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