Introduction

In this blog post, we will explore how to solve the LeetCode problem "3355" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums of length n and a 2D array queries, where queries[i] = [li, ri]. For each queries[i]: Select a subset of indices within the range [li, ri] in nums. Decrement the values at the selected indices by 1. A Zero Array is an array where all elements are equal to 0. Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false. Example 1: Input: nums = [1,0,1], queries = [[0,2]] Output: true Explanation: For i = 0: Select the subset of indices as [0, 2] and decrement the values at these indices by 1. The array will become [0, 0, 0], which is a Zero Array. Example 2: Input: nums = [4,3,2,1], queries = [[1,3],[0,2]] Output: false Explanation: For i = 0: Select the subset of indices as [1, 2, 3] and decrement the values at these indices by 1. The array will become [4, 2, 1, 0]. For i = 1: Select the subset of indices as [0, 1, 2] and decrement the values at these indices by 1. The array will become [3, 1, 0, 0], which is not a Zero Array. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 105 1 <= queries.length <= 105 queries[i].length == 2 0 <= li <= ri < nums.length

Explanation

Here's a breakdown of the approach, followed by the code:

• Difference Array: Use a difference array to track the net decrement applied to each index of nums. Increment the difference at the start index (l) and decrement it at the end index + 1 (r + 1) of each query range.

• Prefix Sum: Calculate the prefix sum of the difference array to obtain the actual decrement value for each index.

• Final Check: After processing all queries, subtract the decrement from the original value at each index. If all elements of nums become zero, return True; otherwise, return False.

• Complexity:

  • Runtime Complexity: O(n + q), where n is the length of nums and q is the number of queries.
  • Storage Complexity: O(n)

Code

    def can_transform_to_zero(nums, queries):
    """
    Checks if it is possible to transform nums into a Zero Array after processing all the queries sequentially.

    Args:
        nums: An integer array.
        queries: A 2D array representing the queries.

    Returns:
        True if it is possible to transform nums into a Zero Array, otherwise False.
    """

    n = len(nums)
    diff = [0] * (n + 1)  # Difference array

    for l, r in queries:
        diff[l] += 1
        diff[r + 1] -= 1

    decrement = [0] * n
    decrement[0] = diff[0]
    for i in range(1, n):
        decrement[i] = decrement[i - 1] + diff[i]

    for i in range(n):
        nums[i] -= decrement[i]

    for num in nums:
        if num != 0:
            return False

    return True