Leetcode: Random Pick Index

Posted on August 6, 2018September 10, 2019 by braindenny

Random Pick Index

Similar Problems:

Given an array of integers 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.

Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge.

Example:

int[] nums = new int[] {1,2,3,3,3};
Solution solution = new Solution(nums);

// pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(3);

// pick(1) should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(1);

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/random-pick-index
## Basic Ideas: Reservior Sampling
##
## Complexity: Time O(n), Space O(1)
import random
class Solution:

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.nums = nums

    def pick(self, target):
        """
        :type target: int
        :rtype: int
        """
        i, count, res = 0, 0, -1
        for i in range(0, len(self.nums)):
            if self.nums[i] == target:
                res = i
                count += 1
                break

        for i in range(res+1, len(self.nums)):
            if self.nums[i] == target:
                count += 1
                random_v = random.randint(1, count)
                if random_v == 1:
                    res = i
        return res

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

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