Given a matrix M of order N*M where Mij represent the occurrence of j in row i. The task is to find the probability of occurrence of given k in last row after applying the given operation:
- Starting from 1st row, we select one element from any column of the entire row, and add it to the same column of next row. We repeating this till the last row.
Examples:
Input: k = 1, M[][] =
{{0, 1},
{1, 1}}
Output:0.666667
Input: k = 1, M[][] =
{{0, 1},
{1, 1}}
Output: 0.333333
Approach :
- Pre-calculate a sum[] array which stores the sum of all elements of 1st row.
- Fill 1st row of dp[][] by the first row of M[][] elements.
- Calculate the probability of selecting an element from each column of a particular row and add the same to the corresponding column.
- Also, update the Sum[] of the row while updating the value of dp[][] matrix.
- Finally, the value of dp[n][k] for given k is the required value for the probability of selecting element k after all iterations.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define n 4#define m 4using namespace std;// Function to calculate probabilityfloat calcProbability(int M[][m], int k){ // declare dp[][] and sum[] float dp[m][n], sum[n]; // precalculate the first row for (int j = 0; j < n; j++) { dp[0][j] = M[0][j]; sum[0] += dp[0][j]; } // calculate the probability for // each element and update dp table for (int i = 1; i < m; i++) { for (int j = 0; j < n; j++) { dp[i][j] += dp[i - 1][j] / sum[i - 1] + M[i][j]; sum[i] += dp[i][j]; } } // return result return dp[n - 1][k - 1] / sum[n - 1];}// Driver codeint main(){ int M[m][n] = { { 1, 1, 0, 3 }, { 2, 3, 2, 3 }, { 9, 3, 0, 2 }, { 2, 3, 2, 2 } }; int k = 3; cout << calcProbability(M, k); return 0;} |
Java
// Java implementation of the above approach class GFG{ final static int n = 4 ; final static int m = 4 ; // Function to calculate probability static float calcProbability(int M[][], int k) { // declare dp[][] and sum[] float dp[][] = new float[m][n] ; float sum[] = new float[n]; // precalculate the first row for (int j = 0; j < n; j++) { dp[0][j] = M[0][j]; sum[0] = sum[0] + dp[0][j]; } // calculate the probability for // each element and update dp table for (int i = 1; i < m; i++) { for (int j = 0; j < n; j++) { dp[i][j] += dp[i - 1][j] / sum[i - 1] + M[i][j]; sum[i] += dp[i][j]; } } // return result return dp[n - 1][k - 1] / sum[n - 1]; } // Driver code public static void main(String []args) { int M[][] = { { 1, 1, 0, 3 }, { 2, 3, 2, 3 }, { 9, 3, 0, 2 }, { 2, 3, 2, 2 } }; int k = 3; System.out.println(calcProbability(M, k)); } }// This code is contributed by Ryuga |
Python3
# Python3 implementation of the # above approachn = 4m = 4# Function to calculate probabilitydef calcProbability(M, k): # declare dp[][] and sum[] dp = [[0 for i in range(n)] for i in range(m)] Sum = [0 for i in range(n)] # precalculate the first row for j in range(n): dp[0][j] = M[0][j] Sum[0] += dp[0][j] # calculate the probability for # each element and update dp table for i in range(1, m): for j in range(n): dp[i][j] += (dp[i - 1][j] / Sum[i - 1] + M[i][j]) Sum[i] += dp[i][j] # return result return dp[n - 1][k - 1] / Sum[n - 1]# Driver codeM = [[ 1, 1, 0, 3 ], [ 2, 3, 2, 3 ], [ 9, 3, 0, 2 ], [ 2, 3, 2, 2 ]] k = 3print(calcProbability(M, k))# This code is contributed# by mohit kumar 29 |
C#
// C# implementation of the above approach using System;class GFG{ static int n = 4 ; static int m = 4 ; // Function to calculate probability static float calcProbability(int[,] M, int k) { // declare dp[][] and sum[] float[,] dp = new float[m,n] ; float[] sum = new float[n]; // precalculate the first row for (int j = 0; j < n; j++) { dp[0, j] = M[0, j]; sum[0] = sum[0] + dp[0, j]; } // calculate the probability for // each element and update dp table for (int i = 1; i < m; i++) { for (int j = 0; j < n; j++) { dp[i, j] += dp[i - 1,j] / sum[i - 1] + M[i, j]; sum[i] += dp[i, j]; } } // return result return dp[n - 1,k - 1] / sum[n - 1]; } // Driver code public static void Main() { int[,] M = { { 1, 1, 0, 3 }, { 2, 3, 2, 3 }, { 9, 3, 0, 2 }, { 2, 3, 2, 2 } }; int k = 3; Console.Write(calcProbability(M, k)); } }// This code is contributed by Ita_c. |
PHP
<?php// PHP implementation of the above approach// Function to calculate probabilityfunction calcProbability($M, $k){ $m = 4; $n = 4; // declare dp[][] and sum[] $dp = array(); $sum = array(); // precalculate the first row for ($j = 0; $j < $n; $j++) { $dp[0][$j] = $M[0][$j]; $sum[0] += $dp[0][$j]; } // calculate the probability for // each element and update dp table for ($i = 1; $i < $m; $i++) { for ($j = 0; $j <$n; $j++) { $dp[$i][$j] += $dp[$i - 1][$j] / $sum[$i - 1] + $M[$i][$j]; $sum[$i] += $dp[$i][$j]; } } // return result return $dp[$n - 1][$k - 1] / $sum[$n - 1];}// Driver code$M = array(array(1, 1, 0, 3), array(2, 3, 2, 3), array(9, 3, 0, 2), array(2, 3, 2, 2));$k = 3;echo calcProbability($M, $k);// This code is contributed // by Arnub Kundu?> |
Javascript
<script>// Javascript implementation of the above approach let n = 4 ; let m = 4 ; // Function to calculate probability function calcProbability(M,k) { // declare dp[][] and sum[] let dp = new Array(m) ; let sum = new Array(n); for(let i = 0; i < dp.length; i++) { dp[i] = new Array(n); for(let j = 0; j < n; j++) { dp[i][j] = 0; } } for(let i = 0; i < n; i++) { sum[i] = 0; } // precalculate the first row for (let j = 0; j < n; j++) { dp[0][j] = M[0][j]; sum[0] = sum[0] + dp[0][j]; } // calculate the probability for // each element and update dp table for (let i = 1; i < m; i++) { for (let j = 0; j < n; j++) { dp[i][j] += dp[i - 1][j] / sum[i - 1] + M[i][j]; sum[i] += dp[i][j]; } } // return result return dp[n - 1][k - 1] / sum[n - 1]; } // Driver code let M = [[ 1, 1, 0, 3 ], [ 2, 3, 2, 3 ], [ 9, 3, 0, 2 ], [ 2, 3, 2, 2 ]]; let k = 3; document.write(calcProbability(M, k)); // This code is contributed by rag2127</script> |
Output
0.201212
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