528. Random Pick with Weight #16
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syoshida20
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May 3, 2025
| * 発想 | ||
| * 重み付きの確率を行う方法。 | ||
| * `w_0`, `w_1`, `w_2`, ...をsumで割り算し`w_0`/sum, `w_1`/sum, `w_2`/sum, ...とした後に、累積和をとる。 | ||
| 0から1のランダムな値がどの重みに属するかを調べることで、重みを考慮したサンプリングできそう。 |
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確率だからよいかと思う一方で、浮動小数点は色々なことが起きるので、自然数で処理したほうが私は好みです。
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- 浮動小数点の懸念事項
- 整数で解く方法
を意識ができておりませんでした。ご指摘、ありがとうございます。
懸念項目の一例を調べました。
- 丸誤差
- 有効数字の桁が落ちる
- 情報が落ちる
- 演算の順序性が保たれない
参考:
| sum += weight_i | ||
| } | ||
| for (let i=0; i<w.length; i++) { | ||
| n const weight_i = w[i] |
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wという変数が重み( weight )を表すことは自明ということですね。
| ## その他の解法 | ||
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| * 二分探索のターゲットを変更する。 | ||
| * 見つけたいターゲット: i-1番目の累積和 <= `random_value` < i番目の累積和となるiの値 (先ほどは、 i-1 < `random_value` <= iで等号の場所を変更した) |
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等号はこっちのほうが自然でしょうね。Math.random は 0 は入って 1 は入らないので。
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- random_valueが1を含む場合には、<=としてあげないと、1が出た際のiが見つからなくなる.
- random_valueが1を含まない場合には、 <=としても1がそもそも出ないとで等号が使われない。
と理解しました。
| * 二分探索において、ターゲットと引き継ぎ条件を考えることで、 | ||
| left,rightの設定や等号の有無について迷わなくなった。 | ||
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| * 前のレビュー依頼から時間が空いてしまったので、もう一度習慣化する。 |
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そうですね。1週間休んでも終わるのが1週間遅れるだけですが、頻度が半分になると時間が倍になりますからね。特に問題ないので進みましょう。
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はい、自分ができるペースで、頻度を意識してコーディング練習会に臨もうと思います。
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問題URL
https://leetcode.com/problems/random-pick-with-weight/
問題文
You are given a 0-indexed array of positive integers w where w[i] describes the weight of the ith index.
You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w).
For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).
Example 1:
Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]
Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.
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 is returning the second element (index = 1) that has a probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.
Since this is a randomization problem, multiple answers are allowed.
All of 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 <= 104
1 <= w[i] <= 105
pickIndex will be called at most 104 times.