L5. Power Exponentiation | Maths Playlist

Notes/Codes/Problem links under day 8 of A2Z DSA Course: takeuforward.org/strivers-a2z...
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[@pardhi8959]

This man is the top person in youtube to provide high quality content

[@tusharmahajan7754]

for overflow scenario's just use an unsigned right shift n>>>=1 is equal to n/2. why unsigned not signed right shift let say in case of number like -2147483648 if u take an abs it will -2147483648 same number again and then if you divide by 2 or signed right shift this number will be going forever negative, reason in signed right shift it replaces left vacated bits with 1 eventually you number will become -1111111....11111 in form of bits and forever loop, in this in case of unsigned right shift >>> it will fill vacated left values to zero so it will never overflow and eventually become 0 all the other code keep same make n = n/2 to n>>>=1 you will cover all cases :)

[@trailblazer555]

I'm improving my logical thinking for problem solving by your teaching only.

[@stith_pragya]

UNDERSTOOD....Thank You So Much for this wonderful video................🙏🏻🙏🏻🙏🏻🙏🏻

[@kunal7433]

Brother please next string playlist if possible 🙂

[@studentchampion718]

please also complete String Playlist.(This topic is more import for third college placement and it is more demanding topic for all guys.

[@drishtirai864]

For LEETCODE : (Covering all the Corner Cases)
class Solution {
public double myPow(double x, int n) {
double ans = 1;
double oriNum = n;
if(x == 0 || x == 1) return x;
if(n < 0){
x = 1/x;
n = -(n+1); //for Integer.MIN_VALUE
ans = ans * x;
}

while(n > 0) {
if(n % 2 == 1) {
ans = ans * x;
n = n-1;
}
else {
n = n/2;
x = x * x;
}
}
return ans;
}
}

[@clanguage7730]

int a,x; // a is base and x is power
cin>>a>>x;
int ans =1;

while(x>0){
if(x&1)ans = ans*a;
a = a*a;
x>>=1;
}
cout

[@moneyonline5299]

class Solution {
public:
double myPow(double x, int n) {
int pow = abs(n);
double ans = 1;
while(pow > 0)
{
if(pow %2 == 0)
{
x=x*x;
pow /=2;
}
else
{
ans *=x;
pow = pow -1;
}
}
if(n < 0)
ans = 1/ans;

return ans;
}
};

[@reddygopichand2002]

Understood ❤

[@ArpitaChoudhury_00]

Thank_You✨

[@AkshitChaudhary-vx8iw]

Here is the Code for the negative power : public static double pow(int x, int n) {
// Handle negative exponent
if (n < 0) {
// Calculate positive exponent result
return 1.0 / pow(x, -n);
}

int result = 1;
while (n > 0) {
if (n % 2 == 1) {
result *= x;
n--;
} else {
n /= 2;
x *= x;
}
}

return result;
}

[@saisardesai5548]

x and result should be double too

[@abhinavnarula7300]

I have a small doubt, why does the last approach work, if someone can provide the intitution it will be super helpful.

[@abhinanda7049]

understood

[@dhruvvekariya2342]

Can you discuss about corner case

[@user-co9ir1wg6y]

Understood

[@AkshitChaudhary-vx8iw]

@striver Code will not work for negative power! BTW Thanks sir for the video❤

[@Adarsh_agrahari]

❤❤

[@hritikminmuley1397]

// for posititve/negative powers
public static double exponent(int x, int n) {
double ans = 1;
int m = n;
if(n 0) {
if (n % 2 == 1) {
ans = ans * x;
n = n - 1;
} else {
n = n / 2;
x = x * x;
}
}
if(m

[@ryuu5768]

bro isme int overflow hoga if we multiply n*-1

[@hritikminmuley1397]

@@ryuu5768 That is for converting the power to positive.

[@ryuu5768]

@@hritikminmuley1397 hn but wo out of int limit hojyega for a test case for n=-214........ Something

[@ryuu5768]

Bhai wahi if negative interger h to -2147 ..... P positive krne p int flow hora

[@user-kp1xu3mp2j]

then store it in a long

[@sysfailureexe6038]

US

[@deepalikumari5319]

Code for Negative powers is not running.please help

[@AkshitChaudhary-vx8iw]

yes because our while will never run because n is negative so our code will simply return 1/ ans (which is 1);

[@deepalikumari5319]

@@AkshitChaudhary-vx8iw how can we fix it?

[@bishalkundu7592]

@@deepalikumari5319 return 1 / pow(x, -n)

[@Manish-rr3nc]

Bhai code kaha hai striver A - Z sheet mein lec 4 mein toh nhi dikh raha please batado

[@AkshitChaudhary-vx8iw]

Here is the Code for the negative power : public static double pow(int x, int n) {
// Handle negative exponent
if (n < 0) {
// Calculate positive exponent result
return 1.0 / pow(x, -n);
}

int result = 1;
while (n > 0) {
if (n % 2 == 1) {
result *= x;
n--;
} else {
n /= 2;
x *= x;
}
}

return result;
}

[@chiragbansod8252]

understood

[@hardikpatel352]

understood

[@shaiksoofi3741]

understood

[@havefunwithshort]

understood