JavaScript - Position of the rightmost set bit of a number
JavaScript Bit Manipulation: Exercise-12 with Solution
Write a JavaScript program to find the position of the rightmost set bit of a given number. Set bits in a binary number is represented by 1.
Test Data:
(34) -> 2
Explanation:
34 in binary is 100010
Set bits in the 2nd position of the said binary format.
(104) -> 4
Explanation:
104 in binary is 1101000
Set bits in the 4th position of the said binary format.
Sample Solution:
JavaScript Code:
const turn_On_Kth_Bit = (n, k) => {
if (typeof n!= "number") {
return 'It must be number!'
}
if ((n & 1) != 0) {
return 1;
}
n = n ^ (n & (n - 1));
pos = 0;
while (n != 0)
{
n = n >> 1;
pos++;
}
return pos;
}
n = 34
console.log(n + " in binary is " + n.toString(2))
position = turn_On_Kth_Bit(n);
console.log("Position of the rightmost set bit of the said number: " + position)
n = 104
console.log(n + " in binary is " + n.toString(2))
position = turn_On_Kth_Bit(n);
console.log("Position of the rightmost set bit of the said number: " + position)
Sample Output:
34 in binary is 100010 Position of the rightmost set bit of the said number: 2 104 in binary is 1101000 Position of the rightmost set bit of the said number: 4
Flowchart:

Live Demo:
See the Pen javascript-bit-manipulation-exercise-12 by w3resource (@w3resource) on CodePen.
* To run the code mouse over on Result panel and click on 'RERUN' button.*
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JavaScript: Tips of the Day
Returns the symmetric difference between two arrays, after applying the provided function to each array element of both
Example:
const tips_symmetricDifference = (x, y, fn) => { const sA = new Set(x.map(v => fn(v))), sB = new Set(y.map(v => fn(v))); return [...x.filter(x => !sB.has(fn(x))), ...y.filter(x => !sA.has(fn(x)))]; }; console.log(tips_symmetricDifference([3.5, 5.5], [5.5, 7.5], Math.floor));
Output:
[3.5, 7.5]
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