Count Ways to Build Good Strings - Leetcode 2466 - Python
HTML-код
- Опубликовано: 21 июл 2024
- Solving Leetcode 2466 - Count Ways to build Good strings, today's daily leetcode problem on May 12.
🚀 neetcode.io/ - A better way to prepare for Coding Interviews
🥷 Discord: / discord
🐦 Twitter: / neetcode1
🐮 Support the channel: / neetcode
⭐ BLIND-75 PLAYLIST: • Two Sum - Leetcode 1 -...
💡 DYNAMIC PROGRAMMING PLAYLIST: • House Robber - Leetco...
Problem Link: leetcode.com/problems/count-w...
0:00 - Read the problem
0:30 - Intuition
2:50 - Explaining Decision Tree
7:48 - Coding Memoization
10:40 - Coding Tabulation
leetcode 2466
#neetcode #leetcode #python
3:18 Ans is "NO", it is length of string not the string, so it can be different. "000" and "100" result can be different;
regardless of previous string formulation as long as they have the same length, the decision is the same, either choose 0 or 1, that's why caches work here.
Can we solve using a combinatorics approach ?
thanks for the daily
Can we do this problem using brute force and applying permutation and combination. Because in this problem all we have to find is the combination of the good string?
If you go backwards in the iterative solution, the result is dp[0]. However, you can't just set dp[high] = 1 and keep everything else 0.
Also, in the general case, max(dp[low:high + 1]) might be > 1.
i think my mind broke when you went from backtracking to linked list lol
Thank you so much man.
What's the time complexity?
This was a really good question
one of the test cases blows up the javascript stack even with memoization. low is 100000, high is 100000, zero is 2 and one is 8.
spent an hour with permutation approach, and it turned out to be so simple, muahh %(
Bro is so emotional when explaining the problem.
Thank you!
Great exoplanation as usual-
java solution
class Solution {
Integer[] dp = null;
int mod= 1000000007;
private int helper(int low, int high, int zero, int one, int currLen){
if(currLen>high) return 0;
if(dp[currLen]!=null) return dp[currLen];
int res=0;
// if within range we wll include this string
if(currLen>=low){
res=1;
}else{
res=0;
}
res+= helper(low,high,zero,one,currLen+zero)%mod
+helper(low,high,zero,one,currLen+one)%mod;
dp[currLen]=res%mod;
return dp[currLen];
}
public int countGoodStrings(int low, int high, int zero, int one) {
dp=new Integer[high+1];
return helper(low,high,zero,one,0);
}
}
Thank you so much
Hey NeetCode or anyone reading this, I really need your help I cannot find single resource for good explanation of this problem:
2673. Make Costs of Paths Equal in a Binary Tree (It has no editorial on Leetcode)
Everyone has almost used the same solution
int minIncrements(int n, vector& cost) {
int res = 0;
function dfs = [&](int i) {
if (i >= cost.size()) return 0;
int a = dfs(2 * i + 1), b = dfs(2 * i + 2);
res += abs(a - b);
return cost[i] + max(a, b);
};
dfs(0);
return res;
}
But NOOOOO one explains clearly whyy do we have to return cost[i] + max(a, b). Any help from anyone in the comments section will be appreciated.
You explained it really well but the code in python is very different when compared to java or c++
This is my code in java ( tabulation)
class Solution {
int mod = 1000000007;
int[] dp ;
public int countGoodStrings(int low, int high, int zero, int one) {
dp = new int[high + Math.max(high , low) + 5];
// filling base case array
for(int i = 0 ; i < high + 5 ; i++){
if(i >= low && i high) dp[i] = 0;
}
for(int i = high ; i >= 0 ; i--){
int left = dp[i+ zero] % mod;
int right = dp[i + one] % mod;
dp[i] = dp[i]+ (left+ right) % mod;
}
return dp[0];
}
}
Java Solution using HashMap:
public int countGoodStrings(int low, int high, int zero, int one) {
int mod = (int) 1e9+7;
HashMap dp = new HashMap();
for(int i=high; i>=0; --i){
if(i>=low) dp.put(i, (1+dp.getOrDefault(i+zero, 0)+dp.getOrDefault(i+one, 0))%mod);
else dp.put(i, (dp.getOrDefault(i+zero, 0)+dp.getOrDefault(i+one, 0))%mod);
}
return dp.get(0);
}
1st comment and big fan
bruh say feel good until tmr xDD
do you know that fb doesn't ask DP at all? stop doing these fake thumbnails pls