#3091

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Apply Operations to Make Sum of Array Greater Than or Equal to k

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Given an array nums, find the minimum number of operations to make the sum of elements greater than or equal to k.

math greedy enumeration

Problem Statement

You are given a positive integer k and an array nums. Initially, nums consists of only one element, nums = [1].

You can perform the following operations any number of times: change any number in nums to any positive integer. The goal is to return the minimum number of operations required to make the sum of nums greater than or equal to k.

Examples

Example 1

Input: k = 11

Output: 5

We can do the following operations on the array nums = [1] : The sum of the final array is 4 + 4 + 4 = 12 which is greater than or equal to k = 11 . The total number of operations performed is 3 + 2 = 5 .

Example 2

Input: k = 1

Output: 0

The sum of the original array is already greater than or equal to 1 , so no operations are needed.

Constraints

1 <= k <= 105

Solution Approach

Greedy Choice

Start by maximizing the sum of the array in the fewest moves. Perform increase operations first to boost the sum as quickly as possible.

Duplicate Handling

After increasing values, duplicate operations can be applied. These will add fewer elements to the sum but help reach the target efficiently.

Invariant Validation

Ensure that the changes made to nums maintain an optimal approach. Reaching the target sum requires checking if each operation aligns with the desired outcome.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity of the solution will depend on the final approach chosen. Efficient greedy algorithms should result in logarithmic or linear complexity with respect to k, while handling array space optimally.

What Interviewers Usually Probe

Look for an understanding of greedy strategies in optimizing sum with minimal operations.
Check the candidate’s ability to prioritize operations effectively.
Evaluate how well the candidate validates their approach using mathematical principles and invariant rules.

Common Pitfalls or Variants

Common pitfalls

Misprioritizing operations, leading to excess operations before reaching k.
Failing to account for the efficient handling of duplicate operations.
Neglecting invariant validation that ensures all operations move toward the target sum.

Follow-up variants

What if nums starts with more elements than just [1]?
What if the array is very large, requiring different optimization strategies?
How can this problem be approached with dynamic programming for different constraints?

FAQ

What is the core pattern for solving 'Apply Operations to Make Sum of Array Greater Than or Equal to k'?

The problem follows a greedy pattern where you apply the increase operations first, followed by handling duplicates to minimize the number of operations.

Can we use dynamic programming to solve this problem?

While dynamic programming could be an option, greedy choice combined with invariant validation is often the optimal and simpler solution for this problem.

How do we optimize the number of operations in this problem?

Maximize the array sum first by applying increase operations, and then focus on duplicates to achieve the minimum number of operations.

What are the challenges in handling large arrays in this problem?

For large arrays, the challenge lies in managing space and time complexity while ensuring the greedy approach is still optimal.

How does GhostInterview assist in preparing for this problem?

GhostInterview provides targeted explanations and step-by-step guidance on greedy algorithms, helping you optimize operations to reach the target sum with minimal effort.

terminal

Solution

Solution 1: Enumeration

We should put the copy operation (i.e., operation $2$) at the end to reduce the number of operations.

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class Solution:
    def minOperations(self, k: int) -> int:
        ans = k
        for a in range(k):
            x = a + 1
            b = (k + x - 1) // x - 1
            ans = min(ans, a + b)
        return ans

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