Introduction

In this blog post, we will explore how to solve the LeetCode problem "2903" 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 a 0-indexed integer array nums having length n, an integer indexDifference, and an integer valueDifference. Your task is to find two indices i and j, both in the range [0, n - 1], that satisfy the following conditions: abs(i - j) >= indexDifference, and abs(nums[i] - nums[j]) >= valueDifference Return an integer array answer, where answer = [i, j] if there are two such indices, and answer = [-1, -1] otherwise. If there are multiple choices for the two indices, return any of them. Note: i and j may be equal. Example 1: Input: nums = [5,1,4,1], indexDifference = 2, valueDifference = 4 Output: [0,3] Explanation: In this example, i = 0 and j = 3 can be selected. abs(0 - 3) >= 2 and abs(nums[0] - nums[3]) >= 4. Hence, a valid answer is [0,3]. [3,0] is also a valid answer. Example 2: Input: nums = [2,1], indexDifference = 0, valueDifference = 0 Output: [0,0] Explanation: In this example, i = 0 and j = 0 can be selected. abs(0 - 0) >= 0 and abs(nums[0] - nums[0]) >= 0. Hence, a valid answer is [0,0]. Other valid answers are [0,1], [1,0], and [1,1]. Example 3: Input: nums = [1,2,3], indexDifference = 2, valueDifference = 4 Output: [-1,-1] Explanation: In this example, it can be shown that it is impossible to find two indices that satisfy both conditions. Hence, [-1,-1] is returned. Constraints: 1 <= n == nums.length <= 100 0 <= nums[i] <= 50 0 <= indexDifference <= 100 0 <= valueDifference <= 50

Explanation

Here's the breakdown of the solution:

• Iterate and Check: The core idea is to iterate through all possible pairs of indices (i, j) in the nums array and check if they satisfy the given conditions: abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference.
• Early Exit: If a suitable pair (i, j) is found, immediately return the pair as an array [i, j]. This avoids unnecessary further iterations.

• Handle No Solution: If the outer loop completes without finding a valid pair, it means no such pair exists, so return [-1, -1].

• Time Complexity: O(n^2), where n is the length of the nums array. This is due to the nested loops that iterate through all possible pairs of indices.

• Space Complexity: O(1). The solution uses a constant amount of extra space, regardless of the input size.

Code

    def findIndices(nums, indexDifference, valueDifference):
    """
    Finds two indices i and j in the nums array that satisfy the given conditions.

    Args:
        nums (list[int]): A 0-indexed integer array.
        indexDifference (int): An integer representing the minimum index difference.
        valueDifference (int): An integer representing the minimum value difference.

    Returns:
        list[int]: An integer array answer = [i, j] if there are two such indices,
                   and answer = [-1, -1] otherwise.
    """
    n = len(nums)

    for i in range(n):
        for j in range(n):
            if abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference:
                return [i, j]

    return [-1, -1]