First Missing Positive | LeetCode 41 | C++, Java, Python | BITWISE operation

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September LeetCoding Challenge | Problem 30 | First Missing Positive | 30 September,
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[@husler7424]

I'm damn sure no better solution is available than this. Thanks a ton, sir ji! I was about to give up but found your video. Keep continue this series.

[@shubhamsamdani3503]

Hi, Did you stoped making newer videos on Leetcode problems?

[@KnowledgeCenter]

Not stopped. But, will be adding selectively.

[@yashagrawal3272]

Great explanation Sir.Thanks

[@pqsk]

Nice. I finally get it. Your visuals are perfect.

[@KnowledgeCenter]

Glad it helped.

[@ramanujachanduri2253]

Really loved your explanation among all the other CZcams channels present. I was only halfway through this video and it gave me enough confidence and clarity to solve this question by my own. Keep working on the good stuff 😄

[@satyammishra-xb3hs]

Sir , Excellent video , but sir I did not understand why do we have to do x=abs(arr[i ]) ??
As we already put 1 for the negative values In the above loop ? So why we are again checking for positive??

[@Music-tp8gg]

In between while traversing we are making the numbers -ve by multiplying with -1. So we take abs

[@satyakiranmayibhamidimarri3160]

so good!

[@KnowledgeCenter]

Thanks.

[@avinashmaurya7819]

Sir ,I write the same code as you, it ran but my program didn't submit

[@rishidhawan9323]

@Knowledge Center, It will be great if you can add leetcode problem "Word Break"

[@ankoor]

Word Break can be solved using recursion + memoization:
memo = {}
wordDict = set(wordDict)
def solve(s, wordDict):
if s in wordDict:
return True
if s in memo:
return memo[s]
n = len(s)
for i in range(1, n+1):
if (s[:i] in wordDict) and solve(s[i:], wordDict):
memo[s] = True
return True
memo[s] = False
return False
solve(s, wordDict)

[@Raj_Patel21]

Awesome

[@KnowledgeCenter]

Thanks.

[@giit85]

But it will not work if there has already negative number or 0 in array

[@shavisharma6581]

Please do make video on Leetcode Challenge September 29 - Word Break.

[@beaglesnlove580]

How the heck can you figure this out without seeing the dang solution for O(1) time. Crazy

[@fffooccc9801]

why can't we use mex here ?

[@mysteriousocean3206]

nice explanation

[@KnowledgeCenter]

Thanks.

[@joydeeprony89]

there is an issue in your logic
updated your logic, just checking the input array has 1 or not, in that case return 1 as an output
public int FirstMissingPositive(int[] nums)
{
int smallestPositive = int.MaxValue;
int length = nums.Length;
// Step 1 -
// loop entire array and mark the cells as 1 wherever cell value is N
for (int i = 0; i < length; i++)
{
if(nums[i] > 0)
{
smallestPositive = Math.Min(smallestPositive, nums[i]);
}
if (nums[i] length)
{
nums[i] = 1;
}
}
if (smallestPositive != 1)
{
return 1;
}
// Step 2 -
// loop entire array and for each cell value -1 position, update to negetive to mark that value is present in range of 1-N
for (int i = 0; i < length; i++)
{
int val = Math.Abs(nums[i]);
if(nums[val - 1] > 0)
nums[val - 1] *= -1;
}
// Step 3 -
// loop entire array, any value as positive , that position + 1 is the missing element, if not found then N+1 is the missing element(eg- (1,2,3,4,5)=>O/p-6)
for (int i = 0; i < length; i++)
{
if (nums[i] > 0) return i + 1;
}
return length + 1;
}

[@joydeeprony89]

not working for input - 7,8,9, 11, 12

[@letslive3782]

It will work as he wrote a condition to check if 1 is present or not..if 1 is not present then 1 is already returned

[@jax40001]

I think the problem really starts after 6 minutes

[@akashyadav3211]

4 for loops 😐...I did it using one while loop and one for loop.....

[@kishorbhadula7816]

I did the same way could you elaborate your approach?

[@ss-xh3hf]

why are we converting negavtives into 1 .

[@panman7]

Because we can't have negative index

[@kaleem8623]

It's very annoying questions