LeetCode in Kotlin

398. Random Pick Index

Medium

Given an integer array nums with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.

Implement the Solution class:

• Solution(int[] nums) Initializes the object with the array nums.
• int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i’s, then each index should have an equal probability of returning.

Example 1:

Input

["Solution", "pick", "pick", "pick"] 
[[[1, 2, 3, 3, 3]], [3], [1], [3]]

Output: [null, 4, 0, 2]

Explanation:

Solution solution = new Solution([1, 2, 3, 3, 3]); 
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning. 
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1. 
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.

Constraints:

• 1 <= nums.length <= 2 * 104
• -231 <= nums[i] <= 231 - 1
• target is an integer from nums.
• At most 104 calls will be made to pick.

Solution

import kotlin.random.Random

@Suppress("kotlin:S2245")
class Solution(nums: IntArray) {
    // O(n) time | O(n) space
    private val map: MutableMap<Int, MutableList<Int>>

    init {
        map = HashMap()
        for (i in nums.indices) {
            map.computeIfAbsent(
                nums[i]
            ) { ArrayList() }.add(i)
        }
    }

    fun pick(target: Int): Int {
        val list: List<Int> = map[target]!!
        return list[Random.nextInt(list.size)]
    }
}

/*
 * Your Solution object will be instantiated and called as such:
 * var obj = Solution(nums)
 * var param_1 = obj.pick(target)
 */

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