Sign of An Array - Leetcode 1822

Sign of An Array - Leetcode 1822 - Python

Рет қаралды 6,865

Күн бұрын

Пікірлер: 28

@DeathSugar Жыл бұрын

You just can set initial value to 1 and flip it's sign every time you meet negative and just return it at the end without conditions

@abisheknair3545 Жыл бұрын

n=1 for i in nums: if i==0: return 0 if i

@sabu4539 8 ай бұрын

works fine - var arraySign = function(nums) { let product = 1; for(let i = 0; i < nums.length; i++){ product *= nums[i] / 2; if(product == 0) return 0; } return (product > 0) ? 1 : -1 };

@youknownothing_ Жыл бұрын

well i came up with this solution. i had also used kind of the same approach. class Solution { public int arraySign(int[] nums) { int res = 1; for(int n: nums) { if(n == 0) return 0; if(n

@steeve1 Жыл бұрын

literally the exact code I came up with. Finally feel like I am making progress. #feelsgoodman

@NeetCodeIO Жыл бұрын

Nice job 👏

@carlmascarenhas4124 Жыл бұрын

same, i came up with the solution on my own 🤯🤯🥲🥲

@flatmapper Жыл бұрын

Just class Solution: def arraySign(self, nums): result = 1 for num in nums: if num == 0: return 0 if num < 0: result *= -1 return result

@d.h.y Жыл бұрын

The channel we've needed.

@BonerPauler Жыл бұрын

my approach would be why dont we loop a function through each element and determine its sign that calculates number of negatives found in the array the we can deducethe sign of the array...there will a lot of function calls but if we could keep it short i think thatsbetter then multiplying 6:10

@Sibearian_ Жыл бұрын

U could also do a true or false for neg 6:36. So if we encounter a negative number we can negate it.

@ericnl Жыл бұрын

yep, that's what i did too!

@theOPindian Жыл бұрын

leetcode is bugged rn, everyone is getting a 5% solution lol

@jayjanodia9445 Жыл бұрын

Oh wow and here I was thinking it was due to my slow internet conmection

@novascotia2015 Жыл бұрын

class Solution { public int arraySign(int[] nums) { int sign = 1; // Initialize the sign to positive for(int num : nums) { if(num == 0) { return 0; // If the element is 0, the product is 0 } else if(num < 0) { sign ^= 1; // Flip the sign bit if the element is negative } } return sign * 2 - 1; // Convert the sign bit to a sign (-1 or 1) } }

@BurhanAijaz Жыл бұрын

my code: class Solution: def arraySign(self, nums: List[int]) -> int: pro=1 for i in nums: if i==0: return i elif i

@shrit1401 Жыл бұрын

isnt the problem saying to create a function, signFunc()?

@animeshdubey941 2 ай бұрын

Thanks a lot.

@arshalakhan567 Жыл бұрын

Hey! Thanks for amazing content. Any idea when hiring will start at FAANG?

@novascotia2015 Жыл бұрын

class Solution { public int arraySign(int[] nums) { int negativeCount = 0; // Initialize the count of negative numbers to 0 for(int num : nums) { if(num == 0) { return 0; // If the element is 0, the product is 0 } else if(num < 0) { negativeCount++; // Increment the count if the element is negative } } if(negativeCount % 2 == 0) { return 1; // Return 1 if the number of negative numbers is even } else { return -1; // Return -1 if the number of negative numbers is odd } } }