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1230. Toss Strange Coins

Description

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.

Solutions

Solution 1: Dynamic Programming

Let $f[i][j]$ represent the probability of having $j$ coins facing up in the first $i$ coins, and initially $f[0][0]=1$. The answer is $f[n][target]$.

Consider $f[i][j]$, where $i\ge 1$. If the current coin is facing down, then $f[i][j]=(1-p)\times f[i-1][j]$; If the current coin is facing up and $j>0$, then $f[i][j]=p\times f[i-1][j-1]$. Therefore, the state transition equation is:

$$f[i][j]=\{\begin{array}{ll}(1-p)\times f[i-1][j],& j=0\\ (1-p)\times f[i-1][j]+p\times f[i-1][j-1],& j>0\end{array}$$

where $p$ represents the probability of the $i$-th coin facing up.

We note that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1]$, so we can optimize the two-dimensional space into one-dimensional space.

The time complexity is $O(n\times target)$, and the space complexity is $O(target)$. Where $n$ is the number of coins.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public double probabilityOfHeads(double[] prob, int target) {
          int n = prob.length;
          double[][] f = new double[n + 1][target + 1];
          f[0][0] = 1;
          for (int i = 1; i <= n; ++i) {
              for (int j = 0; j <= Math.min(i, target); ++j) {
                  f[i][j] = (1 - prob[i - 1]) * f[i - 1][j];
                  if (j > 0) {
                      f[i][j] += prob[i - 1] * f[i - 1][j - 1];
                  }
              }
          }
          return f[n][target];
      }
  }
  

• class Solution {
  public:
      double probabilityOfHeads(vector<double>& prob, int target) {
          int n = prob.size();
          double f[n + 1][target + 1];
          memset(f, 0, sizeof(f));
          f[0][0] = 1;
          for (int i = 1; i <= n; ++i) {
              for (int j = 0; j <= min(i, target); ++j) {
                  f[i][j] = (1 - prob[i - 1]) * f[i - 1][j];
                  if (j > 0) {
                      f[i][j] += prob[i - 1] * f[i - 1][j - 1];
                  }
              }
          }
          return f[n][target];
      }
  };
  

• class Solution:
      def probabilityOfHeads(self, prob: List[float], target: int) -> float:
          n = len(prob)
          f = [[0] * (target + 1) for _ in range(n + 1)]
          f[0][0] = 1
          for i, p in enumerate(prob, 1):
              for j in range(min(i, target) + 1):
                  f[i][j] = (1 - p) * f[i - 1][j]
                  if j:
                      f[i][j] += p * f[i - 1][j - 1]
          return f[n][target]
  
  

• func probabilityOfHeads(prob []float64, target int) float64 {
  	n := len(prob)
  	f := make([][]float64, n+1)
  	for i := range f {
  		f[i] = make([]float64, target+1)
  	}
  	f[0][0] = 1
  	for i := 1; i <= n; i++ {
  		for j := 0; j <= i && j <= target; j++ {
  			f[i][j] = (1 - prob[i-1]) * f[i-1][j]
  			if j > 0 {
  				f[i][j] += prob[i-1] * f[i-1][j-1]
  			}
  		}
  	}
  	return f[n][target]
  }
  

• function probabilityOfHeads(prob: number[], target: number): number {
      const n = prob.length;
      const f = new Array(n + 1).fill(0).map(() => new Array(target + 1).fill(0));
      f[0][0] = 1;
      for (let i = 1; i <= n; ++i) {
          for (let j = 0; j <= target; ++j) {
              f[i][j] = f[i - 1][j] * (1 - prob[i - 1]);
              if (j) {
                  f[i][j] += f[i - 1][j - 1] * prob[i - 1];
              }
          }
      }
      return f[n][target];
  }
  
  

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