Rust OJ

Problem Statement: LMAO EA0XS9GN7D2

Problem Description

LMAO EA0XS9GN7D2

This problem is a variation of the classic "Largest Minimum Number" problem, where you are given an array of integers and you need to find the minimum value of the smallest element in each subarray of size k.

Given an array of n elements, you have to find the minimum value among all subarrays of size k. For example:


5, 3, 4, 2, 6, 7, 8, 9, 10
k = 3
Output: 3

The answer here is 3, since the minimum in every subarray of size 3 is 3.

Solution Approach

To solve this problem, we can use a sliding window technique. Here's how it works:

  1. Iterate through the array using a sliding window approach.
  2. For each window of size k, compute the minimum value.
  3. Keep track of the minimum value across all windows.

We can implement this by using a priority queue (heap) to keep track of the minimum values efficiently.

Implementation


use std::collections::VecDeque;
use std::cmp::min;

fn main() {
    let nums = vec![5, 3, 4, 2, 6, 7, 8, 9, 10];
    let k = 3;
    
    let mut min_values = VecDeque::new();
    let mut current_min = f64::MAX;
    
    for i in 0..nums.len() {
        // Add the current element to the heap
        min_values.push_back(nums[i]);
        
        // If the heap has more than k elements, remove the largest one
        if min_values.len() > k {
            current_min = min(current_min, nums[i]);
            min_values.pop_front();
        }
        
        // Keep track of the minimum value
        if min_values.len() == k {
            println!("Minimum value in window {}: {}", i - k + 1, current_min);
        }
    }
}

This solution uses a priority queue to maintain the minimum values of each window, ensuring efficient computation.

Notes on the Problem

  • The algorithm runs in O(n log k) time due to the heap operations.
  • The space complexity is O(k) for the heap.

License

Copyright © 2023 Rust OJ Team