Task Scheduler | Easy Intuition | Using Heap | Leetcode 621 | codestorywithMIK

This is the 14th Video of our Playlist "Heap (Priority Queue) : Popular Interview Problems".

In this video we will try to solve a very good and famous heap problem : Task Scheduler | Easy Intuition | Using Heap | Leetcode 621 | codestorywithMIK
NOTE - I will make a separate video for discussing another very good Greedy Approach for this problem.
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases. 
Also, please note that my Github solution link below contains both C++ as well as JAVA code.

Problem Name : Task Scheduler | Easy Intuition | Using Heap | Leetcode 621 | codestorywithMIK
Company Tags  : META

Approach Summary : Simply use a greedy approach and get rid of tasks having max frequency so that we don't have to deal with them alone at last.Story frequency in max-heap. Start processing n+1 tasks(in descending order form max-heap) everytime and reduce the frequency of tasks by 1. If they are still remaining, push them back to heap.And keep calculating the time for processing the tasks. Once the heap is empty. Return the total time taken.
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 Timelines
00:00 - Introduction
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[gouravdutta]

Thanks sir for the awesome explanation.
Here is the Java Code for the above explanation :
class Solution {
public int leastInterval(char[] tasks, int n) {
HashMap<Character, Integer> map = new HashMap<>();
for(char ch : tasks){
map.put(ch, map.getOrDefault(ch, 0) + 1);
}
PriorityQueue<Integer> pq = new
for(Map.Entry<Character, Integer> entry : map.entrySet()){
pq.add(entry.getValue());
}
int countInterval = 0;
while(!pq.isEmpty()){
ArrayList<Integer> temp = new ArrayList<>();
for(int i=1; i<=n+1; i++){
if(!pq.isEmpty()){
int freq = pq.peek();
pq.poll();
freq--;
temp.add(freq);
}
}
for(int val : temp){
if(val > 0){
pq.add(val);
}
}
if(!pq.isEmpty()){
countInterval += n+1;
}
else{
countInterval += temp.size();
}
}
return countInterval;
}
}

[FanIQQuiz]

You are just different from other YouTubers . Loved the explanation

[vedbhanushali]

what a intuition thanks sir, never thought of that.

[rakeshgarlapati]

Awesome explanation sir, just i want to correct at the complexities part(t = tasks.size()) :
time: O(t) + O(26log26) + O(tlog26) = O(t)
space: O(n) [the worst case to store in temp vector]
thankyou

[abcdef-hce]

easiest explanation sir 👌. intuition gets into head .

[akshaychavan]

Spent almost 5 hours on this question, and couldn't figure out solution for the edge cases.
But, this solution indeed makes a lot of sense! As always, love your explaination <3

[DevOpskagyaan]

Mind blowing. Your videos exactly contain what i am always looking for

[inderlalchandani]

I am thinking about one more approach where a sort the frequency and use the formula if n<(length-i) then t+= arr[i]*(length-i) else t+= arr[i]*(length-i) + (n-i)*arr[i] by traversing the loop and decreasing prefix sum by adding arr[i] whichever next element I will traverse is it correct

[ManishKumar-kbwx]

As always bhaiyaa Amazing Explanation ✨

[lostcyrus]

Good explanation as always. Please upload the gfg question also maximum paths in tree

[gauravbanerjee]

Thanks a lot bhaiya ❤❤ & congrats for 50k subs 🥳🥳

[abhisheksinghdangi]

greeedy wala method bhi btao bhaiya, wo bhi seekhna h

[ajinkyakhete]

Good Explanation bhai also provide greedy 😄

[mohdalizilani]

could u upload soln of gfg potd, similar type of problem came in my OT Thoughspot but i was unable to solve. problem name is : possible paths in a tree

[TasniaTahsinNujhat]

hlw, can you please lemme know when there is no element to store we store an idle time in the vector/array list, where is exactly the logic used in the code to store that? I cant get that thing