Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #1665: Minimum Initial Energy to Finish Tasks

In this guide, we solve Leetcode #1665 Minimum Initial Energy to Finish Tasks 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.

Leetcode

Problem Statement

You are given an array tasks where tasks[i] = [actuali, minimumi]: actuali is the actual amount of energy you spend to finish the ith task. minimumi is the minimum amount of energy you require to begin the ith task.

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Greedy, Array, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Example

Input: tasks = [[1,2],[2,4],[4,8]] Output: 8 Explanation: Starting with 8 energy, we finish the tasks in the following order: - 3rd task. Now energy = 8 - 4 = 4. - 2nd task. Now energy = 4 - 2 = 2. - 1st task. Now energy = 2 - 1 = 1. Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.

Python Solution

class Solution: def minimumEffort(self, tasks: List[List[int]]) -> int: ans = cur = 0 for a, m in sorted(tasks, key=lambda x: x[0] - x[1]): if cur < m: ans += m - cur cur = m cur -= a return ans

Complexity

The time complexity is O(n×log⁡n)O(n\times \log n)O(n×logn), where nnn is the number of tasks. The space complexity is O(1)O(1)O(1).

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.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy