This technique shows how a nested for loop in few problems can be converted to single for loop and hence reducing the time complexity.

Let’s start with a problem for illustration where we can apply this technique –

Given an array of integers of size‘n’. Our aim is to calculate the maximum sum of‘k’consecutive elements in the array. Input : arr[] = {100, 200, 300, 400} k = 2 Output : 700 Input : arr[] = {1, 4, 2, 10, 23, 3, 1, 0, 20} k = 4 Output : 39 We get maximum sum by adding subarray {4, 2, 10, 23} of size 4. Input : arr[] = {2, 3} k = 3 Output : Invalid There is no subarray of size 3 as size of whole array is 2.

So, let’s analyze the problem with **Brute Force Approach**. 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.

Consider the below C++ implementation :

// O(n*k) solution for finding maximum sum of // a subarray of size k #include <iostream> using namespace std; // Returns maximum sum in a subarray of size k. int maxSum(int arr[], int n, int k) { // Initialize result int max_sum = INT_MIN ; // Consider all blocks starting with i. for (int i=0; i<n-k+1; i++) { int current_sum = 0; for (int j=0; j<k; j++) current_sum = current_sum + arr[i+j]; // Update result if required. max_sum = max(current_sum , max_sum ); } return max_sum; } // Driver code int main() { int arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20}; int k = 4; int n = sizeof(arr)/sizeof(arr[0]); cout << maxSum(arr, n, k); return 0; }

Output :

24

It can be observed from the above code that the time complexity is **O(k*n)** as it contains two nested loops.

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**Window Sliding Technique**

The technique can be best understood with the window pane in bus, consider a window of length **n** and the pane which is fixed in it of length **k**. Consider, initially the pane is at extreme left i.e., at 0 units from the left. Now, co-relate the window with array arr[] of size n and plane with current_sum of size k elements. Now, if we apply force on the window such that it moves a unit distance ahead. The pane will cover next **k** consecutive elements.

Consider an array **arr[]** = {5 , 2 , -1 , 0 , 3} and value of **k** = 3 and **n** = 5

**Applying sliding window technique **:

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

The below representation will make it clear how the window slides over the array.

This is the initial phase where we have calculated the initial window sum starting from index 0 . At this stage the window sum is 6. Now, we set the maximum_sum as current_window i.e 6.

Now, we slide our window by a unit index. Therefore, now it discards 5 from the window and adds 0 to the window. Hence, we will get our new window sum by subtracting 5 and then adding 0 to it. So, our window sum now becomes 1. Now, we will compare this window sum with the maximum_sum. As it is smaller we wont the change the maximum_sum.

Similarly, now once again we slide our window by a unit index and obtain the new window sum to be 2. Again we check if this current window sum is greater than the maximum_sum till now. Once, again it is smaller so we don’t change the maximum_sum.

Therefore, for the above array our maximum_sum is 6.

**C++ code for the above description :**

// O(n) solution for finding maximum sum of // a subarray of size k #include <iostream> using namespace std; // Returns maximum sum in a subarray of size k. int maxSum(int arr[], int n, int k) { // k must be greater if (n < k) { cout << "Invalid"; return -1; } // Compute sum of first window of size k int max_sum = 0; for (int i=0; i<k; i++) max_sum += arr[i]; // Compute sums of remaining windows by // removing first element of previous // window and adding last element of // current window. int window_sum = max_sum; for (int i=k; i<n; i++) { window_sum += arr[i] - arr[i-k]; max_sum = max(max_sum, window_sum); } return max_sum; } // Driver code int main() { int arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20}; int k = 4; int n = sizeof(arr)/sizeof(arr[0]); cout << maxSum(arr, n, k); return 0; }

Output:24

Now, it is quiet obvious that the Time Complexity is linear as we can see that only one loop runs in our code. Hence, our Time Complexity is **O(n)**.

We can use this technique to find max/min k-subarray, XOR, product, sum, etc. Refer sliding window problems for such problems.

This article is contributed by **Kanika Thakral**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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