Count subsequence of length three in a given string
Last Updated :
28 Jul, 2022
Given a string of length n and a subsequence of length 3. Find the total number of occurrences of the subsequence in this string.
Examples :
Input : string = "GFGFGYSYIOIWIN",
subsequence = "GFG"
Output : 4
Explanation : There are 4 such subsequences as shown:
GFGFGYSYIOIWIN
GFGFGYSYIOIWIN
GFGFGYSYIOIWIN
GFGFGYSYIOIWIN
Input : string = "GFGGGZZYNOIWIN",
subsequence = "GFG"
Output : 3
Brute Force Approach: The simplest way to solve this problem is to run three loops for the tree characters of the given subsequences. That starts a loop to find the first character of the subsequence in the string, once this character is found start a loop after this index to find the second character, once this character is found start a loop after this index to find the third character. Maintain a variable to store the number of occurrences of third character, i.e. number of occurrences of the subsequence.
Below is the implementation of above approach:
C++
// C++ program to find number of occurrences of
// a subsequence of length 3
#include<iostream>
using namespace std;
// Function to find number of occurrences of
// a subsequence of length three in a string
int findOccurrences(string str, string substr)
{
// variable to store no of occurrences
int counter = 0;
// loop to find first character
for (int i=0;i<str.length();i++)
{
if (str[i]==substr[0])
{
// loop to find 2nd character
for (int j=i+1;j<str.length();j++)
{
if (str[j]==substr[1])
{
// loop to find 3rd character
for (int k = j+1; k<str.length(); k++)
{
// increment count if subsequence
// is found
if (str[k]==substr[2])
counter++;
}
}
}
}
}
return counter;
}
// Driver code
int main()
{
string str = "GFGFGYSYIOIWIN";
string substr = "GFG";
cout << findOccurrences(str, substr);
return 0;
}
// Java program to find number of occurrences of
// a subsequence of length 3
import java.util.*;
import java.lang.*;
public class GfG{
// Function to find number of occurrences of
// a subsequence of length three in a string
public static int findOccurrences(String str1, String substr1)
{
// variable to store no of occurrences
int counter = 0;
char[] str = str1.toCharArray();
char[] substr = substr1.toCharArray();
// loop to find first character
for (int i=0; i < str1.length(); i++)
{
if (str[i] == substr[0])
{
// loop to find 2nd character
for (int j = i+1; j < str1.length(); j++)
{
if (str[j] == substr[1])
{
// loop to find 3rd character
for (int k = j+1; k < str1.length(); k++)
{
// increment count if subsequence
// is found
if (str[k] == substr[2])
counter++;
}
}
}
}
}
return counter;
}
// Driver function
public static void main(String argc[]){
String str = "GFGFGYSYIOIWIN";
String substr = "GFG";
System.out.println(findOccurrences(str, substr));
}
}
/* This code is contributed by Sagar Shukla */
# Python program to find
# number of occurrences
# of a subsequence of
# length 3
# Function to find number
# of occurrences of a
# subsequence of length
# three in a string
def findOccurrences(str, substr) :
# variable to store
# no of occurrences
counter = 0
# loop to find
# first character
for i in range(0, len(str)) :
if (str[i] == substr[0]) :
# loop to find
# 2nd character
for j in range(i + 1,
len(str)) :
if (str[j] == substr[1]) :
# loop to find
# 3rd character
for k in range(j + 1,
len(str)) :
# increment count if
# subsequence is found
if (str[k] == substr[2]) :
counter = counter + 1
return counter
# Driver Code
str = "GFGFGYSYIOIWIN"
substr = "GFG"
print (findOccurrences(str, substr))
# This code is contributed by
# Manish Shaw(manishshaw1)
// C# program to find number of occurrences of
// a subsequence of length 3
using System;
public class GfG {
// Function to find number of occurrences of
// a subsequence of length three in a string
public static int findOccurrences(string str1,
string substr1)
{
// variable to store no of occurrences
int counter = 0;
//char[] str = str1.toCharArray();
//char[] substr = substr1.toCharArray();
// loop to find first character
for (int i=0; i < str1.Length; i++)
{
if (str1[i] == substr1[0])
{
// loop to find 2nd character
for (int j = i+1; j < str1.Length; j++)
{
if (str1[j] == substr1[1])
{
// loop to find 3rd character
for (int k = j+1; k < str1.Length; k++)
{
// increment count if subsequence
// is found
if (str1[k] == substr1[2])
counter++;
}
}
}
}
}
return counter;
}
// Driver function
public static void Main()
{
string str1 = "GFGFGYSYIOIWIN";
string substr1 = "GFG";
Console.WriteLine(findOccurrences(str1, substr1));
}
}
/* This code is contributed by vt_m */
<?php
// PHP program to find number
// of occurrences of a
// subsequence of length 3
// Function to find number of
// occurrences of a subsequence
// of length three in a string
function findOccurrences($str, $substr)
{
// variable to store
// no of occurrences
$counter = 0;
// loop to find first character
for ($i = 0;$i < strlen($str); $i++)
{
if ($str[$i] == $substr[0])
{
// loop to find 2nd character
for ($j = $i + 1; $j < strlen($str); $j++)
{
if ($str[$j] == $substr[1])
{
// loop to find 3rd character
for ($k = $j + 1; $k < strlen($str); $k++)
{
// increment count if
// subsequence is found
if ($str[$k] == $substr[2])
$counter++;
}
}
}
}
}
return $counter;
}
// Driver Code
$str = "GFGFGYSYIOIWIN";
$substr = "GFG";
echo findOccurrences($str, $substr);
// This code is contributed by aj_36
?>
<script>
// Javascript program to find number
// of occurrences of a subsequence of length 3
// Function to find number of occurrences of
// a subsequence of length three in a string
function findOccurrences(str1, substr1)
{
// Variable to store no of occurrences
let counter = 0;
//char[] str = str1.toCharArray();
//char[] substr = substr1.toCharArray();
// Loop to find first character
for(let i = 0; i < str1.length; i++)
{
if (str1[i] == substr1[0])
{
// Loop to find 2nd character
for(let j = i + 1; j < str1.length; j++)
{
if (str1[j] == substr1[1])
{
// Loop to find 3rd character
for (let k = j + 1; k < str1.length; k++)
{
// Increment count if subsequence
// is found
if (str1[k] == substr1[2])
counter++;
}
}
}
}
}
return counter;
}
// Driver code
let str1 = "GFGFGYSYIOIWIN";
let substr1 = "GFG";
document.write(findOccurrences(str1, substr1));
// This code is contributed by suresh07
</script>
Output :
4
Time Complexity: O(n^3)
Auxiliary Space: O(1)
Efficient Approach: The efficient method to solve this problem is to use the concept of pre-calculation. The idea is for every occurrence of the middle character of subsequence in the string, pre-compute two values. That is, number of occurrences of first character before it and number of occurrences of third character after it and multiply these two values to generate total occurrences for each occurrence of middle character.
Below is the implementation of this idea:
C++
// C++ program to find number of occurrences of
// a subsequence of length 3
#include<iostream>
using namespace std;
// Function to find number of occurrences of
// a subsequence of length three in a string
int findOccurrences(string str, string substr)
{
// calculate length of string
int n = str.length();
// auxiliary array to store occurrences of
// first character
int preLeft[n] = {0};
// auxiliary array to store occurrences of
// third character
int preRight[n] = {0};
if (str[0]==substr[0])
preLeft[0]++;
// calculate occurrences of first character
// upto ith index from left
for (int i=1;i<n;i++)
{
if (str[i]==substr[0])
preLeft[i] = preLeft[i-1] + 1;
else
preLeft[i] = preLeft[i-1];
}
if (str[n-1]==substr[2])
preRight[n-1]++;
// calculate occurrences of third character
// upto ith index from right
for(int i=n-2;i>=0;i--)
{
if (str[i]==substr[2])
preRight[i] = preRight[i+1] + 1;
else
preRight[i] = preRight[i+1];
}
// variable to store total number of occurrences
int counter = 0;
// loop to find the occurrences of middle element
for (int i=1; i<n-1;i++)
{
// if middle character of subsequence is found
// in the string
if (str[i]==str[1])
{
// multiply the total occurrences of first
// character before middle character with
// the total occurrences of third character
// after middle character
int total = preLeft[i-1]*preRight[i+1];
counter += total;
}
}
return counter;
}
// Driver code
int main()
{
string str = "GFGFGYSYIOIWIN";
string substr = "GFG";
cout << findOccurrences(str,substr);
return 0;
}
// Java program to find number of occurrences of
// a subsequence of length 3
import java.util.*;
import java.lang.*;
public class GfG{
// Function to find number of occurrences of
// a subsequence of length three in a string
public static int findOccurrences(String str1, String substr1)
{
// calculate length of string
int n = str1.length();
char[] str = str1.toCharArray();
char[] substr = substr1.toCharArray();
// auxiliary array to store occurrences of
// first character
int[] preLeft = new int[n];
// auxiliary array to store occurrences of
// third character
int[] preRight = new int[n];
if (str[0] == substr[0])
preLeft[0]++;
// calculate occurrences of first character
// upto ith index from left
for (int i = 1; i < n; i++)
{
if (str[i] == substr[0])
preLeft[i] = preLeft[i-1] + 1;
else
preLeft[i] = preLeft[i-1];
}
if (str[n-1] == substr[2])
preRight[n-1]++;
// calculate occurrences of third character
// upto ith index from right
for(int i = n-2; i >= 0; i--)
{
if (str[i] == substr[2])
preRight[i] = preRight[i+1] + 1;
else
preRight[i] = preRight[i+1];
}
// variable to store total number of occurrences
int counter = 0;
// loop to find the occurrences of middle element
for (int i = 1; i < n-1; i++)
{
// if middle character of subsequence is found
// in the string
if (str[i] == str[1])
{
// multiply the total occurrences of first
// character before middle character with
// the total occurrences of third character
// after middle character
int total = preLeft[i-1] * preRight[i+1];
counter += total;
}
}
return counter;
}
// Driver function
public static void main(String argc[]){
String str = "GFGFGYSYIOIWIN";
String substr = "GFG";
System.out.println(findOccurrences(str, substr));
}
/* This code is contributed by Sagar Shukla */
}
# Python 3 program to find number of
# occurrences of a subsequence of length 3
# Function to find number of occurrences of
# a subsequence of length three in a string
def findOccurrences(str1, substr):
# calculate length of string
n = len(str1)
# auxiliary array to store occurrences of
# first character
preLeft = [0 for i in range(n)]
# auxiliary array to store occurrences of
# third character
preRight = [0 for i in range(n)]
if (str1[0] == substr[0]):
preLeft[0] += 1
# calculate occurrences of first character
# upto ith index from left
for i in range(1, n):
if (str1[i] == substr[0]):
preLeft[i] = preLeft[i - 1] + 1
else:
preLeft[i] = preLeft[i - 1]
if (str1[n - 1] == substr[2]):
preRight[n - 1] += 1
# calculate occurrences of third character
# upto ith index from right
i = n - 2
while(i >= 0):
if (str1[i] == substr[2]):
preRight[i] = preRight[i + 1] + 1
else:
preRight[i] = preRight[i + 1]
i -= 1
# variable to store
# total number of occurrences
counter = 0
# loop to find the occurrences
# of middle element
for i in range(1, n - 1):
# if middle character of subsequence
# is found in the string
if (str1[i] == str1[1]):
# multiply the total occurrences of first
# character before middle character with
# the total occurrences of third character
# after middle character
total = preLeft[i - 1] * preRight[i + 1]
counter += total
return counter
# Driver code
if __name__ == '__main__':
str1 = "GFGFGYSYIOIWIN"
substr = "GFG"
print(findOccurrences(str1, substr))
# This code is contributed by
# Surendra_Gangwar
// C# program to find number of occurrences of
// a subsequence of length 3
using System;
public class GfG {
// Function to find number of occurrences of
// a subsequence of length three in a string
public static int findOccurrences(string str1,
string substr1)
{
// calculate length of string
int n = str1.Length;
// char[] str = str1.toCharArray();
// char[] substr = substr1.toCharArray();
// auxiliary array to store occurrences of
// first character
int[] preLeft = new int[n];
// auxiliary array to store occurrences of
// third character
int[] preRight = new int[n];
if (str1[0] == substr1[0])
preLeft[0]++;
// calculate occurrences of first character
// upto ith index from left
for (int i = 1; i < n; i++)
{
if (str1[i] == substr1[0])
preLeft[i] = preLeft[i-1] + 1;
else
preLeft[i] = preLeft[i-1];
}
if (str1[n-1] == substr1[2])
preRight[n-1]++;
// calculate occurrences of third character
// upto ith index from right
for(int i = n-2; i >= 0; i--)
{
if (str1[i] == substr1[2])
preRight[i] = preRight[i+1] + 1;
else
preRight[i] = preRight[i+1];
}
// variable to store total number of occurrences
int counter = 0;
// loop to find the occurrences of middle element
for (int i = 1; i < n-1; i++)
{
// if middle character of subsequence is found
// in the string
if (str1[i] == str1[1])
{
// multiply the total occurrences of first
// character before middle character with
// the total occurrences of third character
// after middle character
int total = preLeft[i-1] * preRight[i+1];
counter += total;
}
}
return counter;
}
// Driver function
public static void Main()
{
string str1 = "GFGFGYSYIOIWIN";
string substr1 = "GFG";
Console.WriteLine(findOccurrences(str1, substr1));
}
}
/* This code is contributed by Vt_m */
<?php
// PHP program to find number
// of occurrences of a
// subsequence of length 3
// Function to find number
// of occurrences of a
// subsequence of length
// three in a string
function findOccurrences($str,
$substr)
{
// calculate length of string
$n = strlen($str);
// auxiliary array to store
// occurrences of first character
$preLeft = array(0);
// auxiliary array to store
// occurrences of third character
$preRight = array(0);
if ($str[0] == $substr[0])
$preLeft[0]++;
// calculate occurrences
// of first character
// upto ith index from left
for ($i = 1; $i < $n; $i++)
{
if ($str[$i] == $substr[0])
$preLeft[$i] = $preLeft[$i - 1] + 1;
else
$preLeft[$i] = $preLeft[$i - 1];
}
if ($str[$n - 1] == $substr[2])
$preRight[$n - 1]++;
// calculate occurrences
// of third character
// upto ith index from right
for($i = $n - 2; $i >= 0; $i--)
{
if ($str[$i] == $substr[2])
$preRight[$i] = ($preRight[$i + 1] + 1);
else
$preRight[$i] = $preRight[$i + 1];
}
// variable to store total
// number of occurrences
$counter = 0;
// loop to find the occurrences
// of middle element
for ($i = 1; $i < $n - 1; $i++)
{
// if middle character of
// subsequence is found
// in the string
if ($str[$i] == $str[1])
{
// multiply the total occurrences
// of first character before
// middle character with the
// total occurrences of third
// character after middle character
$total = $preLeft[$i - 1] *
$preRight[$i + 1];
$counter += $total;
}
}
return $counter;
}
// Driver Code
$str = "GFGFGYSYIOIWIN";
$substr = "GFG";
echo findOccurrences($str, $substr);
// This code is contributed by aj_36
?>
<script>
// Javascript program to find
// number of occurrences
// of a subsequence of length 3
// Function to find number of occurrences of
// a subsequence of length three in a string
function findOccurrences(str, substr)
{
// calculate length of string
let n = str.length;
// char[] str = str.toCharArray();
// char[] substr = substr.toCharArray();
// auxiliary array to store occurrences of
// first character
let preLeft = new Array(n);
preLeft.fill(0);
// auxiliary array to store occurrences of
// third character
let preRight = new Array(n);
preRight.fill(0);
if (str[0] == substr[0])
preLeft[0]++;
// calculate occurrences of first character
// upto ith index from left
for (let i = 1; i < n; i++)
{
if (str[i] == substr[0])
preLeft[i] = preLeft[i-1] + 1;
else
preLeft[i] = preLeft[i-1];
}
if (str[n-1] == substr[2])
preRight[n-1]++;
// calculate occurrences of third character
// upto ith index from right
for(let i = n-2; i >= 0; i--)
{
if (str[i] == substr[2])
preRight[i] = preRight[i+1] + 1;
else
preRight[i] = preRight[i+1];
}
// variable to store total
// number of occurrences
let counter = 0;
// loop to find the occurrences
// of middle element
for (let i = 1; i < n-1; i++)
{
// if middle character of subsequence
// is found
// in the string
if (str[i] == str[1])
{
// multiply the total
// occurrences of first
// character before middle
// character with
// the total occurrences
// of third character
// after middle character
let total = preLeft[i-1] * preRight[i+1];
counter += total;
}
}
return counter;
}
let str = "GFGFGYSYIOIWIN";
let substr = "GFG";
document.write(findOccurrences(str, substr));
</script>
Output:
4
Time Complexity: O(n)
Auxiliary Space: O(n)
Similar Reads
Count of unique Subsequences of given String with lengths in range [0, N] Given a string S of length N, the task is to find the number of unique subsequences of the string for each length from 0 to N. Note: The uppercase letters and lowercase letters are considered different and the result may be large so print it modulo 1000000007. Examples: Input: S = "ababd"Output: Num
14 min read
Count the strings that are subsequence of the given string Given a string S and an array arr[] of words, the task is to return the number of words from the array which is a subsequence of S. Examples: Input: S = “programmingâ€, arr[] = {"prom", "amin", "proj"}Output: 2Explanation: "prom" and "amin" are subsequence of S while "proj" is not) Input: S = “geeksf
11 min read
Count of Subsequences of given string X in between strings Y and Z Given three strings, 'X', 'Y' and 'Z', the task is to count the number of subsequences of 'X' which is lexicographically greater than or equal to 'Y' and lexicographically lesser than or equal to 'Z'. Examples: Input: X = "abc", Y = "a", Z = "bc"Output: 6Explanation: The subsequences of X which are
15+ min read
Count binary Strings that does not contain given String as Subsequence Given a number N and string S, count the number of ways to create a binary string (containing only '0' and '1') of size N such that it does not contain S as a subsequence. Examples: Input: N = 3, S = "10".Output: 4Explanation: There are 8 strings possible to fill 3 positions with 0's or 1's. {"000",
15+ min read
Count of Substrings having Sum equal to their Length Given a numeric string str, the task is to calculate the number of substrings with the sum of digits equal to their length. Examples: Input: str = "112112" Output: 6 Explanation: Substrings "1", "1", "11", "1", "1", "11" satisfy the given condition. Input: str = "1101112" Output: 12 Naive Approach:
8 min read
Count of subsequences of length 4 in form (x, x, x+1, x+1) | Set 2 Given a large number in form of string str of size N, the task is to count the subsequence of length 4 whose digit are in the form of (x, x, x+1, x+1).Example: Input: str = "1515732322" Output: 3 Explanation: For the given input string str = "1515732322", there are 3 subsequence {1122}, {1122}, and
10 min read