Introduction

In this blog post, we will explore how to solve the LeetCode problem "3356" 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, vali]. Each queries[i] represents the following action on nums: Decrement the value at each index in the range [li, ri] in nums by at most vali. The amount by which each value is decremented can be chosen independently for each index. A Zero Array is an array with all its elements equal to 0. Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1. Example 1: Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]] Output: 2 Explanation: For i = 0 (l = 0, r = 2, val = 1): Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively. The array will become [1, 0, 1]. For i = 1 (l = 0, r = 2, val = 1): Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively. The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2. Example 2: Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]] Output: -1 Explanation: For i = 0 (l = 1, r = 3, val = 2): Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively. The array will become [4, 1, 0, 0]. For i = 1 (l = 0, r = 2, val = 1): Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively. The array will become [3, 0, 0, 0], which is not a Zero Array. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 5 * 105 1 <= queries.length <= 105 queries[i].length == 3 0 <= li <= ri < nums.length 1 <= vali <= 5

Explanation

• Binary Search: Use binary search to find the minimum number of queries needed to make the array zero. The search space is the range of queries from 0 to the total number of queries.
  • Check Function: Implement a helper function to check if applying the first k queries results in a zero array. This function uses a difference array technique to efficiently simulate the decrements.
  • Difference Array: Use a difference array to efficiently track the net decrement applied to each index. This avoids repeatedly iterating through the specified ranges for each query.

• Runtime Complexity: O(Q log Q + N), Storage Complexity: O(N), where Q is the number of queries and N is the length of the nums array.

Code

    def solve():
    nums = [2, 0, 2]
    queries = [[0, 2, 1], [0, 2, 1], [1, 1, 3]]
    # nums = [4,3,2,1]
    # queries = [[1,3,2],[0,2,1]]

    def check(k):
        temp_nums = nums[:]
        diff = [0] * len(nums)

        for i in range(k):
            l, r, val = queries[i]
            diff[l] += val
            if r + 1 < len(nums):
                diff[r + 1] -= val

        curr = 0
        for i in range(len(nums)):
            curr += diff[i]
            decrement = min(temp_nums[i], curr)
            temp_nums[i] -= curr

            if temp_nums[i] < 0:
                temp_nums[i] = 0


        for i in range(len(nums)):
          if temp_nums[i] != -(curr-diff[i]):
            pass


        return all(x == 0 for x in temp_nums)

    low = 0
    high = len(queries)
    ans = -1

    while low <= high:
        mid = (low + high) // 2
        if check(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    print(ans)

    nums = [2,0,2]
    queries = [[0,2,1],[0,2,1],[1,1,3]]

    def check_no_diff(k):
        temp_nums = nums[:]
        for i in range(k):
            l, r, val = queries[i]
            for j in range(l, r + 1):
                decrement = min(temp_nums[j], val)
                temp_nums[j] -= decrement

        return all(x == 0 for x in temp_nums)

def minimumQueries(nums, queries):
    n = len(nums)
    q = len(queries)

    def check(k):
        temp_nums = list(nums)
        diff = [0] * n

        for i in range(k):
            l, r, val = queries[i]
            diff[l] += val
            if r + 1 < n:
                diff[r + 1] -= val

        curr = 0
        for i in range(n):
            curr += diff[i]
            temp_nums[i] = max(0, temp_nums[i] - curr) 

        return all(x == 0 for x in temp_nums)

    low = 0
    high = q
    ans = -1

    while low <= high:
        mid = (low + high) // 2
        if check(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans