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);
class Solution {
Map<Integer,List<Integer>> map;
Random random;
public Solution(int[] nums) {
map = new HashMap<Integer, List<Integer>>();
random = new Random();
for(int i = 0; i < nums.length; i++) {
map.putIfAbsent(nums[i], new ArrayList<Integer>());
map.get(nums[i]).add(i);
}
}
public int pick(int target) {
int occurances = map.get(target).size();
int randomIndex = random.nextInt(occurances);
return map.get(target).get(randomIndex);
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int param_1 = obj.pick(target);
*/class Solution {
int[] nums;
Random random;
public Solution(int[] nums) {
this.nums = nums;
random = new Random();
}
public int pick(int target) {
int count = 0;
int index = 0;
for (int i = 0; i < nums.length; i++) {
if(nums[i] == target) {
count++;
// This if changes the probability of an index i to be selected
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);
*/Tip : To better understand the second implementation, run through some examples. e.g. int[] nums = [1, 2, 3] , target = 3 e.g. int[] nums = [1, 2, 3, 3] , target = 3 e.g. int[] nums = [1, 2, 3, 3, 3] , target = 3
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