Leetcode: Random Pick with Weight

Posted on August 5, 2019October 30, 2019 by braindenny

Identity number which appears exactly once.

Similar Problems:

Given an array w of positive integers, where w[i] describes the weight of index i, write a function pickIndex which randomly picks an index in proportion to its weight.

Note:

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

Example 1:

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

Example 2:

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

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has one argument, the array w. pickIndex has no arguments. Arguments are always wrapped with a list, even if there aren't any.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/random-pick-with-weight
// Basic Ideas: presum + binarysearch
//
// Complexity: Time O(log(n)), Space O(n)
import "math/rand"
type Solution struct {
    presums []int
}


func Constructor(w []int) Solution {
    presums := make([]int, len(w))
    sum := 0
    for i, v := range w {
        sum += v
        presums[i] = sum
    }
    return Solution{presums:presums}
}

func (this *Solution) PickIndex() int {
    target := rand.Intn(this.presums[len(this.presums)-1])+1
    left, right := 0, len(this.presums)
    // result must exist in the list
    for left < right {
        mid := (right-left)/2+left
        if this.presums[mid] < target {
            // search right half
            left = mid+1
        } else {
            right = mid
        }
    }
    return left
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(w);
 * param_1 := obj.PickIndex();
 */

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