# 528. Random Pick with Weight
'''
You 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.
'''
from common import *
import random
'''
Prefix sum with binary search.
O(n) runtime for __init__ and O(log(n)) runtime for pickIndex, O(1) storage as we use in-place storage.
Beat 63% runtime, 50% storage of all Leetcode submissions.
'''
class Solution:

    def __init__(self, w):
        for i in range(1,len(w)):
            w[i] += w[i-1]
        self.w = w

    def pickIndex(self) -> int:
        pick = random.randint(1,self.w[-1])
        i,j = 0,len(self.w)
        while i < j:
            k = (i+j) // 2
            if self.w[k] >= pick:
                j = k
            else:
                i = k+1
        return i

# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

# Tests.
assert_call_sequence(globals(),["Solution","pickIndex"],[[[1]],[]],[[None,0]])
assert_call_sequence(globals(),["Solution","pickIndex","pickIndex"],[[[1,3]],[],[]],[[None,1,1],[None,1,0],[None,0,1],[None,0,0]])



# 528. Random Pick with Weight
'''
You 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.
'''
from common import *
import random
'''
Prefix sum with binary search.
O(n) runtime for __init__ and O(log(n)) runtime for pickIndex, O(1) storage as we use in-place storage.
Beat 63% runtime, 50% storage of all Leetcode submissions.
'''
class Solution:

    def __init__(self, w):
        for i in range(1,len(w)):
            w[i] += w[i-1]
        self.w = w

    def pickIndex(self) -> int:
        pick = random.randint(1,self.w[-1])
        i,j = 0,len(self.w)
        while i < j:
            k = (i+j) // 2
            if self.w[k] >= pick:
                j = k
            else:
                i = k+1
        return i

# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

# Tests.
assert_call_sequence(globals(),["Solution","pickIndex"],[[[1]],[]],[[None,0]])
assert_call_sequence(globals(),["Solution","pickIndex","pickIndex"],[[[1,3]],[],[]],[[None,1,1],[None,1,0],[None,0,1],[None,0,0]])



# 528. Random Pick with Weight You 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. Prefix sum with binary search. O(n) runtime for __init__ and O(log(n)) runtime for pickIndex, O(1) storage as we use in-place storage. Beat 63% runtime, 50% storage of all Leetcode submissions. # Your Solution object will be instantiated and called as such: # obj = Solution(w) # param_1 = obj.pickIndex() # Tests.