http://bookshadow.com/weblog/2016/09/11/leetcode-random-pick-index/
https://discuss.leetcode.com/topic/58295/share-my-c-solution-o-lg-n-to-pick-o-nlg-n-for-sorting
Pre-process sorting for O(nlg(n))
Pick in O(lg(n)) using binary search
O(n) space to store value/index pairs
X. O(1) init, O(n) pick
https://discuss.leetcode.com/topic/58301/simple-reservoir-sampling-solution
https://discuss.leetcode.com/topic/58467/java-o-n-variant-of-reservoir-sampling
https://discuss.leetcode.com/topic/58356/o-n-for-java-any-other-good-idea
def __init__(self, nums): """ :type nums: List[int] :type numsSize: int """ size = len(nums) self.next = [0] * (size + 1) self.head = collections.defaultdict(int) for i, n in enumerate(nums): self.next[i + 1] = self.head[n] self.head[n] = i + 1 def pick(self, target): """ :type target: int :rtype: int """ cnt = 0 idx = self.head[target] while idx > 0: cnt += 1 idx = self.next[idx] c = int(random.random() * cnt) idx = self.head[target] for x in range(c): idx = self.next[idx] return idx - 1
X. Offline
https://discuss.leetcode.com/topic/58394/simple-java-solution-o-n
This is appropriate when this api is called many times
https://github.com/mintycc/OnlineJudge-Solutions/blob/master/Leetcode/398_Random_Pick_Index.java
1. reservoir sampling: n个数⾥里里⾯面随机选k个数，要求概率相等。
2. 变种：给⼀一⼀一个vector，求最⼤大元素的index。如果有多个最⼤大元素，
均匀地随机返回任意⼀一⼀一个index。⽐比⽐比如:[1, 2, 3, 3]，随机返回2，3，
每个的概率是50%。解法:扫 ⼀一遍记录最⼤大值和index，返回的时候⽤用随机
数rand()⼀一⼀一下（只记录最⼤大值的所有index即可）
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.
The array size can be very large. Solution that uses too much extra space will not pass the judge.
Example:
https://discuss.leetcode.com/topic/58322/what-on-earth-is-meant-by-too-much-memory- Like mine, O(N) memory, O(N) init, O(1) pick.
- Like @dettier's Reservoir Sampling. O(1) init, O(1) memory, but O(N) to pick.
- Like @chin-heng's binary search: O(N) memory, O(N lg N) init, O(lg N) pick.
public Solution(int[] nums) {
for (int i=0; i<nums.length; i++) {
int num = nums[i];
if (!indexes.containsKey(num))
indexes.put(num, new ArrayList<Integer>());
indexes.get(num).add(i);
}
}
public int pick(int target) {
List<Integer> indexes = this.indexes.get(target);
int i = (int) (Math.random() * indexes.size());
return indexes.get(i);
}
private Map<Integer, List<Integer>> indexes = new HashMap<>();https://discuss.leetcode.com/topic/58295/share-my-c-solution-o-lg-n-to-pick-o-nlg-n-for-sorting
Pre-process sorting for O(nlg(n))
Pick in O(lg(n)) using binary search
O(n) space to store value/index pairs
typedef pair<int, int> pp; // <value, index>
static bool comp(const pp& i, const pp& j) { return (i.first < j.first); }
vector<pp> mNums;
Solution(vector<int> nums) {
for(int i = 0; i < nums.size(); i++) {
mNums.push_back(pp({nums[i], i}));
}
sort(mNums.begin(), mNums.end(), comp);
}
int pick(int target) {
pair<vector<pp>::iterator, vector<pp>::iterator> bounds = equal_range(mNums.begin(), mNums.end(), pp({target,0}), comp);
int s = bounds.first - mNums.begin();
int e = bounds.second - mNums.begin();
int r = e - s;
return mNums[s + (rand() % r)].second;
}X. O(1) init, O(n) pick
https://discuss.leetcode.com/topic/58301/simple-reservoir-sampling-solution
So basically you may return -1 at the end? Correct me if I am wrong, I think the question requires a result which has been guaranteed.
Oops, I got it. Actually
random.nextInt(++counter) this will guarantee that at least the first one will be picked u
this is because we need to save current element with probability 1 / (total # of encountered candidates). So it's 100% for first target element, 50% for the second one, and so on...
public class Solution {
int[] nums;
Random rnd;
public Solution(int[] nums) {
this.nums = nums;
this.rnd = new Random();
}
public int pick(int target) {
int result = -1;
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != target)
continue;
if (rnd.nextInt(++count) == 0)
result = i;
}
return result;
}
}https://discuss.leetcode.com/topic/58467/java-o-n-variant-of-reservoir-sampling
https://discuss.leetcode.com/topic/58356/o-n-for-java-any-other-good-idea
private int[] nums;
public Solution(int[] nums) {
this.nums = nums;
}
private Random r = new Random();
public int pick(int target) {
int ret = -1;
if (nums == null) {
return ret;
}
int upbound = 1;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == target) {
if (r.nextInt(upbound) == 0) {
ret = i;
}
upbound++;
}
}
return ret;
}def __init__(self, nums): """ :type nums: List[int] :type numsSize: int """ size = len(nums) self.next = [0] * (size + 1) self.head = collections.defaultdict(int) for i, n in enumerate(nums): self.next[i + 1] = self.head[n] self.head[n] = i + 1 def pick(self, target): """ :type target: int :rtype: int """ cnt = 0 idx = self.head[target] while idx > 0: cnt += 1 idx = self.next[idx] c = int(random.random() * cnt) idx = self.head[target] for x in range(c): idx = self.next[idx] return idx - 1
X. Offline
https://discuss.leetcode.com/topic/58394/simple-java-solution-o-n
This is appropriate when this api is called many times
static int[] nums;
public Solution(int[] nums) {
this.nums=nums;
}
public int pick(int target) {
List<Integer> mm = new ArrayList<Integer>();
for(int i=0; i<nums.length; i++){
if(nums[i]==target)
mm.add(i);
}
Random rn=new Random();
return mm.get(rn.nextInt(mm.size()));
}https://github.com/mintycc/OnlineJudge-Solutions/blob/master/Leetcode/398_Random_Pick_Index.java
1. reservoir sampling: n个数⾥里里⾯面随机选k个数，要求概率相等。
2. 变种：给⼀一⼀一个vector，求最⼤大元素的index。如果有多个最⼤大元素，
均匀地随机返回任意⼀一⼀一个index。⽐比⽐比如:[1, 2, 3, 3]，随机返回2，3，
每个的概率是50%。解法:扫 ⼀一遍记录最⼤大值和index，返回的时候⽤用随机
数rand()⼀一⼀一下（只记录最⼤大值的所有index即可）
