Rotate Array - Leetcode 189 - Python

0:00 - Read the problem
1:35 - Drawing Explanation
6:33 - Coding Explanation

leetcode 189

#rotate #array #python
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[Pakoda-hcej]

def rotate(nums, k):
def rev(l, r):
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l +=1
r -=1

rev(0, len(nums) - 1)
rev(0, k-1)
rev(k, len(nums)-1)
return nums

# saving you all to write while loop three times

[liegon]

Thanks you, I was stuck trying to do the O(1) solution. Great explanation!

[happyhacker]

What a great solution and explanation! I have nothing to say but "thank you, man”！

[davidm]

I got stuck at 37/38 cases and I hit time limit exceeded. I was trying to do O(1) space. Your approach was so different than mine 😂 this one had me stumped

[sidthakur]

Thanks for the explanation :)

Also appreciate the chapters in the vid! Makes it super easy to navigate

[MsSkip]

Great explanation again :) Though I can't keep myself not thinking, how could I think this and code within 30 minutes?

[fredlala]

This is the best I found, very clear explanation, thank you!

[Mukeshsainaidu]

My bruteforce approach:


class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
for i in range(k):
l = nums[len(nums)-1]
nums.pop()
nums.insert(0, l)

[bartekbinda]

I think this solution is even easier
k = k % len(nums)
if k != 0:
nums[:] = nums[-k:] + nums[:len(nums)-k]

You can just slice last k elem and first len(nums)-k elems
It's also O(n) in time but takes O(n) space, although leet code shows the same amount of space for both (maybe because of caching?)

[RanjuRao]

Cannot thank enough for all your videos!

[yusufadel]

Updated versrion with helper function >> rev()
```
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
k = k % len(nums)

def rev(l=0, r=len(nums) - 1):
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l, r = l + 1, r -1
rev()
rev(r=k-1)
rev(l=k)
```

[chandravijaya]

Thank you, sir; I learned a lot from your tutorials.

[yazoncui]

There is a second way to do it. Let's say you have 1, 2, 3, 4, 5. and k = 2. Step 1. move index 0 number to k+0 position(2). the array now will be _, 2, 1, 4, 5 and you use a temp int to store the index k+i number which is 3. Step 2.instead of moving index 1 number, you move k+2 number. u shift 3 to 3+k position. now you temp int will be 5. and array will be _, 2, 1, 4, 3. With 3 more steps you will get your result. You are only using 1 temp integer so the space is O(1) and the time is O(n)

[dnyanesh.s]

Pop and insert using a for loop. Done😌

[hemantsikarwar]

in the bound condition : k=k%len(nums), you said if k is greater than the len(nums) then this condition to be applied but since we haven't made that condition prior so how is k not changing for cases where k is smaller than the len(nums); can somebody explain me this, thankyou!

[gihan]

swift version -
func rotate(_ nums: inout [Int], _ k: Int) {
var nItems = nums.count
var k = k % nItems
var targetSlice = nums[nItems - k..<nItems] + nums[0..<nItems - k]
nums = Array(targetSlice)


}

[KunalVerma-gipq]

for space O(n), the shifting formula is, ans[i] = nums[(i + size - k)%size];

[priyamf]

Came up with another soln this too is a constant space soln and works for rotating the array to the left too for some problems (just in case) by slight changes

code -

public static int[] nums, int k){
int n = nums.length;
k=k%n;
int[] temp = new int[k];
int m=0;
for(int i=n-k;i<n;i++){
temp[m] = nums[i];
m++;
}

for(int i=n-1;i>=k;i--){
nums[i] = nums[i-k];
}

int j=0;
for (int i = 0; i < k; i++) {
nums[i] = temp[i];
}
return nums;
}

k=k%n was wild throwing a runtime exception lol

[amogchandrashekar]

Nice explaination as always!
Requesting you to please solve cherry pickup, stone game dp solution or bitmasking dp questions.

[thamaraiselvan]

i did this.
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
while k >0:
ele = nums.pop()
nums.insert(0, ele)
k -= 1