JavaScript Algorithms - 17 - Binary Search Solution

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Binary Search Solution
JavaScript Algorithms
Algorithms with JavaScript

[ilyesilyes]

best algorithms course so far please do more problem solving playlists

[stevenedwards]

I think it's useful to include a line that converts whatever array you passed into this into a sorted array, to account for the user possibly entering in a nonsorted array. Something like "arr.sort((a, b) => a - b);"

[daymenpollet]

Ur middle index can and should be a const since each time the loop iterates the const mid is made again.

[mamlzy]

Cant wait for RTK Query Course from you Vizwaz!.. thank you so much!

[metaltank]

I guess my below solution is BigO- O(1) and even if the array is not sorted it is still returning the index,
Solution :-
function binaryArray(arr, target){
let oddNum = arr.length
if(oddNum % 2 !== 0){
for (let i = 0; i < arr.length; i++) {
if(arr.includes(target)){
let center = Math.ceil(arr.length/2) - 1
let arrayCenter = arr[center]

if(arrayCenter > target){
if(arr[i] === target) return i
}

if(arrayCenter === target){
return arrayCenter
}

if(arrayCenter < target){
if(arr[i] === target) return i
}
} else{
return -1
}

}
} else {
console.log("length is even")
return -1
}
}

console.log("1st case :", binaryArray([1, 2, 6, 10, 25, 69], 10))
console.log("2nd case :", binaryArray([1, 2, 5, 10, 25], 100))
console.log("3rd case :", binaryArray([1, 2, 4, 10, 25, 78, 1000], 1000))
console.log("4th case :", binaryArray([1, 2, 4, 10, 25, 78, 1000], 10))
console.log("5th case :", binaryArray([1, 2, 4, 10, 25, 78, 1000], 78))
console.log("6th case :", binaryArray([1, 2, 4, 10, 25, 78, 1000, 56, 2000, 69552, 352455, 2585545, 25505478], 2585545))

sir why should I not use this approach of above, then please tell me, how can I improve.

[anandrajagopal]

Hi Sir,
Can I use below statement for sorting for this Binary search.

Let sort = array.sort((a, b) => a-b);

[farhannawwafal_mipa_km]

Thank you sir! This video very help me to understand a concept of algorithm :)

[BrazilDan]

You explain things clearly! Thank you!

[israamoqbel]

That was clear and helpful, thanks so much!

[Dan-ress]

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]

function binarySearch(t) {
const half = Math.floor(arr.length / 2)

if (t === half) return half-1

if (t < half) {
for (let i = 0; i < arr[half +1]; i++) {
if (arr[i] === t) {
return i
}
}
}


if (t > half) {
for (let i = half; i < arr.length; i++) {
if (arr[i] === t) {
return i
}
}
}
}

[Xraxus_]

The word 'pointer' gave me a PTSD of coding in C 😂

[shanu]

//This is a binary search but can also handle unsorted array as i'have sorted the array in the fuction. can anyone say whether it will increase time complexity or it will remain same as the solution in the video..

const binarySearch = (arr, target) => {
//sorting the array:

arrSorted = arr.sort((a, b) => a - b);
console.log(arrSorted);

let leftIndex = 0;
let rightIndex = arrSorted.length - 1;
if (arrSorted.length === 0) {
console.log("Array is empty.");
return -1;
}

while (leftIndex <= rightIndex) {
let middleIndex = Math.floor((leftIndex + rightIndex) / 2);
if (target === arrSorted[middleIndex]) {
console.log(`The index of ${target} is ${middleIndex}`);
return middleIndex;
} else if (target < arrSorted[middleIndex]) {
rightIndex = middleIndex - 1;
} else if (target > arrSorted[middleIndex]) {
leftIndex = middleIndex + 1;
}
}
console.log(`${target} is not in the array.`);
return -1;
};

console.log(binarySearch([], 7));
console.log(binarySearch([1, 2, 3, 7, 5, 6, 4], 1));
console.log(binarySearch([1, 22, 13, 4, 15, -46, 79], -46));
console.log(binarySearch([10, 62, 3, 41, 5, 66, 37], 11));

[felipeluiz]

why some right and left if left is always zero?

[gibermarregalado]

Can i ask? Its okay to use 'For Loop?

[ankushladani]

Thank You sir
Can we connect on Linkedin?

[RAVIRAJLADHA]

what if the count of the array is even, how to find the middle element?

[Ayoubmajid-uuyv]

function checkIfArrIsSortedInAsc(arr) {
for (let index in arr) {
if (arr[index] > arr[+index + 1]) return false;
}
return true;
}
function checkIfArrIsSortedInDesc(arr) {
for (let index in arr) if (arr[index] < arr[+index + 1]) return false;
return true;
}
function checkIfArrSorted(arr) {
return checkIfArrIsSortedInAsc(arr) ||
}
function binarySearch(arr, valueToFind) {
if (!checkIfArrSorted(arr)) return "Invalid Array you should sorted first : \n";
let left = 0;
let right = arr.length - 1;
let mid = (left + right) / 2;

while (left <= right) {
if (arr[mid] == valueToFind) return mid;
if {
if (arr[mid] < valueToFind) left = mid + 1;
else right = mid - 1;
} else {
if (arr[mid] > valueToFind) left = mid + 1;
else right = mid - 1;
}
mid = Math.floor((left + right) / 2);
}
return -1;
}

let arrAsc = [1, 1, 3, 4, 4, 5, 6];
let arrDesc = arrAsc.slice().reverse();
console.log(binarySearch(arrAsc, 1));
console.log(binarySearch(arrDesc, 1));

[advaithgps]

why can't we use for loop instead of while?

[ahmedthousi]

but here in this question you are not sorting

[saintthomastaquin]

What if the array is [-10, 6, 6, 6, 10] ? Index should be 1 but this algo give 2....