return \$sqrt == floor(\$sqrt);
}

function traversal(\$nums, &\$tree, \$max){
\$count = 0;
var_dump(\$nums);
if(empty(\$nums)){
return 1;
}
foreach(\$tree as \$k => \$v){
// Make a copy
\$nums_copy = \$nums;
if(isset(\$nums_copy[\$k])){
// If exist decrease, if last remove
if(\$nums[\$k] > 1){
\$nums_copy[\$k] --;
} else {
unset(\$nums_copy[\$k]);
}
// move on to children
\$count += \$this -> traversal(\$nums_copy, \$v, \$max - 1);
}
}
return \$count;
}
}```

## 6. 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
``````
``````
/**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function(s, numRows) {
var data = new Array(numRows);
var up = false; // up or down? used in-between tops and bottoms

for(var i = 0, j = 0; i < s.length; i ++){
// init arr elems
if(data[j] === undefined){
data[j] = "";
}
// append cur char
data[j] += s.charAt(i);

if(j == 0){
// On top
up = false; // Change direction
j ++;
} else if(j == numRows-1){
// On bottom
up = true;  // Then go up
j --;
} // in-between, follow prev direction
else if(up){
j --;
} else {
j ++;
}
}
// Join/concat arr of str to single arr
return data.join("");
};
``````

## 5. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Output: "bab"
Note: "aba" is also a valid answer.
Example 2:

Input: "cbbd"
Output: "bb"

``````/**
* @param {string} s
* @return {string}
*/
var longestPalindrome = function(s) {\
// palindromic str for single round
var o = "";
// Longest palindromic str
var max = "";
for (var i = 0; i < s.length; i ++){
var x = i;
var y = i;
// Include current char
o = s[i];

// Collect Same letters on the right
while(s[y+1] == s[y]){
y ++; i ++;
// append same char on the right to p_str
o += s[i];
}

// Then check both sides one by one
// Make sure x & y in range, then compare both ends
while(x > 0
&& y < s.length-1
&& s[x-1] == s[y+1]){
o = s[x-1] + o + s[y+1];
// Separate x & y for one more step
x--; y++;
}

// Save, if longer than current max record
if(o.length > max.length){
max = o;
}

}

return max;
};
``````

## 4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0
Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

``````
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var findMedianSortedArrays = function(nums1, nums2) {
// remove elements from both ends

while ((nums1.length + nums2.length)/2 > 1){
// If any of the arrays is empty, rm the other one
if(nums1.length == 0){
nums2.shift();
nums2.pop();
} else if(nums2.length == 0){
nums1.shift();
nums1.pop();
} else {
if(nums1[0] < nums2[0]){
nums1.shift();
} else {
nums2.shift();
}
// if nums1 is empty
//   OR nums1's tail is less than nums2's, remove nums2
if(nums1.length == 0 ||
nums1[nums1.length -1] < nums2[nums2.length -1]){
nums2.pop();
} else {
nums1.pop();
}
}
}
// only 1 or 2 elements remain now
var r = nums1.concat(nums2);
// send average for two
if(r.length > 1){
return (r[0]+r[1])/2;
} else {
// or just the last element
return r[0];
}
};
``````

## 3. Longest Substring Without Repeating

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

``````/**
* @param {string} s
* @return {number}
*/
var lengthOfLongestSubstring = function(s) {
var max = 0;
var a = [];
for (var i = 0; i < s.length; i ++){
// If element exists, remove any other elements before it,
// and itself
if(a.includes(s[i])){
a.splice(0, a.indexOf(s[i]) + 1);
}
// Append new element
a.push(s[i]);
// Update result
if(max < a.length){
max = a.length;
}
}
return max;
};
``````