# 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]], , , ]
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 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
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).