528. Random Pick with Weight

ou are given an array of positive integers w where w[i] describes the weight of ith index (0-indexed).

We need to call the function pickIndex() which randomly returns an integer in the range [0, w.length - 1]. pickIndex() should return the integer proportional to its weight in the w array. For example, for w = [1, 3], the probability of picking the index 0 is 1 / (1 + 3) = 0.25 (i.e 25%) while the probability of picking the index 1 is 3 / (1 + 3) = 0.75 (i.e 75%).

More formally, the probability of picking index i is w[i] / sum(w).

 

Example 1:

Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]

Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. Since there is only one single element on the array the only option is to return the first element.

Example 2:

Input
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output
[null,1,1,1,1,0]

Explanation
Solution solution = new Solution([1, 3]);
solution.pickIndex(); // return 1. It's returning the second element (index = 1) that has probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It's returning the first element (index = 0) that has probability of 1/4.

Since this is a randomization problem, multiple answers are allowed so the following outputs can be considered correct :
[null,1,1,1,1,0]
[null,1,1,1,1,1]
[null,1,1,1,0,0]
[null,1,1,1,0,1]
[null,1,0,1,0,0]
......
and so on.

 

Constraints:

• 1 <= w.length <= 10000
• 1 <= w[i] <= 10^5
• pickIndex will be called at most 10000 times.

 

Answer: I just copied answer with top vote. Good idea to convert scale of sum to scale 1. Please checkout my comment below. 

Btw, Random r = new Random(); r.nextInt(i); // return integer between [0, i)

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class Solution { // explain probabilities[i]: // Math.random() is to randomly pick an value from 0 to 1 // We need to convert from scale 1 to scale sum // so that any value of Math.random() corresponds to 1 portion of scale sum // in below formula, w[i] is the original value of w // probabilities[i] - probabilities[i - 1] == w[i]/sum // which means in portion i, its length is w[i]/sum, in the scale of 0 to 1 // And this is the weighted probability in scale of 0 to 1 private double[] probabilities; public Solution(int[] w) { double sum = 0; this.probabilities = new double[w.length]; for(int weight : w) sum += weight; for(int i = 0; i < w.length; i++){ w[i] += (i == 0) ? 0 : w[i - 1]; probabilities[i] = w[i]/sum; } } public int pickIndex() { // Arrays.binarySearch returns the index of 1st bigger element return Math.abs(Arrays.binarySearch(this.probabilities, Math.random())) - 1; } } /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(w); * int param_1 = obj.pickIndex(); */

254. Factor Combinations

 Numbers can be regarded as product of its factors. For example,

8 = 2 x 2 x 2;
  = 2 x 4.

Write a function that takes an integer n and return all possible combinations of its factors.

Note:

1. You may assume that n is always positive.
2. Factors should be greater than 1 and less than n.

Example 1:

Input: 1
Output: []

Example 2:

Input: 37
Output:[]

Example 3:

Input: 12
Output:
[
  [2, 6],
  [2, 2, 3],
  [3, 4]
]

Example 4:

Input: 32
Output:
[
  [2, 16],
  [2, 2, 8],
  [2, 2, 2, 4],
  [2, 2, 2, 2, 2],
  [2, 4, 4],
  [4, 8]
]

My answer: didn't figure out solution, only think of sqrt() to restrict choices. Please take a look at the key point below.

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class Solution { public List<List<Integer>> getFactors(int n) { List<List<Integer>> result = new ArrayList<List<Integer>>(); helper(2, n , new ArrayList<Integer>(), result); return result; } private void helper(int start, int n, List<Integer> singleResult, List<List<Integer>> result) { // we use i * i < n to make sure the 1st factor of n is in ascending order // and it can avoid duplicate factor // because the larger factor corresponding the smaller factor won't be picked for (int i = start; i * i <= n; i ++) { if (n % i == 0) {// find i as valid factor singleResult.add(i); singleResult.add(n/i); result.add(new ArrayList<Integer>(singleResult)); // backtrack singleResult as we are using i as starting factor singleResult.remove(singleResult.size() - 1); // continue to get factors for n/i // KEY POINT: new start is i // because factor smaller than i must have already shown up in // previous recursion, no need to go over again helper(i, n/i, singleResult, result); // backtrack singleResult as we keep using i as starting factor singleResult.remove(singleResult.size() - 1); } } } }