Is Toss Strange Coins easy or hard?

Toss Strange Coins is rated Medium on LeetCode. The difficulty comes from recognizing the probability-based DP transition and implementing the state updates correctly without overwriting needed values.

Toss Strange Coins Python/Java solution

The typical implementation uses dynamic programming. Initialize a DP array where dp[0] = 1, iterate through each coin probability, and update states in reverse order to maintain correctness. The same logic works in Python, Java, C++, Go, and TypeScript.

What is the best approach for Toss Strange Coins?

The best approach uses dynamic programming with a space-optimized 1D DP array. It computes the probability distribution of head counts as coins are processed. Each step updates probabilities for getting j heads using the current coin's head and tail probabilities. The time complexity is O(n × target) and space complexity is O(target).

How to solve Toss Strange Coins in O(n × target)?

Define a DP array where dp[j] represents the probability of getting exactly j heads so far. For each coin with probability p, update the array from target down to 1 using dp[j] = dp[j] * (1 - p) + dp[j-1] * p, and update dp[0] *= (1 - p). This builds the probability distribution incrementally.

Is Toss Strange Coins asked at Google/Amazon/Meta?

Probability and dynamic programming problems like Toss Strange Coins commonly appear in interviews at companies such as Google, Amazon, and Meta. Variants that require computing probability distributions or expected outcomes using DP are especially common.

What data structure is used in Toss Strange Coins?

The main structure is a dynamic programming array that stores probabilities for each possible number of heads. The optimized solution uses a 1D array of size target + 1 to track probability states while iterating through the coins.

What is the time complexity of Toss Strange Coins?

The optimal dynamic programming solution runs in O(n × target) time, where n is the number of coins and target is the required number of heads. For each coin, the algorithm updates probabilities for all head counts up to the target.

Description Hints 4 Solution Video 3:59

1230. Toss Strange Coins

Medium58.1% acceptance1 companyPremium elsewhere, free here

You have some coins. The i-th coin has a probability prob[i] of facing heads when tossed.

Return the probability that the number of coins facing heads equals target if you toss every coin exactly once.

Example 1:

Input: prob = [0.4], target = 1
Output: 0.40000

Example 2:

Input: prob = [0.5,0.5,0.5,0.5,0.5], target = 0
Output: 0.03125

Constraints:

1 <= prob.length <= 1000
0 <= prob[i] <= 1
0 <= target <= prob.length
Answers will be accepted as correct if they are within 10^-5 of the correct answer.

Asked at 1 company

Twitch

Topics

Hints open one at a time

Hint 1

What about solving the problem with DP?

Hint 2

Use DP with two states dp[pos][cnt], where pos represents the pos-th coin and cnt is the number of heads seen so far.

Hint 3

You can do the transitions with a little bit math.

Hint 4

For the base case, when pos == n return (cnt == target) to filter out the invalid scenarios.

3:59

Still stuck?Read 2 approaches, or watch a walkthrough.

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1class Solution:
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Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[0.4]
1

Input

[0.5,0.5,0.5,0.5,0.5]
0

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