Count distinct occurrences as a subsequence
Last Updated :
23 Jul, 2025
Given two strings pat and txt, where pat is always shorter than txt, count the distinct occurrences of pat as a subsequence in txt.
Examples:
Input: txt = abba, pat = aba
Output: 2
Explanation: pat appears in txt as below two subsequences.
[abba], [abba]
Input: txt = banana, pat = ban
Output: 3
Explanation: pat appears in txt as below three subsequences.
[banana], [banana], [banana]
Input: txt = geeks, pat = ge
Output: 2
Explanation: pat appears in txt as below two subsequences.
[geeks], [geeks]
[Naive Approach ] Using Recursion - O(2^n) Time and O(n) Space
We can easily identify the recursive nature of this problem. If we match the current character in the pattern with the current character in the text, we have two options: we can either include this character and move to the next character in both the pattern and the text, or we can skip the current character in the text and try to match the same character in the pattern with the next character in the text.
This leads to the following recursive relationship:
1. If the characters match (i.e., txt[i-1] == pat[j-1]), the count of distinct subsequences can be expressed as the sum of:
- The count of subsequences by including this character from both txt and pat (subCountRec(i-1, j-1, txt, pat)).
- The count of subsequences by excluding the character from txt (subCountRec(i-1, j, txt, pat)).
2. If the characters do not match, we can only exclude the character from txt, which gives us:
- The count of subsequences without considering the current character in txt (subCountRec(i-1, j, txt, pat)).
C++
// C++ program to Count distinct
// occurrences as a subsequence
#include <bits/stdc++.h>
using namespace std;
// i and j are current starting
// addresses in txt and pat respectively
int subCountRec(int i, int j, string &txt, string &pat)
{
if (j == 0)
{
// Empty pattern is found in
// all suffixes
return 1;
}
if (i == 0)
{
// No more characters in
// txt to match pat
return 0;
}
if (txt[i - 1] == pat[j - 1])
{
// Count both cases
return subCountRec(i - 1, j - 1, txt, pat) + subCountRec(i - 1, j, txt, pat);
}
return subCountRec(i - 1, j, txt, pat);
}
int subseqCount(string &txt, string &pat)
{
return subCountRec(txt.size(), pat.size(), txt, pat);
}
int main()
{
string pat = "aba";
string txt = "abba";
cout << subseqCount(txt, pat) << endl;
return 0;
}
// Java program to Count distinct
// occurrences as a subsequence
import java.util.Scanner;
class GfG {
// i and j are current starting addresses in
// txt and pat respectively
static int subCountRec(int i, int j, String txt,
String pat)
{
if (j == 0) {
// Empty pattern is found in all suffixes
return 1;
}
if (i == 0) {
// No more characters in txt to match pat
return 0;
}
if (txt.charAt(i - 1) == pat.charAt(j - 1)) {
// Count both cases
return subCountRec(i - 1, j - 1, txt, pat)
+ subCountRec(i - 1, j, txt, pat);
}
return subCountRec(i - 1, j, txt, pat);
}
static int subseqCount(String txt, String pat)
{
return subCountRec(txt.length(), pat.length(), txt,
pat);
}
public static void main(String[] args)
{
String pat = "aba";
String txt = "abba";
System.out.println(subseqCount(txt, pat));
}
}
# Python program to Count distinct
# occurrences as a subsequence
def subCountRec(i, j, txt, pat):
if j == 0:
# Empty pattern is found in all suffixes
return 1
if i == 0:
# No more characters in txt to match pat
return 0
if txt[i - 1] == pat[j - 1]:
# Count both cases
return subCountRec(i - 1, j - 1, txt, pat) \
+ subCountRec(i - 1, j, txt, pat)
return subCountRec(i - 1, j, txt, pat)
def subseqCount(txt, pat):
return subCountRec(len(txt), len(pat), txt, pat)
if __name__ == "__main__":
pat = "aba"
txt = "abba"
print(subseqCount(txt, pat))
// C# program to Count distinct
// occurrences as a subsequence
using System;
class GfG {
// i and j are current starting addresses in txt
// and pat respectively
static int SubCountRec(int i, int j, string txt,
string pat)
{
if (j == 0) {
// Empty pattern is found in all suffixes
return 1;
}
if (i == 0) {
// No more characters in txt to match pat
return 0;
}
if (txt[i - 1] == pat[j - 1]) {
// Count both cases
return SubCountRec(i - 1, j - 1, txt, pat)
+ SubCountRec(i - 1, j, txt, pat);
}
return SubCountRec(i - 1, j, txt, pat);
}
static int subseqCount(string txt, string pat)
{
return SubCountRec(txt.Length, pat.Length, txt,
pat);
}
static void Main(string[] args)
{
string pat = "aba";
string txt = "abba";
Console.WriteLine(subseqCount(txt, pat));
}
}
// JavaScript program to count distinct occurrences of pat
// as a subsequence in txt
function subCountRec(i, j, txt, pat)
{
// If the pattern is empty, there is exactly one way to
// match it (by choosing nothing)
if (j === 0)
return 1;
// If txt is empty but pat is not, it's impossible to
// form the subsequence
if (i === 0)
return 0;
if (txt[i - 1] === pat[j - 1]) {
// If the last characters match, count both
// including and excluding the character
return subCountRec(i - 1, j - 1, txt, pat)
+ subCountRec(i - 1, j, txt, pat);
}
// If the last characters don't match, exclude the
// current character from txt
return subCountRec(i - 1, j, txt, pat);
}
function subseqCount(txt, pat)
{
return subCountRec(txt.length, pat.length, txt, pat);
}
// Test cases
let txt = "abba";
let pat = "aba";
console.log(subseqCount(txt, pat));
[Better Approach 1 ] Using Top-Down DP (Recursion) - O(m*n) Time and O(m*n) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.
Overlapping Subproblems:
While employing a recursive approach to this problem, we notice that certain subproblems are computed multiple times, leading to redundancy. For example, when calculating countSubsequences(i, j), we might find need to calculate countSubsequences(i-1, j-1) and countSubsequences(i, j-1) multiple times for the same indices. This overlapping of computations makes it clear that a simple recursive approach would be inefficient.
- The recursive solution involves changing of two parameters, the current index in the pattern (i) and the current index in the text (j). We need to track both parameters, so we create a 2D array of size (m+1) x (n+1) because the value of i will be in the range [0,m] and j in the range [0,n], where m is the length of the pattern and n is the length of the text.
- We initialize the 2D array with -1 to indicate that no subproblems have been computed yet.
- We check if the value at memo[i][j] is -1. If it is, we proceed to compute else we return the stored result.
C++
// C++ program to Count distinct occurrences as
// a subsequence using memoization
#include <bits/stdc++.h>
using namespace std;
int countSubsequences(int i, int j, string &pat, string &txt, vector<vector<int>> &memo)
{
if (i == 0)
return 1;
if (j == 0)
return 0;
// If already computed, return the result
if (memo[i][j] != -1)
return memo[i][j];
// If last characters don't match
if (pat[i - 1] != txt[j - 1])
{
memo[i][j] = countSubsequences(i, j - 1, pat, txt, memo);
}
else
{
// Both characters match
memo[i][j] =
countSubsequences(i, j - 1, pat, txt, memo) + countSubsequences(i - 1, j - 1, pat, txt, memo);
}
return memo[i][j];
}
int subseqCount(string &txt, string &pat)
{
int m = pat.length(), n = txt.length();
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D vector for memoization
vector<vector<int>> memo(m + 1, vector<int>(n + 1, -1));
return countSubsequences(m, n, pat, txt, memo);
}
int main()
{
string pat = "aba";
string txt = "abba";
cout << subseqCount(txt, pat) << endl;
return 0;
}
// Java program to Count distinct occurrences as
// a subsequence using memoization
import java.util.Arrays;
class GfG {
static int subseqCount(String txt, String pat)
{
int m = pat.length(), n = txt.length();
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D array for memoization
int[][] memo = new int[m + 1][n + 1];
// Initialize the memoization array with -1
for (int[] row : memo) {
Arrays.fill(row, -1);
}
return countSubsequences(m, n, pat, txt, memo);
}
// Helper function for memoization
static int countSubsequences(int i, int j, String pat,
String txt, int[][] memo)
{
if (i == 0)
return 1;
if (j == 0)
return 0;
// If already computed, return the result
if (memo[i][j] != -1)
return memo[i][j];
// If last characters don't match
if (pat.charAt(i - 1) != txt.charAt(j - 1)) {
memo[i][j] = countSubsequences(i, j - 1, pat,
txt, memo);
}
else {
// Both characters match
memo[i][j] = countSubsequences(i, j - 1, pat,
txt, memo)
+ countSubsequences(
i - 1, j - 1, pat, txt, memo);
}
return memo[i][j];
}
public static void main(String[] args)
{
String pat = "aba";
String txt = "abba";
System.out.println(subseqCount(txt, pat));
}
}
# Python program to Count distinct occurrences as
# a subsequence using memoization
def countSubsequences(i, j, pat, txt, memo):
if i == 0:
return 1
if j == 0:
return 0
# If already computed, return the result
if memo[i][j] != -1:
return memo[i][j]
# If last characters don't match
if pat[i - 1] != txt[j - 1]:
memo[i][j] = countSubsequences(i, j - 1, pat, txt, memo)
else:
# Both characters match
memo[i][j] = (countSubsequences(i, j - 1, pat, txt, memo)
+ countSubsequences(i - 1, j - 1, pat, txt, memo))
return memo[i][j]
def subseqCount(txt, pat):
m, n = len(pat), len(txt)
# pat can't appear as a subsequence in txt
if m > n:
return 0
# Create a memoization table
memo = [[-1 for _ in range(n + 1)] for _ in range(m + 1)]
return countSubsequences(m, n, pat, txt, memo)
if __name__ == "__main__":
pat = "aba"
txt = "abba"
print(subseqCount(txt, pat))
// C# program to Count distinct occurrences as
// a subsequence using memoization
using System;
class GfG {
static int CountSubsequences(int i, int j, string pat,
string txt, int[, ] memo)
{
if (i == 0)
return 1;
if (j == 0)
return 0;
// If already computed, return the result
if (memo[i, j] != -1)
return memo[i, j];
// If last characters don't match
if (pat[i - 1] != txt[j - 1]) {
memo[i, j] = CountSubsequences(i, j - 1, pat,
txt, memo);
}
else {
// Both characters match
memo[i, j] = CountSubsequences(i, j - 1, pat,
txt, memo)
+ CountSubsequences(
i - 1, j - 1, pat, txt, memo);
}
return memo[i, j];
}
static int subseqCount(string txt, string pat)
{
int m = pat.Length, n = txt.Length;
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D array for memoization
int[, ] memo = new int[m + 1, n + 1];
// Initialize the memoization array with -1
for (int i = 0; i <= m; i++)
for (int j = 0; j <= n; j++)
memo[i, j] = -1;
return CountSubsequences(m, n, pat, txt, memo);
}
static void Main(string[] args)
{
string pat = "aba";
string txt = "abba";
Console.WriteLine(subseqCount(txt, pat));
}
}
// Javascript program to Count distinct occurrences as
// a subsequence using memoization
function countSubsequences(i, j, pat, txt, memo)
{
if (i === 0)
return 1;
if (j === 0)
return 0;
// If already computed, return the result
if (memo[i][j] !== -1)
return memo[i][j];
// If last characters don't match
if (pat[i - 1] !== txt[j - 1]) {
memo[i][j]
= countSubsequences(i, j - 1, pat, txt, memo);
}
else {
// Both characters match
memo[i][j]
= countSubsequences(i, j - 1, pat, txt, memo)
+ countSubsequences(i - 1, j - 1, pat, txt,
memo);
}
return memo[i][j];
}
function subseqCount(txt, pat)
{
const m = pat.length;
const n = txt.length;
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D array for memoization
const memo = Array.from({length : m + 1},
() => Array(n + 1).fill(-1));
return countSubsequences(m, n, pat, txt, memo);
}
const pat = "aba";
const txt = "abba";
console.log(subseqCount(txt, pat));
[Better Approach 2] Using Bottom-Up DP (Tabulation) - O(m*n) Time and O(m*n) Space
The approach is similar to the previous one, just instead recursion, we iteratively build up the solution by calculating in bottom-up manner. Maintain 2D table dp of size (m + 1) x (n + 1), where m is the length of the pattern and n is the length of the text.
This table will store the number of distinct subsequences of the pattern in the text at various indices.
- For all j from 0 to n, set dp[0][j] = 1. This represents the fact that an empty pattern is a subsequence of any text (including an empty text).
- For i > 0, set dp[i][0] = 0. This indicates that a non-empty pattern cannot be found in an empty text.
Iterate through the table filling it out from 1 to m for the pattern and 1 to n for the text. For each i (from 1 to m) and j (from 1 to n):
if pat[i - 1] == txt[j - 1], then we have two choices:
- Include this character in the subsequence, which gives dp[i - 1][j - 1].
- Exclude this character, which gives dp[i][j - 1]
If the characters do not match, we can only exclude the current character of txt, dp[i][j]=dp[i][j-1].
C++
// C++ program to Count distinct occurrences as
// a subsequence using tabulation
#include <bits/stdc++.h>
using namespace std;
int subseqCount(string &txt, string &pat)
{
int m = pat.length(), n = txt.length();
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D vector initialized with 0
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
// Initializing first row with all 1s. An empty
// string is a subsequence of all.
for (int j = 0; j <= n; j++)
dp[0][j] = 1;
// Fill mat[][] in bottom up manner
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
// If last characters don't match, then value
// is same as the value without last character
// in txt.
if (pat[i - 1] != txt[j - 1])
dp[i][j] = dp[i][j - 1];
else
// Value is obtained considering two cases.
// a) All substrings without last character in txt
// b) All substrings without last characters in both.
dp[i][j] = (dp[i][j - 1] + dp[i - 1][j - 1]);
}
}
return dp[m][n];
}
int main()
{
string pat = "aba";
string txt = "abba";
cout << subseqCount(txt, pat) << endl;
return 0;
}
// Java program to Count distinct occurrences as
// a subsequence using tabulation
class GfG {
static int subseqCount(String txt, String pat)
{
int m = pat.length();
int n = txt.length();
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D array initialized with 0
int[][] dp = new int[m + 1][n + 1];
// Initializing first row with all 1s. An empty
// string is a subsequence of all.
for (int j = 0; j <= n; j++) {
dp[0][j] = 1;
}
// Fill mat[][] in bottom-up manner
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// If last characters don't match
if (pat.charAt(i - 1)
!= txt.charAt(j - 1)) {
dp[i][j] = dp[i][j - 1];
}
else {
// Value is obtained considering two
// cases. a) All substrings without last
// character in txt b) All substrings
// without last characters in both.
dp[i][j]
= (dp[i][j - 1] + dp[i - 1][j - 1]);
}
}
}
return dp[m][n];
}
public static void main(String[] args)
{
String pat = "aba";
String txt = "abba";
System.out.println(subseqCount(txt, pat));
}
}
# Python program to Count distinct occurrences as
# a subsequence using tabulation
def subseqCount(txt, pat):
m = len(pat)
n = len(txt)
# pat can't appear as a subsequence
# in txt
if m > n:
return 0
# Create a 2D list initialized with 0
dp = [[0] * (n + 1) for _ in range(m + 1)]
# Initializing first row with all 1s. An empty string
# is a subsequence of all.
for j in range(n + 1):
dp[0][j] = 1
# Fill mat[][] in bottom-up manner
for i in range(1, m + 1):
for j in range(1, n + 1):
# If last characters don't match, then value
# is the same as the value without the last character in txt.
if pat[i - 1] != txt[j - 1]:
dp[i][j] = dp[i][j - 1]
else:
# Value is obtained considering two cases:
# a) All substrings without last character in txt
# b) All substrings without last characters in both.
dp[i][j] = (dp[i][j - 1] + dp[i - 1][j - 1])
return dp[m][n]
if __name__ == "__main__":
pat = "aba"
txt = "abba"
print(subseqCount(txt, pat))
// C# program to Count distinct occurrences as
// a subsequence using tabulation
using System;
class GfG {
static int subseqCount(string txt, string pat)
{
int m = pat.Length;
int n = txt.Length;
// pat can't appear as a subsequence in txt
if (m > n)
return 0;
// Create a 2D array initialized with 0
int[, ] dp = new int[m + 1, n + 1];
// Initializing first row with all 1s. An empty
// string is a subsequence of all.
for (int j = 0; j <= n; j++) {
dp[0, j] = 1;
}
// Fill mat[][] in bottom-up manner
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// If last characters don't match
if (pat[i - 1] != txt[j - 1]) {
dp[i, j] = dp[i, j - 1];
}
else {
// Value is obtained considering two
// cases: a) All substrings without last
// character in txt b) All substrings
// without last characters in both.
dp[i, j]
= (dp[i, j - 1] + dp[i - 1, j - 1]);
}
}
}
return dp[m, n];
}
static void Main(string[] args)
{
string pat = "aba";
string txt = "abba";
Console.WriteLine(subseqCount(txt, pat));
}
}
// Javascript program to Count distinct occurrences as
// a subsequence using tabulation
function subseqCount(txt, pat)
{
const m = pat.length;
const n = txt.length;
// pat can't appear as a subsequence
// in txt
if (m > n)
return 0;
// Create a 2D array initialized with 0
const dp = Array.from({length : m + 1},
() => Array(n + 1).fill(0));
// Initializing first row with all 1s. An empty
// string is a subsequence of all.
for (let j = 0; j <= n; j++) {
dp[0][j] = 1;
}
// Fill mat[][] in bottom-up manner
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
// If last characters don't match
if (pat[i - 1] !== txt[j - 1]) {
dp[i][j] = dp[i][j - 1];
}
else {
// Value is obtained considering two cases:
// a) All substrings without last character
// in txt b) All substrings without last
// characters in both.
dp[i][j]
= (dp[i][j - 1] + dp[i - 1][j - 1]);
}
}
}
return dp[m][n];
}
const pat = "aba";
const txt = "abba";
console.log(subseqCount(txt, pat));
[Expected Approach ] Using Space Optimized DP – O(m*n) Time and O(m) Space
Since dp[i][j] accesses elements of the current and previous rows only, we can optimize auxiliary space just by using two rows only reducing space from m*n to 2*m.
So we maintain two arrays prev and curr to store previous and current rows of dp[][].
if pat[i - 1] == txt[j - 1], then we have two choices:
- Include this character in the subsequence, which gives prev[j-1] is added to curr[j]
- Exclude this character, which means prev[j] is added to curr[j]
If the characters do not match, we can only exclude the current character which means curr[j] = prev[j]
C++
// C++ program to Count distinct occurrences
// as a subsequence using space optimized dp
#include <bits/stdc++.h>
using namespace std;
// Function to count distinct occurrences of pat in txt
int subseqCount(string &txt, string &pat)
{
int m = pat.length(), n = txt.length();
// If pattern is longer than the text, return 0
if (m > n)
return 0;
// Create two 1D arrays for dynamic programming
vector<int> prev(m + 1, 0);
vector<int> curr(m + 1, 0);
// Base case: An empty pattern can be
// formed from any text
prev[0] = 1;
// Iterate over each character in the text
for (int i = 1; i <= n; i++)
{
// Base case: An empty pattern is always
// a subsequence
curr[0] = 1;
// Iterate over each character in the pattern
for (int j = 1; j <= m; j++)
{
// If characters match, include or
// exclude the current character
if (txt[i - 1] == pat[j - 1])
{
curr[j] = (prev[j - 1] + prev[j]);
}
else
{
curr[j] = prev[j];
}
}
// Update prev array for
// the next iteration
prev = curr;
}
return prev[m];
}
int main()
{
string pat = "aba";
string txt = "abba";
cout << subseqCount(txt, pat) << endl;
return 0;
}
// Java program to Count distinct occurrences
// as a subsequence using space optimized dp
import java.util.Arrays;
class GfG {
// Function to count distinct occurrences of pat in txt
static int subseqCount(String txt, String pat)
{
int m = pat.length();
int n = txt.length();
// If pattern is longer than the
// text, return 0
if (m > n)
return 0;
// Create two 1D arrays for dynamic
// programming
int[] prev = new int[m + 1];
int[] curr = new int[m + 1];
// Base case: An empty pattern can be
// formed from any text
prev[0] = 1;
// Iterate over each character in the text
for (int i = 1; i <= n; i++) {
// Base case: An empty pattern is always a
// subsequence
curr[0] = 1;
// Iterate over each character in the pattern
for (int j = 1; j <= m; j++) {
// If characters match, include or exclude
// the current character
if (txt.charAt(i - 1)
== pat.charAt(j - 1)) {
curr[j] = (prev[j - 1] + prev[j]);
}
else {
curr[j] = prev[j];
}
}
// Update prev array for the next iteration
System.arraycopy(curr, 0, prev, 0, curr.length);
}
return (int)prev[m];
}
public static void main(String[] args)
{
String pat = "aba";
String txt = "abba";
System.out.println(subseqCount(txt, pat));
}
}
# Python program to Count distinct occurrences
# as a subsequence using space optimized dp
def subseqCount(txt, pat):
m, n = len(pat), len(txt)
# If pattern is longer than the
# text, return 0
if m > n:
return 0
# Create two 1D arrays for
# dynamic programming
prev = [0] * (m + 1)
curr = [0] * (m + 1)
# Base case: An empty pattern can
# be formed from any text
prev[0] = 1
# Iterate over each character in the text
for i in range(1, n + 1):
curr[0] = 1
# Iterate over each character in the pattern
for j in range(1, m + 1):
# If characters match, include or
# exclude the current character
if txt[i - 1] == pat[j - 1]:
curr[j] = (prev[j - 1] + prev[j])
else:
curr[j] = prev[j]
# Update prev array for the
# next iteration
prev = curr.copy()
return prev[m]
if __name__ == "__main__":
pat = "aba"
txt = "abba"
print(subseqCount(txt, pat))
// C# program to Count distinct occurrences
// as a subsequence using space optimized dp
using System;
class GfG {
public static int subseqCount(string txt, string pat)
{
int m = pat.Length, n = txt.Length;
// If pattern is longer than the
// text, return 0
if (m > n)
return 0;
// Create two arrays for dynamic
// programming
int[] prev = new int[m + 1];
int[] curr = new int[m + 1];
// Base case: An empty pattern can
// be formed from any text
prev[0] = 1;
// Iterate over each character in the text
for (int i = 1; i <= n; i++) {
curr[0] = 1;
// Iterate over each character in the pattern
for (int j = 1; j <= m; j++) {
// If characters match, include or exclude
// the current character
if (txt[i - 1] == pat[j - 1]) {
curr[j] = (prev[j - 1] + prev[j]);
}
else {
curr[j] = prev[j];
}
}
// Update prev array for the next
// iteration
Array.Copy(curr, prev, m + 1);
}
// Return the count of distinct subsequences
// matching the complete pattern
return prev[m];
}
static void Main()
{
string pat = "aba";
string txt = "abba";
Console.WriteLine(subseqCount(txt, pat));
}
}
// Javascript program to Count distinct occurrences
// as a subsequence using space optimized dp
function subseqCount(txt, pat)
{
const m = pat.length;
const n = txt.length;
// If pattern is longer than the text, return 0
if (m > n)
return 0;
// Create two arrays for dynamic programming
const prev = new Array(m + 1).fill(0);
const curr = new Array(m + 1).fill(0);
// Base case: An empty pattern can
// be formed from any text
prev[0] = 1;
// Iterate over each character in the text
for (let i = 1; i <= n; i++) {
curr[0] = 1;
// Iterate over each character in the pattern
for (let j = 1; j <= m; j++) {
// If characters match, include or exclude the
// current character
if (txt[i - 1] === pat[j - 1]) {
curr[j] = (prev[j - 1] + prev[j]);
}
else {
curr[j] = prev[j];
}
}
// Update prev array for the next iteration
for (let k = 0; k <= m; k++) {
prev[k] = curr[k];
}
}
// Return the count of distinct subsequences matching
// the complete pattern
return prev[m];
}
const txt = "aba";
const pat = "abba";
console.log(subseqCount(txt, pat));
Distinct occurrences | DSA Problem
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