Given a string S of length N and an integer P(1?P?N) denoting the position of a character in the string. The task is to find the number of occurrences of the character present at position P up to P-1 index.
Examples:
Input : S = “ababababab”, P = 9
Output : 4
Character at P is ‘a’. Number of occurrences of ‘a’ upto 8th index is 4Input : S = “geeksforgeeks”, P = 9
Output : 1
Naive Approach:A naive approach is to iterate over the string till Position-1 searches for a similar character. Whenever a similar character occurs increment the counter variable by one.
Below is the implementation of the above approach:
// CPP program to find the number of occurrences // of a character at position P upto p-1 #include <bits/stdc++.h> using namespace std;
// Function to find the number of occurrences // of a character at position P upto p-1 int Occurrence(string s, int position)
{ int count = 0;
for ( int i = 0; i < position - 1; i++)
if (s[i] == s[position - 1])
count++;
// Return the required count
return count;
} // Driver code int main()
{ string s = "ababababab" ;
int p = 9;
// Function call
cout << Occurrence(s, p);
return 0;
} |
// Java program to find the number of occurrences // of a character at position P upto p-1 class GFG
{ // Function to find the number of occurrences // of a character at position P upto p-1 static int Occurrence(String s, int position)
{ int count = 0 ;
for ( int i = 0 ; i < position - 1 ; i++)
if (s.charAt(i) == s.charAt(position - 1 ))
count++;
// Return the required count
return count;
} // Driver code public static void main(String[] args)
{ String s = "ababababab" ;
int p = 9 ;
// Function call
System.out.println(Occurrence(s, p));
} } // This code is contributed by 29AjayKumar |
# Python3 program to find the # number of occurrences of # a character at position P upto p-1 # Function to find the number of occurrences # of a character at position P upto p-1 def Occurrence(s, position):
count = 0
for i in range (position - 1 ):
if (s[i] = = s[position - 1 ]):
count + = 1
# Return the required count
return count
# Driver code s = "ababababab" ;
p = 9
# Function call print (Occurrence(s, p))
# This code is contributed by Mohit Kumar |
// C# program to find the number of occurrences // of a character at position P upto p-1 using System;
class GFG
{ // Function to find the number of occurrences // of a character at position P upto p-1 static int Occurrence(String s, int position)
{ int count = 0;
for ( int i = 0; i < position - 1; i++)
if (s[i] == s[position - 1])
count++;
// Return the required count
return count;
} // Driver code public static void Main(String[] args)
{ String s = "ababababab" ;
int p = 9;
// Function call
Console.WriteLine(Occurrence(s, p));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript program to find the number of occurrences // of a character at position P upto p-1 // Function to find the number of occurrences // of a character at position P upto p-1 function Occurrence(s,position)
{ let count = 0;
for (let i = 0; i < position - 1; i++)
if (s[i] == s[position - 1])
count++;
// Return the required count
return count;
} // Driver code let s = "ababababab" ;
let p = 9; // Function call document.write(Occurrence(s, p)); // This code is contributed by unknown2108 </script> |
4
Time Complexity: O(N) for each query.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach: In case, if there are multiple such queries and we are given a unique index P for every query then an efficient approach is to use a frequency array for storing the character count in each iteration of the string.
Below is the implementation of the above approach :
// CPP program to find the number of occurrences // of a character at position P upto p-1 #include <bits/stdc++.h> using namespace std;
// Function to find the number of occurrences // of a character at position P upto p-1 int countOccurrence(string s, int position)
{ int alpha[26] = { 0 }, b[s.size()] = { 0 };
// Iterate over the string
for ( int i = 0; i < s.size(); i++) {
// Store the Occurrence of same character
b[i] = alpha[( int )s[i] - 97];
// Increase its frequency
alpha[( int )s[i] - 97]++;
}
// Return the required count
return b[position - 1];
} // Driver code int main()
{ string s = "ababababab" ;
int p = 9;
// Function call
cout << countOccurrence(s, p);
return 0;
} |
// Java program to find the number of occurrences // of a character at position P upto p-1 import java.util.*;
class GFG
{ // Function to find the number of occurrences // of a character at position P upto p-1 static int countOccurrence(String s, int position)
{ int []alpha = new int [ 26 ];
int []b = new int [s.length()];
// Iterate over the string
for ( int i = 0 ; i < s.length(); i++)
{
// Store the Occurrence of same character
b[i] = alpha[( int )s.charAt(i) - 97 ];
// Increase its frequency
alpha[( int )s.charAt(i) - 97 ]++;
}
// Return the required count
return b[position - 1 ];
} // Driver code public static void main(String[] args)
{ String s = "ababababab" ;
int p = 9 ;
// Function call
System.out.println(countOccurrence(s, p));
} } // This code is contributed by 29AjayKumar |
# Python3 program to find the number of occurrences # of a character at position P upto p-1 # Function to find the number of occurrences # of a character at position P upto p-1 def countOccurrence(s, position):
alpha = [ 0 ] * 26
b = [ 0 ] * len (s)
# Iterate over the string
for i in range ( 0 , len (s)):
# Store the Occurrence of same character
b[i] = alpha[ ord (s[i]) - 97 ]
# Increase its frequency
alpha[ ord (s[i]) - 97 ] = alpha[ ord (s[i]) - 97 ] + 1
# Return the required count
return b[position - 1 ]
# Driver code s = "ababababab"
p = 9
# Function call print (countOccurrence(s, p))
# This code is contributed by Sanjit_Prasad |
// C# program to find the number of occurrences // of a character at position P upto p-1 using System;
class GFG
{ // Function to find the number of occurrences // of a character at position P upto p-1 static int countOccurrence(String s, int position)
{ int []alpha = new int [26];
int []b = new int [s.Length];
// Iterate over the string
for ( int i = 0; i < s.Length; i++)
{
// Store the Occurrence of same character
b[i] = alpha[( int )s[i] - 97];
// Increase its frequency
alpha[( int )s[i] - 97]++;
}
// Return the required count
return b[position - 1];
} // Driver code public static void Main(String[] args)
{ String s = "ababababab" ;
int p = 9;
// Function call
Console.WriteLine(countOccurrence(s, p));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript program to find the number of occurrences // of a character at position P upto p-1 // Function to find the number of occurrences // of a character at position P upto p-1 function countOccurrence(s, position)
{ let alpha = new Array(26);
for (let i = 0; i < 26; i++)
{
alpha[i] = 0;
}
let b = new Array(s.length);
// Iterate over the string
for (let i = 0; i < s.length; i++)
{
// Store the Occurrence of same character
b[i] = alpha[s[i].charCodeAt(0) - 97];
// Increase its frequency
alpha[s[i].charCodeAt(0) - 97]++;
}
// Return the required count
return b[position - 1];
} // Driver code let s = "ababababab" ;
p = 9; // Function call document.write(countOccurrence(s, p)); // This code is contributed by patel2127 </script> |
4
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.