398 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.

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class Solution:

    def __init__(self, nums: List[int]):
        self.data = collections.defaultdict(list)
        for i, v in enumerate(nums):
            self.data[v].append(i)

    def pick(self, target: int) -> int:
        return random.choice(self.data[target])


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


'''
Reservoir sampling (TLE)
'''
class Solution:

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

    def pick(self, target: int) -> int:
        count, idx = 0, 0
        for i in range(len(self.data)):
            if self.data[i] == target:
                count += 1
                if random.randint(0, count-1) == 0:
                    idx = i
        return idx