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# Coding Patterns: Sliding Window

The Sliding window is a problem-solving technique that aims to reduce the use of nested loop and replace it with a single loop, thereby reducing the time complexity. These problems are easy to solve using a brute force approach in O(n2) or O(n3) but using the 'sliding window' technique, we can reduce the time complexity to O(n).

## Problem: Maximum Average Subarray:

LeetCode 643 - Maximum Average Subarray I [easy]

You are given an integer array `nums` consisting of `n` elements, and an integer `k`.

Brute Force Solution:

We start with first index and sum till k-th element. We do it for all possible consecutive blocks or groups of k elements. This method requires nested for loop, the outer for loop starts with the starting element of the block of k elements and the inner or the nested loop will add up till the k-th element.

class Solution:
def findMaxAverage(self, nums: List[int], k: int) -> float:

max_value = 0

for i in range(len(nums) - k + 1):

sum = 0

for j in range(i, i + k):

sum += nums[j]

max_value = max(max_value, sum)

return max_value / k

Time ComplexityO(n*k) where n is the number of elements in the input array.

Sliding Window Solution:

class Solution:
def findMaxAverage(self, nums: List[int], k: int) -> float:

# Compute sum of first window of size k
max_sum = sum(nums[:k])
result = max_sum      # first sum available
for i in range(k, len(nums)):
max_sum += nums[i] - nums[i - k]
result = max(max_sum, result)
return result / k

1. We compute the sum of the first k elements out of n terms using a linear loop and store the sum in the variable max_sum.
2. Then we will graze linearly over the array till it reaches the end and simultaneously keep track of the maximum sum.
3. To get the current sum of a block of k elements just subtract the first element from the previous block and add the last element of the current block.

Time Complexity is linear as we can see that only one loop runs in our code. Hence, our Time Complexity is O(n).

Problem:  Substring with Concetenation of All words:

Leetcode - 30. Substring with Concatenation of All Words

You are given a string `s` and an array of strings `words`. All the strings of `words` are of the same length.

concatenated substring in `s` is a substring that contains all the strings of any permutation of `words` concatenated.

• For example, if `words = ["ab","cd","ef"]`, then `"abcdef"``"abefcd"``"cdabef"``"cdefab"``"efabcd"`, and `"efcdab"` are all concatenated strings. `"acdbef"` is not a concatenated substring because it is not the concatenation of any permutation of `words`.

Return the starting indices of all the concatenated substrings in `s`. You can return the answer in any order.

Solution:

class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
n=len(words) * len(words[0])
n1=len(words[0])
arr=[]
words.sort()
words=''.join(words)
for i in range(0,len(s)-n+1):
temp=s[i:i+n]
parts=[temp[i:i+n1] for i in range(0,len(temp),n1)]
parts.sort()
parts=''.join(parts)
if(parts==words):
arr.append(i)
return arr

How to identify?

• The problem involves a data structure that is ordered and iterable like arraysstrings, etc.
• The problem is asking to find a subrange in an array/string, contiguous longestshortestaverage, max/min k-subarray, XOR, product, sum, or target value.
• There is an apparent naive or brute force solution that runs in O(N2)O(2N), or some other large time complexity.

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