Leetcode 398: Random Pick Index

398. Random Pick Index

  • Total Accepted: 6047
  • Total Submissions: 16791
  • Difficulty: Medium
  • Contributors: Admin

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);

 

Code

class Solution(object):

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

    def pick(self, target):
        """
        :type target: int
        :rtype: int
        """
        count = 0
        for idx, v in enumerate(self.nums):
            if v == target:
                count += 1
                if random.randrange(count) == 0:
                    res = idx
        return res


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

 

Idea

Reservoir Sampling. O(n) time and O(1) space. n is the length of nums.

 

Reference

https://discuss.leetcode.com/topic/58301/simple-reservoir-sampling-solution

 

If someone requires O(1) pick speed, one can construct a dictionary recording indices in __init__. See: https://discuss.leetcode.com/topic/58635/o-n-constructor-o-1-pick-two-ways

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