How do you solve Zero Array Transformation IV on LeetCode?

Zero Array Transformation IV (LeetCode #3489) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming can optimize cumulative operations on arrays.

What is the time complexity of Zero Array Transformation IV?

The optimal solution for Zero Array Transformation IV runs in O(n) time and uses O(n) space. The complexity is linear as we only traverse the nums array and queries once, updating values efficiently.

What is the brute force solution for Zero Array Transformation IV?

The brute force approach for Zero Array Transformation IV is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible combination of queries to see if we can reduce all elements of nums to zero. This approach checks each query one by one, leading to many redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Zero Array Transformation IV?

Zero Array Transformation IV uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Zero Array Transformation IV?

Explain your thought process clearly. Discuss edge cases and how your solution handles them. Practice similar problems to reinforce concepts.

What are common mistakes when solving Zero Array Transformation IV?

Not resetting the cumulative array for each query. Overlooking edge cases where no queries can satisfy the condition.

#3489

Zero Array Transformation IV

Medium

ArrayDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a dynamic programming approach to track the cumulative effect of queries on each index. This allows us to determine if we can reach zero for all elements efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create an array to store the cumulative decrements for each index based on the queries.
2Step 2: Iterate through each query and update the cumulative decrements for the specified range.
3Step 3: Check if the cumulative decrements can satisfy the original nums values for each index.

solution.py10 lines

1def min_queries(nums, queries):
2    n = len(nums)
3    dp = [0] * n
4    for k in range(len(queries)):
5        l, r, val = queries[k]
6        for i in range(l, r + 1):
7            dp[i] += val
8        if all(dp[i] >= nums[i] for i in range(n)):
9            return k + 1
10    return -1

ℹ

Complexity note: The complexity is linear as we only traverse the nums array and queries once, updating values efficiently.

1Dynamic programming can optimize cumulative operations on arrays.
2Understanding the effect of each query helps in efficiently determining the result.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.