JavaScript Want more resource

[Veronica]

We can use binary search to find the target element in an array. First, we need to sort the array in ascending order. Then, we can use binary search to find the target element in the sorted array. The time complexity of this approach is O(nlogn), where n is the number of elements in the Array. Came across from this Resource.

Can anyone help me with more resources?

Thanks in advance. 😊

 

[Ordep-20]

Yes most case

try this code above, if you dont want programing into HTML file you can use the JSFiddle OR Codepen its online IDE for development and tests

let recursiveFunction = function (arr, x, start, end) {
     
    // Base Condition
    if (start > end) return false;
 
    // Find the middle index
    let mid=Math.floor((start + end)/2);
 
    // Compare mid with given key x
    if (arr[mid]===x) return true;
       
    // If element at mid is greater than x,
    // search in the left half of mid
    if(arr[mid] > x)
        return recursiveFunction(arr, x, start, mid-1);
    else
 
        // If element at mid is smaller than x,
        // search in the right half of mid
        return recursiveFunction(arr, x, mid+1, end);
}
 
// Driver code
let arr = [1, 3, 5, 7, 8, 9];
let x = 5;
 
if (recursiveFunction(arr, x, 0, arr.length-1))
    window.alert("Element found!<br>");
else window.alert("Element not found!<br>");
 
x = 6;
 
if (recursiveFunction(arr, x, 0, arr.length-1))
   window.alert("Element found!<br>");
else window.alert("Element not found!<br>");
</script>

 

[Ordep-20]

Here binary

var a = [
    1,
    2,
    4,
    6,
    1,
    100,
    0,
    10000,
    3
];

a.sort(function (a, b) {
    return a - b;
});

window.alert('a,', a);

function binarySearch(arr, i) {
    var mid = Math.floor(arr.length / 2);
   window.alert(arr[mid], i);
    
    if (arr[mid] === i) {
        window.alert('match', arr[mid], i);
        return arr[mid];
    } else if (arr[mid] < i && arr.length > 1) {
        window.alert('mid lower', arr[mid], i);
        return binarySearch(arr.splice(mid, Number.MAX_VALUE), i);
    } else if (arr[mid] > i && arr.length > 1) {
        window.alert('mid higher', arr[mid], i);
        return binarySearch(arr.splice(0, mid), i);
    } else {
        window.alert('not here', i);
        return -1;
    }
    
}
var result = binarySearch(a, 100);

window.alert(result);