LeetCode in Kotlin

2903. Find Indices With Index and Value Difference I

Easy

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

Solution

import kotlin.math.abs

class Solution {
    fun findIndices(nums: IntArray, indexDifference: Int, valueDifference: Int): IntArray {
        for (i in nums.indices) {
            for (j in i until nums.size) {
                if (j - i >= indexDifference && abs((nums[i] - nums[j]).toDouble()) >= valueDifference) {
                    return intArrayOf(i, j)
                }
            }
        }
        return intArrayOf(-1, -1)
    }
}

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