Leetcode #398: Random Pick Index
In this guide, we solve Leetcode #398 Random Pick Index in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Reservoir Sampling, Hash Table, Math, Randomized
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
def pick(self, target: int) -> int:
n = ans = 0
for i, v in enumerate(self.nums):
if v == target:
n += 1
x = random.randint(1, n)
if x == n:
ans = i
return ans
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.