LeetCode 398. Random Pick Index

Description

https://leetcode.com/problems/random-pick-index/

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.

Explanation

Use a map to store the element indices. Return an random index if the element occurs more than once.

Python Solution

class Solution:

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

        self.counter = defaultdict(list)
        for i, num in enumerate(self.nums):
            self.counter[num].append(i)
        
    def pick(self, target: int) -> int:
        
        indices = self.counter[target]
        
        if len(indices) == 1:
            return indices[0]
        else:
            return random.choice(indices)
        


# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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