# Find the number of occurrences of a character upto preceding position

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 the position P upto 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 4

Input : S = “geeksforgeeks”, P = 9
Output : 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach :A naive approach is to iterate over the string till Position-1 searching for a similar character. Whenever a similar character occurs increment counter variable by one.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the number of occurrences ` `// of a character at position P upto p-1 ` `#include ` `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

 `// 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

 `# 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#

 `// 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  `

Output:

```4
```

Time Complexity : O(N) for each query.

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 :

## C++

 `// CPP program to find the number of occurrences ` `// of a character at position P upto p-1 ` `#include ` `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

 `// 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

 `# 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#

 `// 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 `

Output:

```4
```

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