DAY 204 - Cutting Binary String | DP, Strings , BitMagic | JAVA | C++ | GFG POTD 16 May

I have tried explaining all approaches possible for today's GFG POTD question. Watch at 1.5 or 2x for a better experience
C++ and Java source code, use ctrl+f to search the question name:
DSA REPOSITORY: github.com/AkshayAnil1080/DSA
Methods:
1.O(str(len)^2), O(str(len))
Code "AKSHAYS10" for 10% off at GFG courses:
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Hello, I am Akshay.
I am a Software Engineer at @Brillio. I have made this channel to help all ready to learn, grow, and do something big in Tech. I create content on Data Structures and Algorithms, to fine-tune the problem-solving skills that will help students and software engineers.
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[@HemantKumar-mn7bv]

In this question, the constraint is given that the length of string

[@devennandapurkar7786]

DP ka solution suchjta hi nahi hai yaar 😭😭😭😭

[@AkshayAnil0-1]

Krte rho ....aa jaega.... Dp is an advanced dsa topic. Gfg sde sheet ka dp questions solve kr lo . Ho jaega. It will take time

[@Shujaathullakhan]

im guessing TC: O(n*|left String Size|+log5(n))
bool isPowfive(string s){
long long num=0;
for(int i=0; i1){
if(num%5!=0)return false;
num/=5;
}
return true;
}
int cuts(string s){
//code
if(s.size()==0||s[0]=='0')return -1;
if(isPowfive(s))return 1;
int minCut=INT_MAX;
for(int i=1; i

[@pushanmukhopadhyay3379]

public class Solution {
public boolean isPow5(long n) {
if (n == 0)
return false;
while (n > 1) {
if (n % 5 != 0)
return false;
n /= 5;
}
return true;
}
public int func(long sum, int j, String s) {
for (; j < s.length(); j++) {
sum *= 2;
sum += (s.charAt(j) == '1') ? 1 : 0;
if (isPow5(sum)) {
int x = func(sum, j + 1, s);
int y = 1 + func(0, j + 1, s);
return Math.min(x, y);
}
if (sum == 0)
return 100;
}
if (sum == 0)
return 0;
if (isPow5(sum))
return 1;
return 100;
}
public int cuts(String s) {
int temp = func(0, 0, s);
if (temp > 50)
return -1;
return temp;
}
}

[@444yoshitha8]

it is only valid for even-length string right??