Greetings, aspiring coders! Today, we're going to delve deep into the complexities of data structures, specifically how to handle queries efficiently using JavaScript. This is a common problem often encountered in numerous data science and algorithmic problems. So let's gear up to unravel the mysteries of managing sorted sets and get our hands dirty with some interactive problem-solving!
Before delving into the task, let's understand how we can achieve sorted set functionality in JavaScript. Although JavaScript does not have a built-in data structure specifically for sorted sets, we can use arrays and objects to simulate this behavior.
To maintain a sorted set in JavaScript:
We can store elements in an array and keep it sorted upon every insertion or deletion. This approach will involve operations like using binary search to find the correct position for insertion or deletion:
O(N)
in the worst case.O(N)
in the worst case.O(log N)
.Understanding these operations can help us utilize arrays and objects efficiently for our problem.
We are tasked with designing a JavaScript function named processQueries
that can process a series of distinct requests or queries efficiently. The queries comprise a list of two integers — the type of operation and the operand.
There are three types of operations we'll handle:
0
)1
). Whenever this operation is invoked, we can guarantee that the integer exists in the set.2
).The function should return the current size of the set when the operation type is 0
or 1
, and the smallest possible integer when the operation type is 2
. If such an integer does not exist, the function should return -1
.
To understand the step-by-step breakdown of how the function works, let's go through a set of queries in detail.
JavaScript1const queries = [ 2 [0, 10], // Add 10 to the set 3 [2, 10], // Find the smallest integer >= 10 4 [0, 20], // Add 20 to the set 5 [1, 10], // Remove 10 from the set 6 [2, 10] // Find the smallest integer >= 10 7];
First Query: [0, 10]
0
: Add 10
to the set.10
: [10]
[10]
.1
) to results
.results
now: [1]
Second Query: [2, 10]
2
: Find the smallest integer that is greater than or equal to 10
.[10]
10
is 10
.10
to results
.results
now: [1, 10]
Third Query: [0, 20]
0
: Add 20
to the set.20
: [10, 20]
[10, 20]
.2
) to results
.results
now: [1, 10, 2]
Fourth Query: [1, 10]
1
: Remove 10
from the set.10
: [10, 20]
10
from the set. The set becomes [20]
.1
) to results
.results
now: [1, 10, 2, 1]
Fifth Query: [2, 10]
2
: Find the smallest integer that is greater than or equal to 10
.[20]
10
is 20
.20
to results
.results
now: [1, 10, 2, 1, 20]
The function thus processes each query step-by-step and constructs the results
array as [1, 10, 2, 1, 20]
.
To start, we'll initialize our set as an empty array. We'll also create an empty array labeled results
to store the outputs for each request.
JavaScript1function processQueries(queries) { 2 const sortedSet = []; 3 const results = [];
Next, we utilize a for
loop to traverse through all the queries. For an operation type of 0
or 1
, we either add or remove the provided value from our set, maintaining ordered insertion and deletion using a binary search to find the appropriate index. We then append the size of the current set to results
.
First, we'll implement a helper function for binary search to find the appropriate index where the element should be inserted or could be found for removal.
JavaScript1function binarySearchIndex(arr, value) { 2 let left = 0; 3 let right = arr.length - 1; 4 5 while (left <= right) { 6 const mid = left + Math.floor((right - left) / 2); 7 if (arr[mid] < value) { 8 left = mid + 1; 9 } else if (arr[mid] > value) { 10 right = mid - 1; 11 } else { 12 return mid; // Value found 13 } 14 } 15 16 return left; // Appropriate insert position 17}
With the helper function, we can efficiently handle insertion and deletion operations:
JavaScript1function processQueries(queries) { 2 const sortedSet = []; 3 const results = []; 4 5 for (const query of queries) { 6 const operation = query[0]; 7 const value = query[1]; 8 9 if (operation === 0) { 10 const index = binarySearchIndex(sortedSet, value); 11 sortedSet.splice(index, 0, value); 12 results.push(sortedSet.length); 13 } else if (operation === 1) { 14 const index = binarySearchIndex(sortedSet, value); 15 if (index !== sortedSet.length && sortedSet[index] === value) { 16 sortedSet.splice(index, 1); 17 } 18 results.push(sortedSet.length); 19 } 20 } 21 22 return results; 23}
Lastly, when the operation type is 2
, we need to find the minimum bound, i.e., the smallest value greater than or equal to our provided value in the set. We perform this using a binary search algorithm. If such a value does not exist, we append -1
to results
.
JavaScript1function processQueries(queries) { 2 const sortedSet = []; 3 const results = []; 4 5 for (const query of queries) { 6 const operation = query[0]; 7 const value = query[1]; 8 9 if (operation === 0) { 10 const index = binarySearchIndex(sortedSet, value); 11 sortedSet.splice(index, 0, value); 12 results.push(sortedSet.length); 13 } else if (operation === 1) { 14 const index = binarySearchIndex(sortedSet, value); 15 if (index !== sortedSet.length && sortedSet[index] === value) { 16 sortedSet.splice(index, 1); 17 } 18 results.push(sortedSet.length); 19 } else if (operation === 2) { 20 let left = 0; 21 let right = sortedSet.length - 1; 22 let result = -1; 23 24 while (left <= right) { 25 const mid = Math.floor((left + right) / 2); 26 if (sortedSet[mid] >= value) { 27 result = sortedSet[mid]; 28 right = mid - 1; 29 } else { 30 left = mid + 1; 31 } 32 } 33 34 results.push(result); 35 } 36 } 37 return results; 38} 39 40// Example usage 41const queries = [ 42 [0, 10], 43 [2, 10], 44 [0, 20], 45 [1, 10], 46 [2, 10] 47]; 48 49console.log(processQueries(queries)); // Output: [1, 10, 2, 1, 20]
Well done! You've successfully navigated the complexities of sorted set operations and developed an understanding of how to handle various types of queries efficiently using JavaScript. Resolving the problem involved using arrays and maintaining sorted order efficiently by leveraging binary search for insertions, deletions, and lookups.
The next step in your learning journey involves tackling similar challenges on your own using the concepts that you've just learned. Be sure to review this lesson as needed, and always remember: practice and apply these concepts. Happy coding!