What is Sliding Window Algorithm? Examples?
What is Sliding Window Algorithm? Examples?
I think of it as more a technique than an algorithm. It's a technique that could be utilized in various algorithms.
Tianjin Frame Metal supply professional and honest service.
I think the technique is best understood with the following example. Imagine we have this array:
[ 5, 7, 1, 4, 3, 6, 2, 9, 2 ]
How would we find the largest sum of five consecutive elements? Well, we'd first look at 5, 7, 1, 4, 3 and see that the sum is 20. Then we'd look at the next set of five consecutive elements, which is 7, 1, 4, 3, 6. The sum of those is 21. This is more than our previous sum, so 7, 1, 4, 3, 6 is currently the best we've got so far.
Let's see if we could improve. 1, 4, 3, 6, 2? No, that sums to 16. 4, 3, 6, 2, 9? That sums to 24, so now that's the best sequence we've got. Now we move along to the next sequence, 3, 6, 2, 9, 2. That one sums to 22, which doesn't beat our current best of 24. And we've reached the end, so we're done.
The brute force approach to implementing this programmatically is as follows:
const getMaxSumOfFiveContiguousElements = (arr) => {
let maxSum = -Infinity;
let currSum;
for (let i = 0; i <= arr.length - 5; i++) {
currSum = 0;
for (let j = i; j < i + 5; j++) {
currSum += arr[j];
}
maxSum = Math.max(maxSum, currSum);
}
return maxSum;
};
What is the time complexity of this? It's O(n*k). The outer loop is going through n - k + 1 items, but when n is much larger than k, we can forget about the k + 1 part and just call it n items. Then the inner loop is going through k items, so we have O(n*k). Try visualizing it like this:
Can we get this down to just O(n)? Let's return to this array:
[ 5, 7, 1, 4, 3, 6, 2, 9, 2 ]
First we get the sum of 5, 7, 1, 4, 3. Next we need the sum of 7, 1, 4, 3, 6. Visualize it like this, with a "window" surrounding each group of five elements.
What's the difference between the first window and the second window? Well, the second window got rid of the 5 on the left but added a 6 on the right. So since we know the sum of the first window was 20, to get the sum of the second window, we take that 20, subtract out the 5, and add the 6 to get 21. We don't actually have to go through each element in the second window and add them up (7 + 1 + 4 + 3 + 6). That would involve doing repeated and unnecessary work.
Here the sliding window approach ends up being two operations instead of five, since k is 5. That's not a huge improvement, but you can imagine that for larger k (and larger n) it really does help.
Here's how the code would work using the sliding window technique:
const getLargestSumOfFiveConsecutiveElements = (arr) => {
let currSum = getSum(arr, 0, 4);
let largestSum = currSum;
for (let i = 1; i <= arr.length - 5; i++) {
currSum -= arr[i - 1]; // subtract element to the left of curr window
currSum += arr[i + 4]; // add last element in curr window
largestSum = Math.max(largestSum, currSum);
}
return largestSum;
};
const getSum = (arr, start, end) => {
let sum = 0;
for (let i = start; i <= end; i++) {
sum += arr[i];
}
return sum;
};
And that's the gist of the sliding window technique. In other problems you may be doing something more complicated than getting the sum of the elements inside the window. Or the window itself may be of varying size instead of the fixed size of five that we saw here. But this basic application of the sliding window technique should give you a foundation from which you could build off of.
Sliding Window Technique
Sliding Window problems are problems in which a fixed or variable-size window is moved through a data structure, typically an array or string, to solve problems efficiently based on continuous subsets of elements. This technique is used when we need to find subarrays or substrings according to a given set of conditions.
What is Sliding Window Technique?
Sliding Window Technique is a method used to efficiently solve problems that involve defining a window or range in the input data (arrays or strings) and then moving that window across the data to perform some operation within the window. This technique is commonly used in algorithms like finding subarrays with a specific sum, finding the longest substring with unique characters, or solving problems that require a fixed-size window to process elements efficiently.
Let’s take an example to understand this properly, say we have an array of size N and also an integer K. Now, we have to calculate the maximum sum of a subarray having size exactly K. Now how should we approach this problem?
One way to do this by taking each subarray of size K from the array and find out the maximum sum of these subarrays. This can be done using Nested loops which will result into O(N2) Time Complexity.
But can we optimize this approach?
The answer is Yes, instead of taking each K sized subarray and calculating its sum, we can just take one K size subarray from 0 to K-1 index and calculate its sum now shift our range one by one along with the iterations and update the result, like in next iteration increase the left and right pointer and update the previous sum as shown in the below image:
Now follow this method for each iteration till we reach the end of the array:
So, we can see that instead of recalculating the sum of each K sized subarray we are using previous window of size K and using its results we update the sum and shift the window right by moving left and right pointers, this operation is optimal because it take O(1) time to shift the range instead of recalculating.
This approach of shifting the pointers and calculating the results accordingly is known as Sliding window Technique.
There are basically two types of sliding window:
The general steps to solve these questions by following below steps:
Find the size of the window required, say K.
Compute the result for 1st window, i.e. include the first K elements of the data structure.
Then use a loop to slide the window by 1 and keep computing the result window by window.
The general steps to solve these questions by following below steps:
In this type of sliding window problem, we increase our right pointer one by one till our condition is true.
At any step if our condition does not match, we shrink the size of our window by increasing left pointer.
Again, when our condition satisfies, we start increasing the right pointer and follow step 1.
We follow these steps until we reach to the end of the array.
How to Identify Sliding Window Problems:
These problems generally require Finding Maximum/Minimum
Subarray,
Substringswhich satisfy some specific condition.
The size of the subarray or substring ‘
K’will be given in some of the problems.
These problems can easily be solved in O(N
2
) time complexity using nested loops, using sliding window we can solve these in
O(n)Time Complexity.
- Required Time Complexity:
O(N) or O(Nlog(N))
- Constraints:
N <= 10
6
, If N is the size of the Array/String.
Use Cases of Sliding Window Technique:
1. To find the maximum sum of all subarrays of size K:
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 : 700Input : 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.
Naïve Approach: So, let’s analyze the problem with Brute Force Approach. We start with the first index and sum till the kth element. We do it for all possible consecutive blocks or groups of k elements. This method requires a 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 kth element.
Below is the implementation of the above approach:
C++// O(n*k) solution for finding maximum sum of
// a subarray of size k
#include
<bits/stdc++.h>
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
;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// O(n*k) solution for finding maximum sum of
// a subarray of size k
#include
<limits.h>
#include
<math.h>
#include
<stdio.h>
// Find maximum between two numbers.
int
max
(
int
num1
,
int
num2
)
{
return
(
num1
>
num2
)
?
num1
:
num2
;
}
// 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
]);
printf
(
"%d"
,
maxSum
(
arr
,
n
,
k
));
return
0
;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// Java code O(n*k) solution for finding maximum sum of
// a subarray of size k
class
GFG
{
// Returns maximum sum in
// a subarray of size k.
static
int
maxSum
(
int
arr
[]
,
int
n
,
int
k
)
{
// Initialize result
int
max_sum
=
Integer
.
MIN_VALUE
;
// 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
=
Math
.
max
(
current_sum
,
max_sum
);
}
return
max_sum
;
}
// Driver code
public
static
void
main
(
String
[]
args
)
{
int
arr
[]
=
{
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
};
int
k
=
4
;
int
n
=
arr
.
length
;
System
.
out
.
println
(
maxSum
(
arr
,
n
,
k
));
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# code
import
sys
# O(n * k) solution for finding
# maximum sum of a subarray of size k
INT_MIN
=
-
sys
.
maxsize
-
1
# Returns maximum sum in a
# subarray of size k.
def
maxSum
(
arr
,
n
,
k
):
# Initialize result
max_sum
=
INT_MIN
# Consider all blocks
# starting with i.
for
i
in
range
(
n
-
k
+
1
):
current_sum
=
0
for
j
in
range
(
k
):
current_sum
=
current_sum
+
arr
[
i
+
j
]
# Update result if required.
max_sum
=
max
(
current_sum
,
max_sum
)
return
max_sum
# Driver code
arr
=
[
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
]
k
=
4
n
=
len
(
arr
)
print
(
maxSum
(
arr
,
n
,
k
))
# This code is contributed by mits
// C# code here O(n*k) solution for
// finding maximum sum of a subarray
// of size k
using
System
;
class
GFG
{
// Returns maximum sum in a
// subarray of size k.
static
int
maxSum
(
int
[]
arr
,
int
n
,
int
k
)
{
// Initialize result
int
max_sum
=
int
.
MinValue
;
// 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
=
Math
.
Max
(
current_sum
,
max_sum
);
}
return
max_sum
;
}
// Driver code
public
static
void
Main
()
{
int
[]
arr
=
{
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
};
int
k
=
4
;
int
n
=
arr
.
Length
;
Console
.
WriteLine
(
maxSum
(
arr
,
n
,
k
));
}
}
// This code is contributed by anuj_67.
function
maxSum
(
arr
,
n
,
k
)
{
let
max_sum
=
0
;
// Loop from i to k
for
(
let
i
=
0
;
i
<
k
;
i
++
)
{
max_sum
+=
arr
[
i
];
}
let
window_sum
=
max_sum
;
for
(
let
i
=
k
;
i
<
n
;
i
++
)
{
window_sum
=
window_sum
-
arr
[
i
-
k
]
+
arr
[
i
];
max_sum
=
Math
.
max
(
max_sum
,
window_sum
);
}
return
max_sum
;
}
// Driver code
let
arr
=
[
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
];
let
k
=
4
;
let
n
=
arr
.
length
;
console
.
log
(
maxSum
(
arr
,
n
,
k
));
<?php
// code
?>
<?php
// O(n*k) solution for finding maximum sum of
// a subarray of size k
// Returns maximum sum in a subarray of size k.
function
maxSum
(
$arr
,
$n
,
$k
)
{
// Initialize result
$max_sum
=
PHP_INT_MIN
;
// Consider all blocks
// starting with i.
for
(
$i
=
0
;
$i
<
$n
-
$k
+
1
;
$i
++
)
{
$current_sum
=
0
;
for
(
$j
=
0
;
$j
<
$k
;
$j
++
)
$current_sum
=
$current_sum
+
$arr
Additional reading:
10 Questions You Should to Know about spring pipe support
Factors to consider when buying wires and cables | Blog
Gain Insight Into the Difference Between ERW and EFW Pipe
10+ Different Types of Resin & Their Applications - A Guide ...
If you are looking for more details, kindly visit sliding commercial windows.
[
$i
+
$j
];
// Update result if required.
$max_sum
=
max
(
$current_sum
,
$max_sum
);
}
return
$max_sum
;
}
// Driver code
$arr
=
array
(
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
);
$k
=
4
;
$n
=
count
(
$arr
);
echo
maxSum
(
$arr
,
$n
,
$k
);
// This code is contributed by anuj_67.
?>
Output
24
Time complexity: O(k*n) as it contains two nested loops.
Auxiliary Space: O(1)
Applying the sliding window technique:
We compute the sum of the first k elements out of n terms using a linear loop and store the sum in variable
window_sum.
Then we will traverse linearly over the array till it reaches the end and simultaneously keep track of the maximum sum.
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.
The below representation will make it clear how the window slides over the array.
Consider an array arr[] = {5, 2, -1, 0, 3} and value of k = 3 and n = 5
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 won’t 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.
Below is the code for above approach:
C++// 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
)
{
// n 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
;
}
// Java code for
// O(n) solution for finding
// maximum sum of a subarray
// of size k
class
GFG
{
// Returns maximum sum in
// a subarray of size k.
static
int
maxSum
(
int
arr
[]
,
int
n
,
int
k
)
{
// n must be greater
if
(
n
<=
k
)
{
System
.
out
.
println
(
"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
=
Math
.
max
(
max_sum
,
window_sum
);
}
return
max_sum
;
}
// Driver code
public
static
void
main
(
String
[]
args
)
{
int
arr
[]
=
{
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
};
int
k
=
4
;
int
n
=
arr
.
length
;
System
.
out
.
println
(
maxSum
(
arr
,
n
,
k
));
}
}
// This code is contributed
// by prerna saini.
# O(n) solution for finding
# maximum sum of a subarray of size k
def
maxSum
(
arr
,
k
):
# length of the array
n
=
len
(
arr
)
# n must be greater than k
if
n
<=
k
:
print
(
"Invalid"
)
return
-
1
# Compute sum of first window of size k
window_sum
=
sum
(
arr
[:
k
])
# first sum available
max_sum
=
window_sum
# Compute the sums of remaining windows by
# removing first element of previous
# window and adding last element of
# the current window.
for
i
in
range
(
n
-
k
):
window_sum
=
window_sum
-
arr
[
i
]
+
arr
[
i
+
k
]
max_sum
=
max
(
window_sum
,
max_sum
)
return
max_sum
# Driver code
arr
=
[
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
]
k
=
4
print
(
maxSum
(
arr
,
k
))
# This code is contributed by Kyle McClay
// C# code for O(n) solution for finding
// maximum sum of a subarray of size k
using
System
;
class
GFG
{
// Returns maximum sum in
// a subarray of size k.
static
int
maxSum
(
int
[]
arr
,
int
n
,
int
k
)
{
// n must be greater
if
(
n
<=
k
)
{
Console
.
WriteLine
(
"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
=
Math
.
Max
(
max_sum
,
window_sum
);
}
return
max_sum
;
}
// Driver code
public
static
void
Main
()
{
int
[]
arr
=
{
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
};
int
k
=
4
;
int
n
=
arr
.
Length
;
Console
.
WriteLine
(
maxSum
(
arr
,
n
,
k
));
}
}
// This code is contributed by anuj_67.
<
script
>
// Javascript code for
// O(n) solution for finding
// maximum sum of a subarray
// of size k
function
maxSum
(
arr
,
n
,
k
)
{
let
max
=
0
;
let
sum
=
0
;
// find initial sum of first k elements
for
(
let
i
=
0
;
i
<
k
;
i
++
)
{
sum
+=
arr
[
i
];
max
=
sum
;
}
// iterate the array once
// and increment the right edge
for
(
let
i
=
k
;
i
<
n
;
i
++
)
{
sum
+=
arr
[
i
]
-
arr
[
i
-
k
];
// compare if sum is more than max,
// if yes then replace
// max with new sum value
if
(
sum
>
max
)
{
max
=
sum
;
}
}
return
max
;
}
// Driver code
let
arr
=
[
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
];
let
k
=
4
;
let
n
=
arr
.
length
;
document
.
write
(
maxSum
(
arr
,
n
,
k
));
<
/script>
<?php
// O(n) solution for finding maximum sum of
// a subarray of size k
// Returns maximum sum in a
// subarray of size k.
function
maxSum
(
$arr
,
$n
,
$k
)
{
// n must be greater
if
(
$n
<=
$k
)
{
echo
"Invalid"
;
return
-
1
;
}
// Compute sum of first
// window of size k
$max_sum
=
0
;
for
(
$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.
$window_sum
=
$max_sum
;
for
(
$i
=
$k
;
$i
<
$n
;
$i
++
)
{
$window_sum
+=
$arr
[
$i
]
-
$arr
[
$i
-
$k
];
$max_sum
=
max
(
$max_sum
,
$window_sum
);
}
return
$max_sum
;
}
// Driver code
$arr
=
array
(
1
,
4
,
2
,
10
,
2
,
3
,
1
,
0
,
20
);
$k
=
4
;
$n
=
count
(
$arr
);
echo
maxSum
(
$arr
,
$n
,
$k
);
// This code is contributed by anuj_67
?>
Output
24
Time Complexity: O(n), where n is the size of input array arr[].
Auxiliary Space: O(1)
Given an array arr[] of integers and a number X, the task is to find the smallest subarray with a sum greater than the given value.
Approach:
We can solve this problem using Sliding Window Technique and maintaining two pointers: start and end to mark the starting and ending of the window. We can keep incrementing the end pointer till the sum of the window is less than or equal to X. When, the sum of window becomes greater than X, we record the length of the window and start moving the start pointer till the sum of window becomes smaller than or equal to X. Now, when the sum becomes smaller than or equal to X, again start incrementing the end pointer. Keep on moving the start and end pointer till we have reached the end of the array.
Given an array arr[] of non-negative integers and an integer sum, find a subarray that adds to a given sum.
Approach:
The idea is simple as we know that all the elements in subarray are positive so, If a subarray has sum greater than the given sum then there is no possibility that adding elements to the current subarray will be equal to the given sum. So the Idea is to use a similar approach to a sliding window.
Start with an empty subarray.
add elements to the subarray until the sum is less than
x( given sum ).
If the sum is greater than
x, remove elements from the
startof the current subarray.
Approach:
Basically a window of characters is maintained by using two pointers namely start and end. These start and end pointers can be used to shrink and increase the size of window respectively. Whenever the window contains all characters of given string, the window is shrinked from left side to remove extra characters and then its length is compared with the smallest window found so far.
If in the present window, no more characters can be deleted then we start increasing the size of the window using the end until all the distinct characters present in the string are also there in the window. Finally, find the minimum size of each window.
Practice Problems on Sliding Window Technique:
Related Articles:
Like Article
Suggest improvement
Share your thoughts in the comments
Please
Login
to comment...
For more information, please visit best sliding glass doors.
Comments
0