Question
Formatted question description: https://leetcode.ca/all/398.html
398 Random Pick Index
Given an array of integers 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.
Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge.
Example:
int[] nums = new int[] {1,2,3,3,3};
Solution solution = new Solution(nums);
// pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(3);
// pick(1) should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(1);
Algorithm
Define two variables, counter cnt and return result res.
Traverse the entire array,
- If the value of the array is not equal to target, skip directly;
- If it is equal to target, the counter is increased by 1,
Then we randomly generate a number in the range of [0,cnt)
. If this number is 0, we assign res.
Code
Java
import java.util.Random;
public class Random_Pick_Index {
// ref: https://leetcode.com/problems/random-pick-index/discuss/88072/Simple-Reservoir-Sampling-solution
class Solution {
int[] nums;
Random rnd;
public Solution(int[] nums) {
this.nums = nums;
this.rnd = new Random();
}
public int pick(int target) {
int resultIndex = -1;
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != target) {
continue;
}
count++;
if (rnd.nextInt(count) == 0) {
resultIndex = i;
// first time nextInt is always 0, so set as first index
// for the rest, probobility issue
}
}
return resultIndex;
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int param_1 = obj.pick(target);
*/
}