Find the number of players who roll the dice when the dice output sequence is given

0 0 3 minutes read

Given a string S and a number X. There are M players who roll the dice. A player keeps on rolling the dice until he gets a number other than X. In the string S, S[i] represents the number at i^th roll of a dice. The task is to find M. Note that the last character in S will never be X.
Examples:

Input: s = “3662123”, X = 6
Output: 5
First player rolls and gets 3.
Second player rolls and gets 6, 6 and 2.
Third player rolls and gets 1.
Fourth player rolls and gets 2.
Fifth player rolls and gets 3.
Input: s = “1234223”, X = 2
Output: 4

Approach: Iterate in the string and count the characters which are not X. The number of characters which are not X will be the number of players.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of players
int findM(string s, int x)
{
 
    // Initialize cnt as 0
    int cnt = 0;
 
    // Iterate in the string
    for (int i = 0; i < s.size(); i++) {
 
        // Check for numbers other than x
        if (s[i] - '0' != x)
            cnt++;
    }
    return cnt;
}
 
// Driver code
int main()
{
    string s = "3662123";
    int x = 6;
    cout << findM(s, x);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the number of players
static int findM(String s, int x)
{
 
    // Initialize cnt as 0
    int cnt = 0;
 
    // Iterate in the string
    for (int i = 0; i < s.length(); i++)
    {
 
        // Check for numbers other than x
        if (s.charAt(i) - '0' != x)
            cnt++;
    }
    return cnt;
}
 
// Driver code
public static void main(String args[])
{
    String s = "3662123";
    int x = 6;
    System.out.println(findM(s, x));
 
}
}
 
//This code is contributed by
// Surendra_Gangwar

Python3

# Python 3 implementation of the approach
 
# Function to return the number of players
def findM(s, x):
     
    # Initialize cnt as 0
    cnt = 0
 
    # Iterate in the string
    for i in range(len(s)):
         
        # Check for numbers other than x
        if (ord(s[i]) - ord('0') != x):
            cnt += 1
 
    return cnt
 
# Driver code
if __name__ == '__main__':
    s = "3662123"
    x = 6
    print(findM(s, x))
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the number of players
static int findM(String s, int x)
{
 
    // Initialize cnt as 0
    int cnt = 0;
 
    // Iterate in the string
    for (int i = 0; i < s.Length; i++)
    {
 
        // Check for numbers other than x
        if (s[i] - '0' != x)
            cnt++;
    }
    return cnt;
}
 
// Driver code
public static void Main()
{
    String s = "3662123";
    int x = 6;
    Console.Write(findM(s, x));
}
}
 
// This code is contributed by
// mohit kumar

PHP

<?php
// PHP implementation of the approach 
 
// Function to return the number of players 
function findM($s, $x) 
{ 
 
    // Initialize cnt as 0 
    $cnt = 0; 
 
    // Iterate in the string 
    for ($i = 0; $i < strlen($s); $i++) 
    { 
 
        // Check for numbers other than x 
        if (ord($s[$i]) - ord('0') != $x) 
            $cnt++; 
    } 
    return $cnt; 
} 
 
// Driver code 
$s = "3662123"; 
$x = 6; 
 
echo findM($s, $x); 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
// javascript implementation of the approach
 
    // Function to return the number of players
    function findM( s , x) {
 
        // Initialize cnt as 0
        var cnt = 0;
 
        // Iterate in the string
        for (i = 0; i < s.length; i++) {
 
            // Check for numbers other than x
            if (s.charCodeAt(i) - '0'.charCodeAt(0) != x)
                cnt++;
        }
        return cnt;
    }
 
    // Driver code
     
        var s = "3662123";
        var x = 6;
        document.write(findM(s, x));
 
// This code contributed by Rajput-Ji 
</script>

Output:

Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.

Auxiliary Space: O(1), as we are not using any extra.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!