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15 changes: 8 additions & 7 deletions ✅ Pattern 01 : Sliding Window.md
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# Pattern 1: Sliding Window
In many problems dealing with an array (or a <b>LinkedList</b>), we are asked to find or calculate something among all the contiguous subarrays (or sublists) of a given size. For example, take a look at this problem:

### Find Averages of Sub Arrays
### Find Averages of Sub Arrays ✅
https://leetcode.com/problems/maximum-average-subarray-i/

> Given an array, find the average of all contiguous subarrays of size `K` in it.
Expand Down Expand Up @@ -206,7 +206,8 @@ smallestSubarrayWithGivenSum([3, 4, 1, 1, 6], 8)//3
- The time complexity of the above algorithm will be `O(N)`. The outer for loop runs for all elements, and the inner while loop processes each element only once; therefore, the time complexity of the algorithm will be `O(N+N)`), which is asymptotically equivalent to `O(N)`.
- The algorithm runs in constant space `O(1)`.

## Longest Substring with K Distinct Characters (medium)
## Longest Substring with K Distinct Characters (medium) ✅ [solution](https://www.geeksforgeeks.org/problems/longest-k-unique-characters-substring0853/1)

https://leetcode.com/problems/longest-substring-with-at-most-k-distinct-characters/

>Given a string, find the length of the <b>longest substring</b> in it with <b>no more than `K` distinct characters</b>.
Expand Down Expand Up @@ -265,7 +266,7 @@ longestSubstringWithKdistinct("cbbebi", 3)//5, The longest substrings with no mo
- The above algorithms time complexity will be `O(N)`, where `N` is the number of characters in the input string. The outer for loop runs for all characters, and the inner while loop processes each character only once; therefore, the time complexity of the algorithm will be `O(N+N)`, which is asymptotically equivalent to `O(N)`
- The algorithms space complexity is `O(K)`, as we will be storing a maximum of `K+1` characters in the <b>HashMap</b>.

## 🔎 Fruits into Baskets (medium)
## 🔎 Fruits into Baskets (medium) ✅
https://leetcode.com/problems/fruit-into-baskets/

> Given an array of characters where each character represents a fruit tree, you are given <b>two baskets</b>, and your goal is to put the <b>maximum number of fruits in each basket</b>. The only restriction is that <b>each basket can have only one type of fruit</b>.
Expand Down Expand Up @@ -359,7 +360,7 @@ fruitsInBaskets(['A', 'B', 'C', 'B', 'B', 'C'])//5 , We can put 3 'B' in one bas
````
- The above algorithms time complexity will be `O(N)`, where `N` is the number of characters in the input array. The outer `for` loop runs for all characters, and the inner `while` loop processes each character only once; therefore, the time complexity of the algorithm will be `O(N+N)`, which is asymptotically equivalent to `O(N)`.
- The algorithm runs in constant space `O(1)` as there can be a maximum of three types of fruits stored in the frequency map.
### Longest Substring with at most 2 distinct characters
### Longest Substring with at most 2 distinct characters ✅
https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/
> Given a string, find the length of the longest substring in it with at most two distinct characters.

Expand Down Expand Up @@ -402,7 +403,7 @@ lengthOfLongestSubstringTwoDistinct('ccaabbb')//5



## No-repeat Substring (hard)
## No-repeat Substring (hard) ✅
https://leetcode.com/problems/longest-substring-without-repeating-characters/

> Given a string, find the <b>length of the longest substring</b>, which has <b>no repeating characters</b>.
Expand Down Expand Up @@ -450,7 +451,7 @@ nonRepeatSubstring("abccde")//3
- The above algorithms time complexity will be `O(N)`, where `N` is the number of characters in the input string.
- The algorithms space complexity will be `O(K)`, where `K` is the number of distinct characters in the input string. This also means `K<=N`, because in the worst case, the whole string might not have any repeating character, so the entire string will be added to the <b>HashMap</b>. Having said that, since we can expect a fixed set of characters in the input string (e.g., 26 for English letters), we can say that the algorithm runs in fixed space `O(1)`; in this case, we can use a fixed-size array instead of the <b>HashMap</b>.

## Longest Substring with Same Letters after Replacement (hard)
## Longest Substring with Same Letters after Replacement (hard) ✅
https://leetcode.com/problems/longest-repeating-character-replacement/

> Given a string with lowercase letters only, if you are allowed to <b>replace no more than `K` letters</b> with any letter, find the <b>length of the longest substring having the same letters</b> after replacement.
Expand Down Expand Up @@ -504,7 +505,7 @@ lengthOfLongestSubstring("abccde", 1)//3, Replace the 'b' or 'd' with 'c' to hav
- The above algorithms time complexity will be `O(N)`, where `N` is the number of letters in the input string.
- As we expect only the lower case letters in the input string, we can conclude that the space complexity will be `O(26)` to store each letters frequency in the <b>HashMap</b>, which is asymptotically equal to `O(1)`.

## Longest Subarray with Ones after Replacement (hard)
## Longest Subarray with Ones after Replacement (hard) ✅
https://leetcode.com/problems/max-consecutive-ones-iii/

> Given an array containing `0`'s and `1`'s, if you are allowed to <b>replace no more than `K` `0`'s with `1`'s</b>,
Expand Down

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