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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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.

husler

Really loved your explanation among all the other YouTube 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 😄

ramanujachanduri

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

pqsk

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

shubhamsamdani

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??

satyammishra-xbhs

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

beaglesnlove

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

rishidhawan

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 <= 0 or value > N
for (int i = 0; i < length; i++)
{
if(nums[i] > 0)
{
smallestPositive = Math.Min(smallestPositive, nums[i]);
}
if (nums[i] <= 0 || 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;
}

joydeeprony