Leetcode: 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);
Three types of answer:
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Map Solution, O(N) memory, O(N) init, O(1) pick.
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Like @dettier's Reservoir Sampling. O(1) init, O(1) memory, but O(N) to pick.
Reservior Sampling
1 public class Solution { 2 int[] arr; 3 Random random; 4 5 6 public Solution(int[] nums) { 7 arr = nums; 8 random = new Random(); 9 } 10 11 public int pick(int target) { 12 int count = 0; 13 int index = 0; 14 for (int i=0; i<arr.length; i++) { 15 if (arr[i] == target) { 16 count++; 17 if (random.nextInt(count) == 0) { 18 index = i; 19 } 20 } 21 } 22 return index; 23 } 24 } 25 26 /** 27 * Your Solution object will be instantiated and called as such: 28 * Solution obj = new Solution(nums); 29 * int param_1 = obj.pick(target); 30 */
Map solution: MLE for big case
1 public class Solution { 2 3 public Solution(int[] nums) { 4 for (int i=0; i<nums.length; i++) { 5 int num = nums[i]; 6 if (!indexes.containsKey(num)) 7 indexes.put(num, new ArrayList<Integer>()); 8 indexes.get(num).add(i); 9 } 10 } 11 12 public int pick(int target) { 13 List<Integer> indexes = this.indexes.get(target); 14 int i = (int) (Math.random() * indexes.size()); 15 return indexes.get(i); 16 } 17 18 private Map<Integer, List<Integer>> indexes = new HashMap<>(); 19 }