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 = “neveropen”, 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:
C++
// 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-1int 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 codeint 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-1class GFG {// Function to find the number of occurrences// of a character at position P upto p-1static 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 codepublic 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-1def 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 codes = "ababababab";p = 9# Function callprint(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-1using System; class GFG {// Function to find the number of occurrences// of a character at position P upto p-1static 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 codepublic static void Main(String[] args){ String s = "ababababab"; int p = 9; // Function call Console.WriteLine(Occurrence(s, p));}}// This code is contributed by PrinciRaj1992 |
Javascript
<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-1function 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 codelet s = "ababababab"; let p = 9;// Function calldocument.write(Occurrence(s, p));// This code is contributed by unknown2108</script> |
Output
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 :
C++
// 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-1int 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 codeint 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-1import java.util.*;class GFG {// Function to find the number of occurrences// of a character at position P upto p-1static 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 codepublic 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-1using System; class GFG {// Function to find the number of occurrences// of a character at position P upto p-1static 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 codepublic static void Main(String[] args) { String s = "ababababab"; int p = 9; // Function call Console.WriteLine(countOccurrence(s, p));}}// This code is contributed by PrinciRaj1992 |
Javascript
<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-1function 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 codelet s = "ababababab";p = 9;// Function calldocument.write(countOccurrence(s, p));// This code is contributed by patel2127</script> |
Output
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.
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