# 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).