LeetCode #2809 — HARD

Minimum Time to Make Array Sum At Most x

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:

Choose an index 0 <= i < nums1.length and make nums1[i] = 0.

You are also given an integer x.

Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.

Example 1:

Input: nums1 = [1,2,3], nums2 = [1,2,3], x = 4
Output: 3
Explanation: 
For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. 
For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. 
For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. 
Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.

Example 2:

Input: nums1 = [1,2,3], nums2 = [3,3,3], x = 4
Output: -1
Explanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.

Constraints:

1 <= nums1.length <= 10³
1 <= nums1[i] <= 10³
0 <= nums2[i] <= 10³
nums1.length == nums2.length
0 <= x <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation: Choose an index 0 <= i < nums1.length and make nums1[i] = 0. You are also given an integer x. Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[1,2,3]
[1,2,3]
4

Example 2

[1,2,3]
[3,3,3]
4

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">It can be proven that in the optimal solution, for each index <code>i</code>, we only need to set <code>nums1[i]</code> to <code>0</code> at most once. (If we have to set it twice, we can simply remove the earlier set and all the operations “shift left” by <code>1</code>.)</div>
<div class="_1l1MA">It can also be proven that if we select several indexes <code>i<sub>1</sub>, i<sub>2</sub>, ..., i<sub>k</sub></code> and set <code>nums1[i<sub>1</sub>], nums1[i<sub>2</sub>], ..., nums1[i<sub>k</sub>]</code> to <code>0</code>, it’s always optimal to set them in the order of <code>nums2[i<sub>1</sub>] <= nums2[i<sub>2</sub>] <= ... <= nums2[i<sub>k</sub>]</code> (the larger the increase is, the later we should set it to <code>0</code>).</div>
<div class="_1l1MA">Let’s sort all the values by <code>nums2</code> (in non-decreasing order). Let <code>dp[i][j]</code> represent the maximum total value that can be reduced if we do <code>j</code> operations on the first <code>i</code> elements. Then we have <code>dp[i][0] = 0</code> (for all <code>i = 0, 1, ..., n</code>) and <code>dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + nums2[i - 1] * j + nums1[i - 1])</code> (for <code>1 <= i <= n</code> and <code>1 <= j <= i</code>).</div>
<div class="_1l1MA">The answer is the minimum value of <code>t</code>, such that <code>0 <= t <= n</code> and <code>sum(nums1) + sum(nums2) * t - dp[n][t] <= x</code>, or <code>-1</code> if it doesn’t exist.</div>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2809: Minimum Time to Make Array Sum At Most x
class Solution {
    public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {
        int n = nums1.size();
        int[][] f = new int[n + 1][n + 1];
        int[][] nums = new int[n][0];
        for (int i = 0; i < n; ++i) {
            nums[i] = new int[] {nums1.get(i), nums2.get(i)};
        }
        Arrays.sort(nums, Comparator.comparingInt(a -> a[1]));
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j > 0) {
                    int a = nums[i - 1][0], b = nums[i - 1][1];
                    f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + a + b * j);
                }
            }
        }
        int s1 = 0, s2 = 0;
        for (int v : nums1) {
            s1 += v;
        }
        for (int v : nums2) {
            s2 += v;
        }

        for (int j = 0; j <= n; ++j) {
            if (s1 + s2 * j - f[n][j] <= x) {
                return j;
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #2809: Minimum Time to Make Array Sum At Most x
func minimumTime(nums1 []int, nums2 []int, x int) int {
	n := len(nums1)
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	type pair struct{ a, b int }
	nums := make([]pair, n)
	var s1, s2 int
	for i := range nums {
		s1 += nums1[i]
		s2 += nums2[i]
		nums[i] = pair{nums1[i], nums2[i]}
	}
	sort.Slice(nums, func(i, j int) bool { return nums[i].b < nums[j].b })
	for i := 1; i <= n; i++ {
		for j := 0; j <= n; j++ {
			f[i][j] = f[i-1][j]
			if j > 0 {
				a, b := nums[i-1].a, nums[i-1].b
				f[i][j] = max(f[i][j], f[i-1][j-1]+a+b*j)
			}
		}
	}
	for j := 0; j <= n; j++ {
		if s1+s2*j-f[n][j] <= x {
			return j
		}
	}
	return -1
}

# Accepted solution for LeetCode #2809: Minimum Time to Make Array Sum At Most x
class Solution:
    def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:
        n = len(nums1)
        f = [[0] * (n + 1) for _ in range(n + 1)]
        for i, (a, b) in enumerate(sorted(zip(nums1, nums2), key=lambda z: z[1]), 1):
            for j in range(n + 1):
                f[i][j] = f[i - 1][j]
                if j > 0:
                    f[i][j] = max(f[i][j], f[i - 1][j - 1] + a + b * j)
        s1 = sum(nums1)
        s2 = sum(nums2)
        for j in range(n + 1):
            if s1 + s2 * j - f[n][j] <= x:
                return j
        return -1

// Accepted solution for LeetCode #2809: Minimum Time to Make Array Sum At Most x
/**
 * [2809] Minimum Time to Make Array Sum At Most x
 */
pub struct Solution {}

// submission codes start here

struct Pair {
    num1: i32,
    num2: i32,
}

use std::cmp::max;

impl Solution {
    pub fn minimum_time(nums1: Vec<i32>, nums2: Vec<i32>, x: i32) -> i32 {
        let mut pairs = Vec::with_capacity(nums1.len());
        let sum1: i32 = nums1.iter().sum();
        let sum2: i32 = nums2.iter().sum();
        for (index, value) in nums1.iter().enumerate() {
            pairs.push(Pair {
                num1: *value,
                num2: nums2[index],
            });
        }

        pairs.sort_by(|a, b| a.num2.cmp(&b.num2));

        let mut dp = vec![0; nums1.len() + 1];
        for i in 1..=nums1.len() {
            let (num1, num2) = (pairs[i - 1].num1, pairs[i - 1].num2);
            for j in (1..=i).rev() {
                dp[j] = max(dp[j], dp[j - 1] + num2 * j as i32 + num1);
            }
        }

        for i in 0..=nums1.len() {
            let j = i as i32;

            if sum1 + sum2 * j - dp[i] <= x {
                return j;
            }
        }

        -1
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2809() {
        assert_eq!(Solution::minimum_time(vec![1, 2, 3], vec![1, 2, 3], 4), 3);
        assert_eq!(Solution::minimum_time(vec![1, 2, 3], vec![3, 3, 3], 4), -1);
        assert_eq!(
            Solution::minimum_time(vec![4, 4, 9, 10], vec![4, 4, 1, 3], 16),
            4
        );
    }
}

// Accepted solution for LeetCode #2809: Minimum Time to Make Array Sum At Most x
function minimumTime(nums1: number[], nums2: number[], x: number): number {
    const n = nums1.length;
    const f: number[][] = Array(n + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(0));
    const nums: number[][] = [];
    for (let i = 0; i < n; ++i) {
        nums.push([nums1[i], nums2[i]]);
    }
    nums.sort((a, b) => a[1] - b[1]);
    for (let i = 1; i <= n; ++i) {
        for (let j = 0; j <= n; ++j) {
            f[i][j] = f[i - 1][j];
            if (j > 0) {
                const [a, b] = nums[i - 1];
                f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + a + b * j);
            }
        }
    }
    const s1 = nums1.reduce((a, b) => a + b, 0);
    const s2 = nums2.reduce((a, b) => a + b, 0);
    for (let j = 0; j <= n; ++j) {
        if (s1 + s2 * j - f[n][j] <= x) {
            return j;
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.