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 arraynums
.int pick(int target)
Picks a random indexi
fromnums
wherenums[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 fromnums
.- At most
104
calls will be made topick
.
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).