Introduction

In this blog post, we will explore how to solve the LeetCode problem "3048" 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 two 1-indexed integer arrays, nums and, changeIndices, having lengths n and m, respectively. Initially, all indices in nums are unmarked. Your task is to mark all indices in nums. In each second, s, in order from 1 to m (inclusive), you can perform one of the following operations: Choose an index i in the range [1, n] and decrement nums[i] by 1. If nums[changeIndices[s]] is equal to 0, mark the index changeIndices[s]. Do nothing. Return an integer denoting the earliest second in the range [1, m] when all indices in nums can be marked by choosing operations optimally, or -1 if it is impossible. Example 1: Input: nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1] Output: 8 Explanation: In this example, we have 8 seconds. The following operations can be performed to mark all indices: Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0]. Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0]. Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0]. Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0]. Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0. Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0. Second 7: Do nothing. Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0. Now all indices have been marked. It can be shown that it is not possible to mark all indices earlier than the 8th second. Hence, the answer is 8. Example 2: Input: nums = [1,3], changeIndices = [1,1,1,2,1,1,1] Output: 6 Explanation: In this example, we have 7 seconds. The following operations can be performed to mark all indices: Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2]. Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1]. Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0]. Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0. Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0]. Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0. Now all indices have been marked. It can be shown that it is not possible to mark all indices earlier than the 6th second. Hence, the answer is 6. Example 3: Input: nums = [0,1], changeIndices = [2,2,2] Output: -1 Explanation: In this example, it is impossible to mark all indices because index 1 isn't in changeIndices. Hence, the answer is -1. Constraints: 1 <= n == nums.length <= 2000 0 <= nums[i] <= 109 1 <= m == changeIndices.length <= 2000 1 <= changeIndices[i] <= n

Explanation

Here's the breakdown of the solution:

• Binary Search: We use binary search to find the earliest second at which all indices can be marked. We check if it's possible to mark all indices within a given number of seconds.
• Greedy Strategy: For a given number of seconds, we iterate through changeIndices and use a greedy strategy. We prioritize using the last occurrences of each index in changeIndices to mark them. This ensures that we have maximum flexibility to decrement nums before marking.

• Checking Feasibility: In the check function, we simulate the process of marking indices up to a given second. We keep track of the last occurrence of each index in changeIndices. We then simulate the decrementing process up to that second, prioritizing indices not marked already.

• Time Complexity: O(n log m + m), where n is the length of nums and m is the length of changeIndices. The binary search contributes O(n log m) (the feasibility check is O(n) and it is run log m times). The initial computation of last array contributes O(m). Thus, the overall time complexity is O(n log m + m).

• Space Complexity: O(n + m)

Code

    def earliestSecondToMarkAllIndices(nums, changeIndices):
    n = len(nums)
    m = len(changeIndices)

    if n > len(set(changeIndices)):
        return -1

    def check(seconds):
        last = [-1] * (n + 1)
        for i in range(seconds):
            last[changeIndices[i]] = i

        marked = [False] * n
        ops = 0
        for i in range(seconds):
            idx = changeIndices[i] - 1
            if last[changeIndices[i]] == i:
                marked[idx] = True
            else:
                ops += 1

        needed = 0
        for i in range(n):
            if not marked[i]:
                needed += nums[i]

        remaining_ops = ops
        for i in range(n):
            if not marked[i]:
                if remaining_ops >= nums[i]:
                    remaining_ops -= nums[i]
                else:
                    return False

        return True

    left, right = 1, m
    ans = -1
    while left <= right:
        mid = (left + right) // 2
        if check(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans