1089. Duplicate Zeros - LeetCode Solution

Given a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.
Note that elements beyond the length of the original array are not written.
Do the above modifications to the input array in place, do not return anything from your function.
Example 1:
Input: [1,0,2,3,0,4,5,0]
Output: null
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
LeetCode Problem: leetcode.com/p...
GitHub Solution: github.com/Raj...
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[@gsb22]

I wasted 1 day to understand the solution given on LC and got fed up. Glad I found your video. It's simple, and not confusing.

[@gameofcoders1036]

Thank you for your kind words.

[@SalviPapiii]

I want to let you know that it took me almost 3 days to figure it out since I haven't coded in a couple months... Thank you so much for this! It really helped refresh my memory a lot.

[@gameofcoders1036]

Glad that it helped you. Your comment really motivated me. Please subscribe to my channel to see my latest content.

[@gangster4069]

This solution has a problem, if your input array's last element is zero, this solution will fail. input array's last element with zero is an boundary condition, to handle it decremenet count of zeros by 1, if the last element in the array is zero.

[@nandanimadhukar]

easy explanation to what seemed like a twisted but marked "easy"!

[@satyamgupta6030]

crystal clear explaination and concise code.
thanks for the solution.

[@huseynmikayil6354]

Here is the TypeScript solution without additional insert function:
const duplicateZeros = function (nums: number[]) {
let zeroes: number = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] === 0) zeroes++;
}
let i = nums.length - 1,
j = nums.length - 1 + zeroes;
while (i !== j) {
if (j < nums.length) nums[j] = nums[i];
j--;
if (nums[i] === 0) {
if (j < nums.length) nums[j] = nums[i];
j--;
}
i--;
}
};

[@pradeepanbaluchamy9503]

Good and simple explanation. Thanks

[@MunniDivya]

can u please tell me how did u get this ideA?

[@rohanjuneja1778]

not of use. only algo.no intuition.

[@mohanaravind7656]

Sorry I might be missing something, but I see a `i--` in the code, but not a `j--` when going through the array. How is `j` decremented?

When the insert function is called, she is passing j-- as argument, the -- operator will decrement the variable value itself, not only the value being passed to the function.

[@gameofcoders1036]

Your comment really motivates me. Please subscribe to my channel to see my latest content.

[@MunniDivya]

why did u decrement j pointer when i==0 . I dint understand

[@gameofcoders1036]

I hope you figured it out. Please subscribe to my channel to see my latest content.

[@putinvladimir420]

this code fails if zero is present at the last index.

[@user-cl3un2qx7s]

please explain the maths behind your code / trick !

[@HarshitMaurya]

Best Explanation

[@saikirankorri2180]

Your explanation was good .. thank you

[@gameofcoders1036]

Glad that you liked it. Please subscribe to my channel to see my latest content.

[@joscandynunez2169]

how is this marked easy ?

[@Nature-pb9dh]

Thank you this is great!

[@himanshuchhikara4918]

very helpful solution

[@kyotang8657]

Hi there, would you please explain how you have arrive the condition for the pointer decrements, especially for the case where when the array element is zero we would copy the element to the index of j then decrement j... and give you this example [8,4,5,0,0,7]... I can’t seem to wrap my head around it using the two pointer approach

[@kyotang8657]

Never mind when I stepped through I realize that when you have consecutive Zeros, the termination point of the while loop actually happens at the index of the very next left element .... this is so smart use of two pointers as j will be decremented before i, then i gets decremented then the while loop termination condition will be met

[@gameofcoders1036]

Glad that you figured. Please subscribe to my channel to see my latest content.

[@chankwongyin7455]

very well explained!

[@janhavibhosle683]

Why didn't you shift i till first element?

[@pradeepanbaluchamy9503]

Watch it again. There she mentioned I=j and hence she stopped the shifting.

[@gameofcoders1036]

Yes, once i and j becomes equal, we know that all items are at correct places. So no computation needed.

[@nirajchowdhary7372]

because after this point i and j will always point at same element, so it would be always swapping element with itself till the first element

[@gameofcoders1036]

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[@jorgegonzaloalfaro5378]

insert operation is O(n) time complexity no? so worst case scenario would be o(n^2) . if you disagree can you explain why this is the case?

[@AdrianGonzalezBlogs]

Insert operation is O(1) since it only swaps the array elements values

[@gameofcoders1036]

Your comment really motivates me. Please subscribe to my channel to see my latest content.

[@shuchitam3868]

can you explain the run time of the solution as well?

[@gameofcoders1036]

Hi Gayatri, Time complexity for solution covered in video will be O(n), as we are traversing the array just once.

[@shuchitam3868]

@@gameofcoders1036 Thanks!

[@systemforge]

woah!.. nice

[@hhuynh8195]

your solution might work but I don't understand your explanation at all. T.T I'm sorry I try but the docs you used to explain your idea doesn't really help with visualization. Honestly it actually confuse me more.

[@MunniDivya]

is there any algorithm already present?

[@jasjeetsingh1967]

There are other algorithms as well, and they are available on different websites!

[@gameofcoders1036]

Yes that's right, you can find different algorithms from other websites.

[@gameofcoders1036]

Please subscribe to my channel to see my latest content.

[@speedbreaker5642]

Good one.

[@gameofcoders1036]

Thank you! Cheers!

[@oleksii7691]

Nice solution but awful explanation.

[@Abhishek-Khelge]

I have used ArrayList this is easier this way
List al =new ArrayList();
for(int k : arr){
al.add(k);
}

boolean flag=false;
for(int i=0;i