709984. K-th Largest Sum Contiguous Subarray
Here is a detailed Markdown blog post for the GeeksforGeeks problem "K-th Largest Sum Contiguous Subarray":
K-th Largest Sum Contiguous Subarray
Slug: k-th-largest-sum-contiguous-subarray
Difficulty: Medium
Id: 709984
Topic Tags: Arrays, priority-queue, Data Structures, Algorithms
Company Tags: None
Summary
Given an array of integers and a positive integer K, find the Kth largest sum of contiguous subarrays. A contiguous subarray is a sequence of numbers within the original array that can be accessed by its start index, end index, and length.
In this problem, we need to find the Kth largest sum of such subarrays. We will use a priority queue to efficiently keep track of the top-K sums encountered so far.
Detailed Explanation
To solve this problem, we first initialize an empty priority queue Q. Then, we iterate over each possible contiguous subarray in the given array. For each subarray, we calculate its sum and add it to the priority queue. If the size of the priority queue exceeds K, we remove the smallest element from the queue.
Finally, after processing all subarrays, we retrieve the Kth largest sum from the priority queue. This sum is the desired answer.
Here's a step-by-step breakdown:
- Initialize an empty priority queue Q.
- Iterate over each possible contiguous subarray in the given array.
- For each subarray:
- Calculate its sum.
- Add the sum to the priority queue Q.
- If the size of the priority queue exceeds K, remove the smallest element from the queue.
- After processing all subarrays, retrieve the Kth largest sum from the priority queue.
The time complexity for this solution is O(n log k), where n is the length of the input array and k is the value of K. The space complexity is O(k).
Optimized Solutions
Java
import java.util.PriorityQueue;
import java.util.Queue;
public class Solution {
public int kthLargestSum(int[] nums, int k) {
Queue<Integer> pq = new PriorityQueue<>();
for (int i = 0; i < nums.length; i++) {
for (int j = i; j < nums.length; j++) {
int sum = 0;
for (int x = i; x <= j; x++) {
sum += nums[x];
}
pq.offer(sum);
if (pq.size() > k) {
pq.poll();
}
}
}
return pq.peek();
}
}