Leetcode #2809: Minimum Time to Make Array Sum At Most x
In this guide, we solve Leetcode #2809 Minimum Time to Make Array Sum At Most x in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming, Sorting
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is , where is the length of the array. The space complexity is , where is the length of the array.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.