'Squaring the base and taking half of the exponential' - You summed up the algorithm very well :)

Nice explanation. However there's one edge case added on this LC problem. If given number is negative then take the absolute value but store this as long to avoid overflow. At the end return accordingly.

Thank you soo much sir. Amazing Explanation🙌

Great Explanation!

Nice thanks for sharing

great explanation;

Great!

how can double hold big value in "ans" variable? it will overflow

can you please tell me which software are you using in this video to teach .

Does anyone took techdose DSA course 0:31? How much it costs? What is timings of live classes?

@jawwadakhter5261

Жыл бұрын

Bro techdose ke playlist me videos hai, usme dekho sab bataya hai

nice explanation

@techdose4u

Жыл бұрын

Thanks and welcome

'''class Solution { public double myPow(double x, int n) { //exponentiation double result=1; int m=Math.abs(n); while(m>=1){ if(m%2==1){ result=result*x; } x=x*x; m=m/2; } return (n

@kidoo1567

10 ай бұрын

Use long data type..

@bibhabpanda

10 ай бұрын

after result=result*x; write m=m-1;

@developer_save

9 ай бұрын

// here is the correct one class Solution { public double myPow(double x, int n) { double result = 1.0; long m = Math.abs((long) n); // Convert n to long to handle Integer.MIN_VALUE while (m > 0) { if (m % 2 == 1) { result *= x; } x *= x; m /= 2; } return (n } } //The issue is with the line m = m / 2;. In the exponentiation by squaring algorithm, you typically divide the exponent n by 2 in each iteration. However, in your code, you are modifying m instead of n. As a result, the algorithm won't work correctly for negative exponents.