Count the Number of Powerful Integers | Digit DP | Hard Cases Problem

In this video, I'll talk about how to solve Leetcode Count the Number of Powerful Integers
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[@ARYANMITTAL]

Considering the number of solutions, this problem got, i don't think, this was that easy, that it got 500 AC's, May be Constraints are the cause of it 🥶

[@bishalkundu7592]

Mass cheating 😂 bhaiyaa Ye toh har roz ka baat hai aj kaal

[@sukhpreetsingh5200]

4th que that suffix matching was quite diff

[@harikrushnasuhagiya3925]

Thanks bhai

[@phoddaal7130]

Bhaiya Leading Zeroes wali cheese isme kyon use Karne ki zarurat nahi ho rhi ?

[@gauravkabdwal309]

Great!!

[@SanjayKumar-iu7rq]

for num = 845, and limit = 6,
According to your code
for(int i=0;i

[@yashvij21]

He is wrong there actually it should not pass as tight for 6 . Ex: If maximum number is 8452 with suffix s="52" then 6652 is also valid answer. His code works right but in explanation he made a mistake.

[@vanshbaghel5884]

Bro please don't use red color, it's not lesser visible than others 😁
Rest is good
It is not working for java code:
Code:
static long[][] dp = new long[20][2];
public static long numberOfPowerfulInt(long start, long finish, int limit, String s) {
String r = Long.toString(finish);
String l = Long.toString(start - 1);
for (long[] row: dp) Arrays.fill(row, -1);
long a = solve(r, s, 1, r.length(), limit);
for (long[] row: dp) Arrays.fill(row, -1);
long b = solve(l, s, 1, l.length(), limit);
return a - b;
}
private static long solve(String num, String s, int tight, int n, int limit) {
if (num.length() < s.length()) return 0;
if (dp[n][tight] != -1) return dp[n][tight];
// If tight hai, matlab if end is 3640 and current num is 3, then further nums cannot be greater than 640, else limit k upar hojayega, which will make it invalid. Therefore ub set krdo, ki iske upar nhi jaa skte, iske neeche hee raho.
int ub = tight == 1 ? num.charAt(num.length() - n) - '0' : limit;
long ans = 0;
// Matlab, last number of num hua, toh check with s
if (n == 1) {
// last char check karo
int lastCharOfS = s.charAt(s.length() - 1) - '0';
// Agar ub se greater hai apna nums ka last number, tab 0 return krdo. EG:
if (lastCharOfS > ub) return 0;
return 1;
}
// Means s se peeche k numbers hai, toh limit se chhota hona chahiye, buss itna dhyaan rakho
if (n ub) return 0;
else if (correspondingSChar == ub) {
ans = ans + solve(num, s, 1, n - 1, limit);
return ans;
// Means if corresponding s se smaller hai, ie,
} else return 1;
} else return 1;
} else {
for (int i = 0; i