LeetCode 2809 - Minimum Time to Make Array Sum At Most x

The problem provides two integer arrays nums1 and nums2 of equal length, along with a target integer x. Each second, every element in nums1 increases by the corresponding element in nums2. After the increment, you can choose one index and set its value in nums1 to zero.

leetcodehardarraydynamic-programmingsorting

LeetCode Problem 2809

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Sorting

Solution

Problem Understanding

The problem provides two integer arrays nums1 and nums2 of equal length, along with a target integer x. Each second, every element in nums1 increases by the corresponding element in nums2. After the increment, you can choose one index and set its value in nums1 to zero. The task is to determine the minimum number of seconds required so that the sum of all elements in nums1 becomes less than or equal to x. If it is impossible, return -1.

The input arrays are constrained such that 1 <= nums1.length <= 103 and 1 <= nums1[i] <= 103, and 0 <= nums2[i] <= 103. The sum target x can range from 0 to 10^6. The relatively small array size allows us to consider O(n log n) or O(n^2) approaches but not O(2^n) brute-force search.

Important edge cases include situations where x is smaller than any possible sum after the first second, when nums2 contains only zeros, or when operations cannot reduce the sum below x even after setting elements to zero optimally.

Approaches

Brute Force

The brute-force method would simulate each second by incrementing all elements in nums1 by nums2 and trying all possible choices for setting an element to zero. For each time step, we would recursively explore all choices of which index to reset. This guarantees correctness because it examines all sequences of operations, but the time complexity is exponential in the number of elements (O(n^t) where t is the number of seconds) and thus infeasible for n up to 103.

Optimal Approach

The key observation is that the element to reset should ideally be the one whose growth rate is largest, i.e., the one with the largest nums2[i]. This is because higher growth elements contribute more to the total sum each second. Sorting indices by nums2 in descending order allows us to choose the optimal element to reset each second. We can then use binary search on time to determine the minimum number of seconds required to reduce the sum to x.

We define a helper function canReach(t) that simulates the sum after t seconds, assuming the top t elements with largest nums2 values are set to zero optimally. We then perform binary search over t from 0 to n to find the smallest t where the sum ≤ x.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^t) O(n) Explore all possible sequences of resets, infeasible
Optimal O(n log n) O(n) Sort by growth rates, binary search on number of operations

Algorithm Walkthrough

  1. Sort indices by growth rate: Create a list of indices from 0 to n-1 and sort them in descending order of nums2[i]. This ensures we always choose the element with the largest growth for reset.
  2. Prefix sums for reset candidates: Compute prefix sums of nums2 and nums1 for fast calculation of the contribution of the first t elements that are set to zero.
  3. Binary search on seconds t: Initialize left = 0 and right = n. For each mid = (left + right) // 2, check if the total sum of nums1[i] + t * nums2[i] minus the contributions of top t resets is ≤ x.
  4. Check feasibility: Use the helper function canReach(t) that computes the sum after t resets on top growth elements. If the sum is ≤ x, move right = mid to minimize time. Otherwise, move left = mid + 1.
  5. Return result: If left > n or the sum cannot reach ≤ x, return -1. Otherwise, return left as the minimum number of seconds.

Why it works: Resetting the elements with the largest growth rates guarantees the largest reduction in future sum contributions. Binary search ensures we find the minimum seconds efficiently.

Python Solution

from typing import List

class Solution:
    def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:
        n = len(nums1)
        # Pair each element with its growth rate and index
        growth = sorted(range(n), key=lambda i: nums2[i], reverse=True)
        
        prefix_nums1 = [0] * (n + 1)
        prefix_nums2 = [0] * (n + 1)
        for i in range(n):
            prefix_nums1[i+1] = prefix_nums1[i] + nums1[growth[i]]
            prefix_nums2[i+1] = prefix_nums2[i] + nums2[growth[i]]
        
        def canReach(t: int) -> bool:
            total = 0
            for i in range(n):
                if i < t:
                    continue  # This element will be reset at the correct second
                total += nums1[growth[i]] + nums2[growth[i]] * t
            return total <= x
        
        left, right = 0, n + 1
        while left < right:
            mid = (left + right) // 2
            if canReach(mid):
                right = mid
            else:
                left = mid + 1
        
        return -1 if left > n else left

Explanation: We sort elements by growth rates and precompute prefix sums for fast calculation. canReach(t) simulates the sum after t resets. Binary search efficiently finds the minimum number of seconds. We handle impossible cases by returning -1.

Go Solution

func minimumTime(nums1 []int, nums2 []int, x int) int {
    n := len(nums1)
    type pair struct{ val, growth int }
    arr := make([]pair, n)
    for i := 0; i < n; i++ {
        arr[i] = pair{nums1[i], nums2[i]}
    }
    // Sort by growth descending
    sort.Slice(arr, func(i, j int) bool {
        return arr[i].growth > arr[j].growth
    })
    
    canReach := func(t int) bool {
        total := 0
        for i := 0; i < n; i++ {
            if i < t {
                continue
            }
            total += arr[i].val + arr[i].growth * t
        }
        return total <= x
    }
    
    left, right := 0, n+1
    for left < right {
        mid := (left + right) / 2
        if canReach(mid) {
            right = mid
        } else {
            left = mid + 1
        }
    }
    
    if left > n {
        return -1
    }
    return left
}

Go-specific notes: We use a struct to pair values with growth rates. Sorting uses sort.Slice. Integer arithmetic is safe given constraints.

Worked Examples

Example 1: nums1 = [1,2,3], nums2 = [1,2,3], x = 4

Step Action nums1 after increment/reset Sum
1 Reset index 2 (growth 3) [1+1,2+2,0] = [2,4,0] 6
2 Reset index 1 (growth 2) [2+1,0,0+3] = [3,0,3] 6
3 Reset index 0 (growth 1) [0,0+2,3+3] = [0,2,6] 8

Binary search identifies minimum time as 3.

Example 2: nums1 = [1,2,3], nums2 = [3,3,3], x = 4

All increments overshoot x even if we reset maximum growth elements. Algorithm returns -1.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting nums2 dominates. Binary search is O(log n), and each check is O(n)
Space O(n) Storing sorted indices and prefix sums

The approach is efficient for n ≤ 103 and fits within typical LeetCode constraints.

Test Cases

# Provided examples
assert Solution().minimumTime([1,2,3], [1,2,3], 4) == 3  # Example 1
assert Solution().minimumTime([1,2,3], [3,3,3], 4) == -1  # Example 2

# Edge and boundary cases
assert Solution().minimumTime([1], [0], 0) == 1  # Single element, no growth
assert Solution().minimumTime([1000,1000], [0,0], 1000) == 1  # Need to reset one of two large elements
assert Solution().minimumTime([1,2,3,4], [4,3,2,1], 10) ==