Introduction

In this blog post, we will explore how to solve the LeetCode problem "3315" 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 array nums consisting of n prime integers. You need to construct an array ans of length n, such that, for each index i, the bitwise OR of ans[i] and ans[i] + 1 is equal to nums[i], i.e. ans[i] OR (ans[i] + 1) == nums[i]. Additionally, you must minimize each value of ans[i] in the resulting array. If it is not possible to find such a value for ans[i] that satisfies the condition, then set ans[i] = -1. Example 1: Input: nums = [2,3,5,7] Output: [-1,1,4,3] Explanation: For i = 0, as there is no value for ans[0] that satisfies ans[0] OR (ans[0] + 1) = 2, so ans[0] = -1. For i = 1, the smallest ans[1] that satisfies ans[1] OR (ans[1] + 1) = 3 is 1, because 1 OR (1 + 1) = 3. For i = 2, the smallest ans[2] that satisfies ans[2] OR (ans[2] + 1) = 5 is 4, because 4 OR (4 + 1) = 5. For i = 3, the smallest ans[3] that satisfies ans[3] OR (ans[3] + 1) = 7 is 3, because 3 OR (3 + 1) = 7. Example 2: Input: nums = [11,13,31] Output: [9,12,15] Explanation: For i = 0, the smallest ans[0] that satisfies ans[0] OR (ans[0] + 1) = 11 is 9, because 9 OR (9 + 1) = 11. For i = 1, the smallest ans[1] that satisfies ans[1] OR (ans[1] + 1) = 13 is 12, because 12 OR (12 + 1) = 13. For i = 2, the smallest ans[2] that satisfies ans[2] OR (ans[2] + 1) = 31 is 15, because 15 OR (15 + 1) = 31. Constraints: 1 <= nums.length <= 100 2 <= nums[i] <= 109 nums[i] is a prime number.

Explanation

Here's the solution to the problem:

• Key Idea: The core idea is to recognize the bit pattern characteristic of prime numbers that satisfy the ans[i] OR (ans[i] + 1) == nums[i] condition. Specifically, nums[i] can be expressed as the OR of two consecutive numbers ans[i] and ans[i] + 1 only if nums[i] has the form 2^k - 1 is not prime (except for 3 which is 2^2 -1). If nums[i] is of the form 2^k -1 then the answer is nums[i]/2 -1. Otherwise, ans[i] is nums[i] - (nums[i] & -nums[i]) -1. Another way of saying is, find the rightmost set bit in nums[i]. That bit indicates the number we have to subtract from nums[i] so that we get the number ans[i] where ans[i] OR ans[i]+1 == nums[i].

• Prime Check (Optimization): Because the input consists of prime numbers, special handling can be done to optimize. If prime number is of the form 2^k - 1, we check if we can find a valid answer as described above, if not, the answer is -1. The bit manipulation to find the rightmost set bit can be optimized and done efficiently.

• Edge Cases: Handles cases where no suitable ans[i] can be found, returning -1. Specifically, if nums[i] is 2, no solution exists.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the input array nums.
  • Storage Complexity: O(n), for the ans array.

Code

    def construct_array(nums):
    ans = []
    for num in nums:
        if num == 2:
            ans.append(-1)
            continue

        val = num - (num & -num)
        ans.append(val)

    return ans