Algorithm

void main() {

// booo(n){

// for(var i = 0; i < n.length; i++){

// print('boo');

// }

// }

// booo([1,2,3,4,5]); // space complexity of 0(1) because we only declared var = i and not doing much in the function and we cant control what we put in as in the array

// arrayOfNTimes(n){

// List arrayHi = []; // created new array = data structure , takes space

// // var = i = variables , takes space

// for(var i = 0; i < n; i++){

// arrayHi.add('hi'); // allocation , allocated hi to arrayi , takes space

// }

// return arrayHi;

// }

// print(arrayOfNTimes(6)) ; // 0(n) so this function has space complexity of 0(n)

// find the ist and nth tweets

// compare the items using the dates

// const array =

// [{'date': 2012, 'tweet': 'hi'},

// {'date': 2013, 'tweet': 'dear'},

// {'date': 2015, 'tweet': 'make'},];

// print('jkbdsdnvgv jk. f'.length); //0(1);

//look for a pair of number in the list that give s you the sum of 8 when you add them

// check for best possible solution in terms of time and space complexity

// using linear time would be more efficient rather than nested loop (quadratic)

// List array1 = [1,2,3,9]; // NO

// List array2 = [1,2,4,4]; // YES so to check this

// int sum = 8;

// bool doesTheArrayHasPairSumEqualTo8( int sum, array){

// int low = 0;

// int high = array.length-1;

// bool thereIsSum = false;

// while(low < high){

// if(array[low] + array[high] == sum){

// thereIsSum = true;

// }

// print (array[low] + array[high] );

// low ++;

// high--;

// }

// return thereIsSum;

// }

// print(doesTheArrayHasPairSumEqualTo8(8, array2)) ;

// bool doesTheArrayHasPairSumEqualTo8( int sum, array){

// Set<int> comp = {};

// array.forEach((value ){

// if(comp.contains(value )){

// print( comp);

// return true;

// }

// comp.add((sum - value).toInt());

// });

// return false;

// }

// print(doesTheArrayHasPairSumEqualTo8(sum, array2)) ;

// Given two array create a function that lets a user know (true/false)

// whether these two array contains any common item

const array = ['a', 'b','c','x'];

const array1 = ['z','y', 'i'];

//shoould return false;

const array2 = ['a', 'b','c','x'];

const array3 = ['z','y', 'a'];

//shoould return true;

// what are the input and output

// if the interviewer said its always array the input

// how large can the input get in future

// if its always a minimal size then we dont have to worry about big o timecomplexity or space

// is our goal to be more efficient as possible, is time complexity or space complexity more important to us

// the interviewer might say they just want more efficient function. assuming the array can get very very large

// 2 parameters -- no size limit might be what the interviewer says

// talk about the easiest solution that comes to mind in this case nested loop

//also talk while is no the bbest solution in this case (0(n^2)) quadratic time complexity which we try as muchj as possible in interviews

// can we assume always two parameter

bool checkIfTheTwoListHasCommonItem(arrayX, arrayY){

// int totalLength = list1.length + list2.length;

// List<String> mergedList = arrayX + arrayY;

// List<String> firstArray = arrayX;

// print(mergedList);

bool hasCommon = false;

Set<String> comp = {};

// for(var i = 0; i < mergedList.length; i++){

// if(comp.contains(mergedList[i])){

// hasCommon = true;

// }

// comp.add( mergedList[i]);

// }

for(var i = 0; i < arrayY.length; i++){

if(arrayX.contains(arrayY[i])){

hasCommon = true;

}

// comp.add( mergedList[i]);

}

//Assigning array to object as properties this function is mostly used when trying to improve time complexity

// loop throough first array and create object where properties === items in the array

Map<String, bool> map = {};

for(var i = 0; i< arrayX.length; i++){

if(map[i] != true){

var item = arrayX[i];

map[item] = true;

}

}

print(map);

// loop through the second array and check if the item in the second array exists on created object

for(var j = 0; j< arrayY.length; j++){

if(map [arrayY[j]] == true){

hasCommon = true;

}

}

// 0(a) space complexity;

return hasCommon;

}

// print(checkIfTheTwoListHasCommonItem(array2, array3));

//find out if there is a pair that sums up to the number given

// sum = 8

//items []

void googleInterview(){

List array1 = [1,2,3,9]; // NO

List array2 = [1,2,4,4]; // YES so to check this

int sum = 8;

// naive way

bool checkIfAnyPairSumsUptoTheSum(array, sum){

for(var i = 0; i < array.length -1; i ++ ){

for(var j = 1 + i; j < array.length ; j++ ){

if(array[i] + array[j] == sum) {

print('${(array[i], array[j] )}');

return true;

}

}

}

return false;

}

print(checkIfAnyPairSumsUptoTheSum(array2, sum)) ;

// Better way

bool checkIfAnyPairSumsUptoTheSum1(array, sum){

Set<int> comp = {};

array.forEach((value ){

if(comp.contains(value )){

print( comp);

return true;

}

comp.add((sum - value).toInt());

});

return false;

}

print(checkIfAnyPairSumsUptoTheSum1(array1, sum)) ;

}

// Arrays();

}