Binary Exponentiation | Pow(x,n) | Leetcode #50

[@hiteshbhandari7009]

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

[@Sranju23]

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.

[@Martinnnnnn-e5s]

amazing! I finally understood the iterative approach

[@yogitajoshi7218]

Thank you soo much sir. Amazing Explanation🙌

[@ajaygonepuri2285]

Great Explanation!

[@chakravarthybatna1589]

great explanation;

[@rahulsharma15]

Nice thanks for sharing

[@anshumaan1024]

nice explanation

[@techdose4u]

Thanks and welcome

[@Bkmap]

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

[@franly656]

Great!

[@GameplayRoudie]

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

[@SK-qn5ry]

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

[@jawwadakhter5261]

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

[@ToanPham-wr7xe]

😮

[@prathmeshkhandare6113]

Why we not using x**n Or pow function ?

[@techdose4u]

that will multiply X n times

[@prathmeshkhandare6113]

@@techdose4u thats what we want , and also the time complexity is also less

[@btengaming4899]

where is the code of even exponent

[@techdose4u]

Please do dry run on the given code for both odd & even.

[@rahulchoudhary414]

'''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]

Use long data type..

[@bibhabpanda]

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

[@developer_save]

// 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 < 0) ? 1 / result : result; } } //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.