Description
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);
Solution
HashMap + nextInt, constructor O(n), pick O(1), S(n)
class Solution {
private Map<Integer, List<Integer>> map;
private Random random;
public Solution(int[] nums) {
map = new HashMap<>();
random = new Random();
for (int i = 0; i < nums.length; ++i) {
if (!map.containsKey(nums[i])) {
map.put(nums[i], new ArrayList<>());
}
map.get(nums[i]).add(i);
}
}
public int pick(int target) {
if (!map.containsKey(target)) {
return -1;
}
int index = random.nextInt(map.get(target).size());
return map.get(target).get(index);
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int param_1 = obj.pick(target);
*/
Reservoir sampling, constructor O(1), pick O(n), S(n)
class Solution {
private int[] nums;
private Random random;
public Solution(int[] nums) {
this.nums = nums;
random = new Random();
}
public int pick(int target) {
int index = -1;
int count = 0;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] != target) {
continue;
}
if (random.nextInt(++count) == 0) {
index = i;
}
}
return index;
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int param_1 = obj.pick(target);
*/
