Insertion Sort List - Leetcode 147 - Python

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Problem Link: leetcode.com/problems/inserti...
0:00 - Read the problem
0:35 - Understand Insertion Sort
4:15 - Drawing Explanation
9:37 - Coding Explanation
leetcode 147
This question was identified as a microsoft interview question from here: github.com/xizhengszhang/Leet...
#microsoft #interview #python
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13 июл 2024

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Комментарии : 35

@john_youtube 2 года назад

Just got my job at Amazon and I can say without a doubt you're a major reason for that. Thank you neetcode

@__________________________6910 2 года назад

noice

@henryrussell7392 2 года назад

congrats john

@sumosam1980 2 года назад

Congratulations! That's lots of hard work and time. Best wishes going forward!

@zabona Год назад

tmp = dummy, or you will not sort first element. Dummy is not even used after it's initialized.

@ssenthilnathan3 2 года назад

I just learned insertion sort, and you posted this problem. Very Thankyou !!

@__________________________6910 2 года назад

noice

@alaaal-habashna4722 2 года назад

Thank you! This is very clear.

@danielshvartz9702 7 месяцев назад

worth to mention that prev pointer will always point to the last sorted element so when we do prev.next = curr.next we do it to save the next nodes and to continue from the last sorted node.

@nathnaeldereje5127 Год назад

Great Job, That was so helpful.

@erkhembayarbayart3966 2 года назад

nice explanation! thank u

@shidharthraj9310 2 года назад

Really helpful 😀👍

@rahulmalakar8288 22 дня назад

Beautiful explanation

@sarahbavan3959 11 месяцев назад

you are the best!!!

@poorpanda9033 10 месяцев назад

Thank You

@meme-blower Год назад

super helpful

@ashkanfb 6 месяцев назад

Even if you don't do the extra comparison with the last element, this algorithm is still O(n) best case, but this best case happens when the initial linked list is sorted in reverse order.

@vrajeshbadgujar Год назад

Shukriya !

@chendong2197 Год назад

after initializing the dummy node, which point to head. When assigning head to prev with (prev = head). Are you assigning the dummy node to prev or the old head to prev? 9th line

@cosepeter2197 Год назад

old head is assigned to prev not the dummy node. This done because it's not required to check the first node , so cur node start from the second node

@orellavie6233 2 года назад

Why not checking if temp.next is not none before actually using its .val?

@xiaoyunzhi6900 Год назад

Because cur.val is guaranteed to be smaller than prev.val, tmp will first reach prev before it reaches None

@christiangualteros36 2 года назад

hello. i have a question silly but i don't know how I can proof this example and what I need to call for print in the screem the result. thank you

@anmolbakshi7983 Год назад

start iterating from the head until it becomes null and print each node's value

@christiangualteros36 Год назад

@@anmolbakshi7983 thank you. You're a great person

@venkateshvardhineedi9571 Год назад

ListNode temp = head while(temp != null){ System.out.println(temp.val) temp = temp.next }

@Monkey45139 2 года назад

Hi it’s me again I am your biggest fan plz read this comment u r helping me so much

@AlexSav 2 года назад

11:05 Is't it crash on empty list? Line 9 (head might be None)

@venkateshvardhineedi9571 Год назад

add the condition if(head == null) return head

@wz9815 Год назад

Hello, why my code in java is not correct(Time Limit Exceeded)? ListNode dummy = head; ListNode curr = head.next; ListNode prev = head; ListNode tmp = null; while( curr != null){ if (curr.val >= prev.val){ prev = curr; curr = curr.next; continue; } tmp = head; while(curr.val > tmp.next.val){ tmp = tmp.next; } prev.next = curr.next; curr.next = tmp.next; tmp.next = curr; curr = prev; } return dummy.next;

@zabona Год назад

That's most likely because two first elements are not sorted, and because of a mistake in the video you keep inserting second element before itself. Try changing tmp initialization to tmp = dummy

@user-vu8jp3si2r Год назад

Why this code gives TLE? /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode insertionSortList(ListNode head) { ListNode helpHead = new ListNode(-9999, head); //without set pointer dummy.next is works ListNode curr = head; while (curr != null) { ListNode prev = helpHead; while(prev.next != null && prev.next.val