Given two strings, find the number of times the second string occurs in the first string, whether continuous or discontinuous.
Examples:
Input: string a = "neveropen" string b = "Gks" Output: 4 Explanation: The four strings are - (Check characters marked in bold) GeeksforGeeks GeeksforGeeks GeeksforGeeks GeeksforGeeks
If we carefully analyze the given problem, we can see that it can be easily divided into sub-problems. The idea is to process all characters of both strings one by one starting from either from left or right side. Let us traverse from right corner, there are two possibilities for every pair of character being traversed.
m: Length of str1 (first string)
n: Length of str2 (second string)
If last characters of two strings are same,
1. We consider last characters and get count for remaining
strings. So we recur for lengths m-1 and n-1.
2. We can ignore last character of first string and
recurse for lengths m-1 and n.
else
If last characters are not same,
We ignore last character of first string and
recurse for lengths m-1 and n.
Below is the implementation of above Naive recursive solution –
C++
// A Naive recursive C++ program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuous#include <iostream>using namespace std;// Recursive function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuousint count(string a, string b, int m, int n){ // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and second // string is not empty, return 0 if (m == 0) return 0; // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if (a[m - 1] == b[n - 1]) return count(a, b, m - 1, n - 1) + count(a, b, m - 1, n); else // If last characters are different, ignore // last char of first string and recur for // remaining string return count(a, b, m - 1, n);}// Driver codeint main(){ string a = "neveropen"; string b = "Gks"; cout << count(a, b, a.size(), b.size()) << endl; return 0;} |
Java
// A Naive recursive java program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuousimport java.io.*;class GFG { // Recursive function to find the number of times // the second string occurs in the first string, // whether continuous or discontinuous static int count(String a, String b, int m, int n) { // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and // second string is not empty, return 0 if (m == 0) return 0; // If last characters are same // Recur for remaining strings by // 1. considering last characters of // both strings // 2. ignoring last character of // first string if (a.charAt(m - 1) == b.charAt(n - 1)) return count(a, b, m - 1, n - 1) + count(a, b, m - 1, n); else // If last characters are different, // ignore last char of first string // and recur for remaining string return count(a, b, m - 1, n); } // Driver code public static void main (String[] args) { String a = "neveropen"; String b = "Gks"; System.out.println( count(a, b, a.length(), b.length())) ; }}// This code is contributed by vt_m |
Python 3
# A Naive recursive Python program # to find the number of times the # second string occurs in the first # string, whether continuous or # discontinuous # Recursive function to find the # number of times the second string# occurs in the first string, # whether continuous or discontinuous def count(a, b, m, n): # If both first and second string # is empty, or if second string # is empty, return 1 if ((m == 0 and n == 0) or n == 0): return 1 # If only first string is empty # and second string is not empty, # return 0 if (m == 0): return 0 # If last characters are same # Recur for remaining strings by # 1. considering last characters # of both strings # 2. ignoring last character # of first string if (a[m - 1] == b[n - 1]): return (count(a, b, m - 1, n - 1) + count(a, b, m - 1, n)) else: # If last characters are different, # ignore last char of first string # and recur for remaining string return count(a, b, m - 1, n) # Driver code a = "neveropen"b = "Gks"print(count(a, b, len(a),len(b)))# This code is contributed by ash264 |
C#
// A Naive recursive C# program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuoususing System; class GFG { // Recursive function to find the number of times // the second string occurs in the first string, // whether continuous or discontinuous static int count(string a, string b, int m, int n) { // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and // second string is not empty, return 0 if (m == 0) return 0; // If last characters are same // Recur for remaining strings by // 1. considering last characters of // both strings // 2. ignoring last character of // first string if (a[m - 1] == b[n - 1]) return count(a, b, m - 1, n - 1) + count(a, b, m - 1, n); else // If last characters are different, // ignore last char of first string // and recur for remaining string return count(a, b, m - 1, n); } // Driver code public static void Main () { string a = "neveropen"; string b = "Gks"; Console.Write( count(a, b, a.Length, b.Length) ); }} // This code is contributed by nitin mittal |
PHP
<?php // A Naive recursive PHP program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuous // Recursive function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuousfunction count_1($a, $b, $m, $n){ // If both first and second string is empty, // or if second string is empty, return 1 if (($m == 0 && $n == 0) || $n == 0) return 1; // If only first string is empty and second // string is not empty, return 0 if ($m == 0) return 0; // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if ($a[$m - 1] == $b[$n - 1]) return count_1($a, $b, $m - 1, $n - 1) + count_1($a, $b, $m - 1, $n); else // If last characters are different, ignore // last char of first string and recur for // remaining string return count_1($a, $b, $m - 1, $n);} // Driver code$a = "neveropen";$b = "Gks";echo count_1($a, $b, strlen($a), strlen($b)) ."\n";return 0;?> |
Javascript
<script>// A Naive recursive javascript program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuous // Recursive function to find the number of times // the second string occurs in the first string, // whether continuous or discontinuous function count( a, b , m , n) { // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and // second string is not empty, return 0 if (m == 0) return 0; // If last characters are same // Recur for remaining strings by // 1. considering last characters of // both strings // 2. ignoring last character of // first string if (a[m - 1] == b[n - 1]) return count(a, b, m - 1, n - 1) + count(a, b, m - 1, n); else // If last characters are different, // ignore last char of first string // and recur for remaining string return count(a, b, m - 1, n); } // Driver code var a = "neveropen"; var b = "Gks"; document.write(count(a, b, a.length, b.length));// This code is contributed by Amit Katiyar </script> |
Output
4
The time complexity of above solution is exponential. If we carefully analyze, we can see that many sub-problems are solved again and again. Since same sub-problems are called again, this problem has Overlapping sub-problems property. So the problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming problems, re-computations of same sub-problems can be avoided by constructing a temporary array that stores results of sub-problems.
Space Complexity: O(n)
We can solve it in 2 ways
1) Top-down DP
We can extend the recursion solution and stores the states of dp in 2d vector so that we don’t recompute the sub problems again.
C++
// A Memoization DP C++ program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuous#include <bits/stdc++.h>using namespace std;// Memoization DP function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuousint count(string a, string b, int m, int n, vector<vector<int> >& dp){ // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and second // string is not empty, return 0 if (m == 0) return 0; if (dp[m][n] != -1) { return dp[m][n]; } // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if (a[m - 1] == b[n - 1]) return dp[m][n] = count(a, b, m - 1, n - 1, dp) + count(a, b, m - 1, n, dp); else // If last characters are different, ignore // last char of first string and recur for // remaining string return dp[m][n] = count(a, b, m - 1, n, dp);}// Driver codeint main(){ string a = "neveropen"; string b = "Gks"; vector<vector<int> > dp(a.size() + 1, vector<int>(b.size() + 1, -1)); cout << count(a, b, a.size(), b.size(), dp) << endl; return 0;} |
Java
import java.util.Arrays;public class Main{ // Memoization DP function to find the number of times // the second string occurs in the first string, // whether continuous or discontinuous public static int count(String a, String b, int m, int n, int[][] dp) { // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) return 1; // If only first string is empty and second // string is not empty, return 0 if (m == 0) return 0; if (dp[m][n] != -1) { return dp[m][n]; } // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if (a.charAt(m - 1) == b.charAt(n - 1)) return dp[m][n] = count(a, b, m - 1, n - 1, dp) + count(a, b, m - 1, n, dp); else // If last characters are different, ignore // last char of first string and recur for // remaining string return dp[m][n] = count(a, b, m - 1, n, dp); } // Driver code public static void main(String[] args) { String a = "neveropen"; String b = "Gks"; int[][] dp = new int[a.length() + 1][b.length() + 1]; for (int[] row : dp) Arrays.fill(row, -1); System.out.println(count(a, b, a.length(), b.length(), dp)); }}// This code is contributed by unstoppablepandu. |
Python3
# A Memoization DP Python program to find the number of# times the second string occurs in the first string,# whether continuous or discontinuousdef count(a: str, b: str, m: int, n: int, dp: list) -> int: # If both first and second string is empty, # or if second string is empty, return 1 if (m == 0 and n == 0) or n == 0: return 1 # If only first string is empty and second # string is not empty, return 0 if m == 0: return 0 if dp[m][n] != -1: return dp[m][n] # If last characters are same # Recur for remaining strings by # 1. considering last characters of both strings # 2. ignoring last character of first string if a[m - 1] == b[n - 1]: dp[m][n] = count(a, b, m - 1, n - 1, dp) + count(a, b, m - 1, n, dp) else: # If last characters are different, ignore # last char of first string and recur for # remaining string dp[m][n] = count(a, b, m - 1, n, dp) return dp[m][n]# Driver codea = "neveropen"b = "Gks"dp = [[-1] * (len(b) + 1) for _ in range(len(a) + 1)]print(count(a, b, len(a), len(b), dp))# This code is contributed by phasing17. |
C#
using System;public class GFG{// Memoization DP function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuouspublic static int Count(string a, string b, int m, int n, int[,] dp){// If both first and second string is empty,// or if second string is empty, return 1if ((m == 0 && n == 0) || n == 0)return 1; // If only first string is empty and second // string is not empty, return 0 if (m == 0) return 0; if (dp[m, n] != -1) { return dp[m, n]; } // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if (a[m - 1] == b[n - 1]) return dp[m, n] = Count(a, b, m - 1, n - 1, dp) + Count(a, b, m - 1, n, dp); else // If last characters are different, ignore // last char of first string and recur for // remaining string return dp[m, n] = Count(a, b, m - 1, n, dp);}// Driver codepublic static void Main(){ string a = "neveropen"; string b = "Gks"; int[,] dp = new int[a.Length + 1, b.Length + 1]; for (int i = 0; i <= a.Length; i++) { for (int j = 0; j <= b.Length; j++) { dp[i, j] = -1; } } Console.WriteLine(Count(a, b, a.Length, b.Length, dp));}} |
Javascript
// A Memoization DP JavaScript program to find the number of// times the second string occurs in the first string,// whether continuous or discontinuousfunction count(a, b, m, n, dp) { // If both first and second string is empty, // or if second string is empty, return 1 if ((m == 0 && n == 0) || n == 0) { return 1; } // If only first string is empty and second // string is not empty, return 0 if (m == 0) { return 0; } if (dp[m][n] != -1) { return dp[m][n]; } // If last characters are same // Recur for remaining strings by // 1. considering last characters of both strings // 2. ignoring last character of first string if (a[m - 1] === b[n - 1]) { return (dp[m][n] = count(a, b, m - 1, n - 1, dp) + count(a, b, m - 1, n, dp) ); } else { // If last characters are different, ignore // last char of first string and recur for // remaining string return (dp[m][n] = count(a, b, m - 1, n, dp)); }}// Driver codeconst a = "neveropen";const b = "Gks";const dp = Array.from({ length: a.length + 1 }, () => Array(b.length + 1).fill(-1));console.log(count(a, b, a.length, b.length, dp)); |
Output
4
Time complexity of above solutions is O(MN) excluding recursion stack space.
Auxiliary space used by the program is O(MN).
2) Bottom-Up DP
Below is its implementation using Dynamic Programming –
C++
// A Dynamic Programming based C++ program to find the// number of times the second string occurs in the first// string, whether continuous or discontinuous#include <iostream>using namespace std;// Iterative DP function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuousint count(string a, string b){ int m = a.length(); int n = b.length(); // Create a table to store results of sub-problems int lookup[m + 1][n + 1] = { { 0 } }; // If first string is empty for (int i = 0; i <= n; ++i) lookup[0][i] = 0; // If second string is empty for (int i = 0; i <= m; ++i) lookup[i][0] = 1; // Fill lookup[][] in bottom up manner for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { // If last characters are same, we have two // options - // 1. consider last characters of both strings // in solution // 2. ignore last character of first string if (a[i - 1] == b[j - 1]) lookup[i][j] = lookup[i - 1][j - 1] + lookup[i - 1][j]; else // If last character are different, ignore // last character of first string lookup[i][j] = lookup[i - 1][j]; } } return lookup[m][n];}// Driver codeint main(){ string a = "neveropen"; string b = "Gks"; cout << count(a, b); return 0;} |
Java
// A Dynamic Programming based // Java program to find the // number of times the second // string occurs in the first// string, whether continuous // or discontinuousimport java.io.*;class GFG { // Iterative DP function to // find the number of times // the second string occurs// in the first string, whether// continuous or discontinuousstatic int count(String a, String b){ int m = a.length(); int n = b.length(); // Create a table to store // results of sub-problems int lookup[][] = new int[m + 1][n + 1]; // If first string is empty for (int i = 0; i <= n; ++i) lookup[0][i] = 0; // If second string is empty for (int i = 0; i <= m; ++i) lookup[i][0] = 1; // Fill lookup[][] in // bottom up manner for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { // If last characters are // same, we have two options - // 1. consider last characters // of both strings in solution // 2. ignore last character // of first string if (a.charAt(i - 1) == b.charAt(j - 1)) lookup[i][j] = lookup[i - 1][j - 1] + lookup[i - 1][j]; else // If last character are // different, ignore last // character of first string lookup[i][j] = lookup[i - 1][j]; } } return lookup[m][n];}// Driver Codepublic static void main (String[] args){ String a = "neveropen"; String b = "Gks"; System.out.println(count(a, b));}}// This code is contributed by anuj_67. |
Python3
# A Dynamic Programming based Python # program to find the number of times # the second string occurs in the first # string, whether continuous or discontinuous # Iterative DP function to find the # number of times the second string # occurs in the first string, # whether continuous or discontinuous def count(a, b): m = len(a) n = len(b) # Create a table to store results of sub-problems lookup = [[0] * (n + 1) for i in range(m + 1)] # If first string is empty for i in range(n+1): lookup[0][i] = 0 # If second string is empty for i in range(m + 1): lookup[i][0] = 1 # Fill lookup[][] in bottom up manner for i in range(1, m + 1): for j in range(1, n + 1): # If last characters are same, # we have two options - # 1. consider last characters of # both strings in solution # 2. ignore last character of first string if a[i - 1] == b[j - 1]: lookup[i][j] = lookup[i - 1][j - 1] + lookup[i - 1][j] else: # If last character are different, ignore # last character of first string lookup[i][j] = lookup[i - 1][j] return lookup[m][n]# Driver code if __name__ == '__main__': a = "neveropen" b = "Gks" print(count(a, b)) # this code is contributed by PranchalK |
C#
// A Dynamic Programming based // C# program to find the // number of times the second // string occurs in the first// string, whether continuous // or discontinuoususing System;class GFG { // Iterative DP function to // find the number of times // the second string occurs// in the first string, whether// continuous or discontinuousstatic int count(String a, String b){ int m = a.Length; int n = b.Length; // Create a table to store // results of sub-problems int[,] lookup = new int[m + 1, n + 1]; // If first string is empty for (int i = 0; i <= n; ++i) lookup[0, i] = 0; // If second string is empty for (int i = 0; i <= m; ++i) lookup[i, 0] = 1; // Fill lookup[][] in // bottom up manner for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { // If last characters are // same, we have two options - // 1. consider last characters // of both strings in solution // 2. ignore last character // of first string if (a[i - 1] == b[j - 1]) lookup[i, j] = lookup[i - 1, j - 1] + lookup[i - 1, j]; else // If last character are // different, ignore last // character of first string lookup[i, j] = lookup[i - 1, j]; } } return lookup[m, n];}// Driver Codepublic static void Main(){ String a = "neveropen"; String b = "Gks"; Console.WriteLine(count(a, b));}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// A Dynamic Programming based PHP program to find the// number of times the second string occurs in the first// string, whether continuous or discontinuous// Iterative DP function to find the number of times// the second string occurs in the first string,// whether continuous or discontinuousfunction count1($a, $b){ $m = strlen($a); $n = strlen($b); // Create a table to store results of sub-problems $lookup = array_fill(0, $m + 1, array_fill(0, $n + 1, 0)); // If second string is empty for ($i = 0; $i <= $m; ++$i) $lookup[$i][0] = 1; // Fill lookup[][] in bottom up manner for ($i = 1; $i <= $m; $i++) { for ($j = 1; $j <= $n; $j++) { // If last characters are same, we have two // options - // 1. consider last characters of both strings // in solution // 2. ignore last character of first string if ($a[$i - 1] == $b[$j - 1]) $lookup[$i][$j] = $lookup[$i - 1][$j - 1] + $lookup[$i - 1][$j]; else // If last character are different, ignore // last character of first string $lookup[$i][$j] = $lookup[$i - 1][$j]; } } return $lookup[$m][$n];}// Driver code $a = "neveropen"; $b = "Gks"; echo count1($a, $b);// This code is contributed by chandan_jnu?> |
Javascript
<script>// A Dynamic Programming based // Javascript program to find the // number of times the second // string occurs in the first// string, whether continuous // or discontinuous// Iterative DP function to// find the number of times// the second string occurs// in the first string, whether// continuous or discontinuousfunction count(a, b){ var m = a.length; var n = b.length; // Create a table to store // results of sub-problems var lookup = Array(m + 1); for(var i = 0; i < m + 1; i++) lookup[i] = Array(n + 1).fill(0); // If first string is empty for(i = 0; i <= n; ++i) lookup[0][i] = 0; // If second string is empty for(i = 0; i <= m; ++i) lookup[i][0] = 1; // Fill lookup in // bottom up manner for(i = 1; i <= m; i++) { for(j = 1; j <= n; j++) { // If last characters are // same, we have two options - // 1. consider last characters // of both strings in solution // 2. ignore last character // of first string if (a.charAt(i - 1) == b.charAt(j - 1)) lookup[i][j] = lookup[i - 1][j - 1] + lookup[i - 1][j]; else // If last character are // different, ignore last // character of first string lookup[i][j] = lookup[i - 1][j]; } } return lookup[m][n];}// Driver Codevar a = "neveropen";var b = "Gks";document.write(count(a, b));// This code is contributed by gauravrajput1 </script> |
Output
4
Time complexity of above solutions is O(MN).
Auxiliary space used by the program is O(MN).
Another Method :
This method was contributed by Kunal Hotwani
We can solve this problem by considering the fact that, in the final subsequences that will be chosen in the count, the character at b[i] will always follow the character at b[i – 1] (by the definition of a subsequence).
Using this fact, we count the total number of occurrences of string b, using an array of size equal to the length of string b.
We can create an array of size equal to n (b.size()), the jth element in this array stores the number of subsequences in the string a which are equal to b[0]b[1]…b[j]. We will traverse the string a, each time updating all the values in this array. Let’s call this array count.
While traversing the string a, we will traverse the string b in reverse for each character of a, let’s say we are at the ith character of a (a[i]) and we start the reverse traversal of the b, if the jth character of b (b[j]) is equal to a[i], a[i] can be appended to all the subsequences of a (traversed until now) which are equal to b[0]b[1]…b[j – 1] (number of these type of subsequences is stored in count[j -1]), making subsequences equal to b[0]b[1]…b[j] (number of these type of subsequences is stored in count[j]), thus count[j] can be increased by count[j – 1], when we encounter a character in string b which is equal to a[i].
NOTE : When j is equal to 0, the character a[i] cannot be appended to any subsequences, instead it can start a new subsequence equal to b[0], thus when a[i] is equal to b[0], we increase count[0] by 1 (As it stores the count all the subsequences of type b[0]).
Thus, in this method we are counting all the subsequences of a type b[0]b[1]…b[j] that can created using a particular character of a, a[i].
Finally, after traversing the complete string a, the final answer will be stored in count[n – 1], as we require the number of subsequences of type b[0]b[1]…b[n – 1].
The reason for traversal of b in reverse
Let us consider b = neveropen, we will create the count array of size 5 and start the traversal of b string from front for each a[i].
When a[i] = ‘e’, we will first increase count[1](representing the first e in string b) by count[0], and then increase count[2](representing the second e in string b) by count[1], this not ideal because we always the need the value of count[j – 1] to represent the number of subsequences of type b[0]b[1]…b[j – 1] in the string a[0]a[1]…a[i – 1] as we cannot use the same ‘e’ twice in one subsequence (In this case count[1] will represent a[i] appended to all the subsequences of type b[0] and using this count[1] to update count[2] which will mean we use the same ‘e’ twice).
Below is the code for this approach
C++
#include <bits/stdc++.h>using namespace std;int countWays(string S1, string S2){ int m = S1.size(), n = S2.size(); vector<int> count(n, 0); // Initialization to 0 is done because if a is not traversed yet, // no subsequence of any type b[0]b[1]...b[i] can exist, i.e. the // number is zero. for (int i = 0; i < m; i++) // Traversing the string a { for (int j = n - 1; j >= 0; j--) // Reverse traversal of b { if (S1[i] == S2[j]) { // This ternary expression checks weather j == 0, // gives 1 if j == 0. count[j] += (j == 0 ? 1 : count[j - 1]); } } } return count[n - 1];}int main(){ string S1 = "neveropen", S2 = "neveropen"; cout << countWays(S1, S2) << "\n"; return 0;} |
Java
import java.util.*;class GFG { public static int countWays(String s1, String s2) { int m = s1.length(), n = s2.length(); int[] count = new int[n]; // Initialization to 0 is done because if a is not traversed yet, // no subsequence of any type b[0]b[1]...b[i] can exist, i.e. the // number is zero. for (int i = 0; i < m; i++) { // Traversing the string a for (int j = n - 1; j >= 0; j--) { // Reverse traversal of b if (s1.charAt(i) == s2.charAt(j)) { // This ternary expression checks weather j == 0, // gives 1 if j == 0. count[j] += (j == 0 ? 1 : count[j - 1]); } } } return count[n - 1]; } public static void main(String[] args) { String s1 = "neveropen", s2 = "neveropen"; System.out.println(countWays(s1, s2)); }}// This Code is Contributed by Prasad Kandekar(prasad264) |
Python3
def countWays(S1, S2): m = len(S1) n = len(S2) count = [0] * n # Initialization to 0 is done because if a is not traversed yet, # no subsequence of any type b[0]b[1]...b[i] can exist, i.e. the # number is zero. for i in range(m): # Traversing the string a for j in range(n-1, -1, -1): # Reverse traversal of b if S1[i] == S2[j]: # This ternary expression checks weather j == 0, # gives 1 if j == 0. count[j] += count[j-1] if j != 0 else 1 return count[n-1]S1 = "neveropen"S2 = "neveropen"print(countWays(S1, S2))# This Code is Contributed by Prasad Kandekar(prasad264) |
C#
using System;public class GFG { static int countWays(string S1, string S2) { int m = S1.Length, n = S2.Length; int[] count = new int[n]; // Initialization to 0 is done because if a is not traversed yet, // no subsequence of any type b[0]b[1]...b[i] can exist, i.e. the // number is zero. for (int i = 0; i < m; i++) // Traversing the string a { for (int j = n - 1; j >= 0; j--) // Reverse traversal of b { if (S1[i] == S2[j]) { // This ternary expression checks weather j == 0, // gives 1 if j == 0. count[j] += (j == 0 ? 1 : count[j - 1]); } } } return count[n - 1]; } static void Main(string[] args) { string S1 = "neveropen", S2 = "neveropen"; Console.WriteLine(countWays(S1, S2)); }}// This Code is Contributed by Prasad Kandekar(prasad264) |
Javascript
function countWays(s1, s2) { const m = s1.length, n = s2.length; const count = new Array(n).fill(0); // Initialization to 0 is done because if a is not traversed yet, // no subsequence of any type b[0]b[1]...b[i] can exist, i.e. the // number is zero. for (let i = 0; i < m; i++) { // Traversing the string a for (let j = n - 1; j >= 0; j--) { // Reverse traversal of b if (s1.charAt(i) == s2.charAt(j)) { // This ternary expression checks weather j == 0, // gives 1 if j == 0. count[j] += (j == 0 ? 1 : count[j - 1]); } } } return count[n - 1];}const s1 = "neveropen", s2 = "neveropen";console.log(countWays(s1, s2)); |
Output
9
Time complexity of above solutions is O(MN).
Auxiliary space used by the program is O(N).
This article is contributed by Aditya Goel. If you like neveropen and would like to contribute, you can also write an article using contribute.neveropen.org or mail your article to contribute@neveropen.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
