💥💥 Beats 100% on runtime and memory [EXPLAINED]
IntuitionFigure out how to spend time in a way that minimizes a result, considering both time and starting values, while trying to find an optimal selection of
# Intuition
Figure out how to spend time in a way that minimizes a result, considering both time and starting values, while trying to find an optimal selection of elements from two lists. I realized that sorting the pairs and using recursion with memoization would help efficiently explore the possible options without recalculating overlapping subproblems.

# Approach
Sort the pairs in descending order based on one list, then use recursion with memoization (stored in a HashMap) to calculate the best result for each possible time, considering whether to take or skip each pair.

# Complexity
- Time complexity:
O(n^2) because of the recursive DFS combined with sorting the array, where n is the length of the lists.

- Space complexity:
O(n^2) as well, due to the memoization storage (HashMap), where n is the number of elements in the input lists.

# Code
```rust []
use std::collections::HashMap;

impl Solution {
    pub fn minimum_time(nums1: Vec<i32>, nums2: Vec<i32>, x: i32) -> i32 {
        let n = nums1.len() as i32; // Convert the length to i32
        let mut nums: Vec<(i32, i32)> = nums1.iter()
            .zip(nums2.iter())
            .map(|(&n1, &n2)| (n2, n1))
            .collect();
        
        nums.sort_by(|a, b| b.cmp(a)); // Sort in reverse order

        // Memoization storage using HashMap
        let mut cache: HashMap<(i32, i32), i32> = HashMap::new(); // Change usize to i32 in the cache key

        // DFS function with memoization
        fn dfs(nums: &[(i32, i32)], i: i32, time: i32, cache: &mut HashMap<(i32, i32), i32>) -> i32 {
            if time == 0 {
                return 0;
            }
            if i == nums.len() as i32 {
                return 0;
            }

            // Check the cache
            if let Some(&result) = cache.get(&(i, time)) {
                return result;
            }

            let (speed, start) = nums[i as usize];
            let mut local_res = start + speed * time + dfs(nums, i + 1, time - 1, cache);

            // Skip this number, only if there are enough elements left for time
            if nums.len() as i32 - i > time {
                local_res = local_res.max(dfs(nums, i + 1, time, cache));
            }

            // Store the result in the cache
            cache.insert((i, time), local_res);
            local_res
        }

        let sum1: i32 = nums1.iter().sum();
        let sum2: i32 = nums2.iter().sum();

        for time in 0..=n {
            let result = sum1 + time * sum2 - dfs(&nums, 0, time, &mut cache);
            if result <= x {
                return time;
            }
        }
        
        -1
    }
}


```
['Array', 'Math', 'Binary Search', 'Dynamic Programming', 'Recursion', 'Memoization', 'Sorting', 'Data Stream', 'Rust']