# Count pairs of substrings from a string S such that S1 does not occur after S2 in each pair

• Last Updated : 04 Oct, 2021

Given three strings str, str1 and str2, the task is to count the number of pairs of occurrences of str1 and str2 as a substring in the string str such that in each pair, the starting index of str1 is less than or equal to str2.

Examples:

Input: str = “geeksforgeeksfor”, str1 = “geeks”, str2 = “for”
Output:
Explanation:
Starting indices of str1 are { 0, 8 }
Starting indices of str2 are { 5, 13 }
Possible pairs such that start index of str1 less than or equal to str2 are { (0, 5), (0, 13), (8, 13) }
Therefore, the required output is 3.

Input: str = “geeksforgeeks”, str1 = “geek”, str2 = “for”
Output: 1

Approach: The idea is to find the starting indices of the substring str2 in the string str and at every starting index of string str2 increment the count of pairs by the count of starting indices of str1 up to the ith index of str. Finally, print the total count of pairs obtained. Follow the steps below to solve the problem:

• Initialize a variable, say cntPairs, to store the count of pairs of str1 and str2 such that the starting index of str1 is less than or equal to str2.
• Initialize a variable, say cntStr1, to store the count of substring, str1 in str whose start index is less than or equal to i.
• Iterate over the range [0, N – 1]. For every ith index of the string str check if the substring str1 exists in the string str whose starting index is equal to i or not. If found to be true, then increment the value of cntStr1 by 1.
• For every ith index, check if the substring str2 is present in the string whose starting index is equal to i or not. If found to be true, then increment the value of cntPairs by cntStr1.
• Finally, print the value of cntPairs.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to find the count of pairs of str1``// and str2 in str such that starting index of``// str1 less than or equal to str2``int` `CountSubstring(string str, string str1, string str2)``{` `    ``// Stores count of pairs of str1 and str2``    ``// such that the start index of str1 is``    ``// less than or equal to str2``    ``int` `cntPairs = 0;` `    ``// Stores count of substring str1 in``    ``// str whose start index is up to i``    ``int` `cntStr1 = 0;` `    ``// Iterate over the range [0, N -1]``    ``for` `(``int` `i = 0; i < str.length(); i++) {` `        ``// If substring str1 whose``        ``// start index is i``        ``if` `(str.substr(i, str1.length())``            ``== str1) {` `            ``// Update cntStr1``            ``cntStr1++;``        ``}` `        ``// If substring str2 whose``        ``// start index is i``        ``else` `if` `(str.substr(i, str2.length())``                 ``== str2) {` `            ``// Update cntPairs``            ``cntPairs += cntStr1;``        ``}``    ``}` `    ``return` `cntPairs;``}` `// Driver Code``int` `main()``{` `    ``string str = ``"geeksforgeeksfor"``;``    ``string str1 = ``"geeks"``, str2 = ``"for"``;` `    ``cout << CountSubstring(str, str1, str2);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``public` `class` `GFG {` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``new` `String(``"geeksforgeeksfor"``);``        ``String str1 = ``new` `String(``"geeks"``);``        ``String str2 = ``new` `String(``"for"``);` `        ``System.out.println(CountSubstring(``            ``str, str1, str2));``    ``}` `    ``// Function to find the count of pairs of str1``    ``// and str2 in str such that starting index of``    ``// str1 less than or equal to str2``    ``static` `int` `CountSubstring(String str,``                              ``String str1,``                              ``String str2)``    ``{` `        ``// Stores count of pairs of str1 and str2``        ``// such that the start index of str1 is``        ``// less than or equal to str2``        ``int` `cntPairs = ``0``;` `        ``// Stores count of substring str1 in``        ``// str whose start index is up to i``        ``int` `cntStr1 = ``0``;` `        ``// Stores length of str``        ``int` `N = str.length();` `        ``// Stores length of str1``        ``int` `s1 = str1.length();` `        ``// Stores length of str2``        ``int` `s2 = str2.length();` `        ``// Iterate over the range [0, N -1]``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// If substring str1 whose``            ``// starting index is i``            ``if` `(((i + s1) <= N)``                ``&& (str.substring(``                        ``i, (i + s1)))``                           ``.compareTo(str1)``                       ``== ``0``) {` `                ``// Update cntStr1``                ``cntStr1++;``            ``}` `            ``// If substring str2 whose``            ``// starting index is i``            ``else` `if` `(((i + s2) <= N)``                     ``&& (str.substring(``                             ``i, (i + s2)))``                                ``.compareTo(str2)``                            ``== ``0``) {` `                ``// Update cntPairs``                ``cntPairs += cntStr1;``            ``}``        ``}` `        ``return` `cntPairs;``    ``}``}`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the count of pairs of str1``# and str2 in str such that starting index of``# str1 less than or equal to str2``def` `CountSubstring(``str``, str1, str2):``    ` `    ``# Stores count of pairs of str1 and str2``    ``# such that the start index of str1 is``    ``# less than or equal to str2``    ``cntStr1 ``=` `0``    ` `    ``# Stores count of substring str1 in``    ``# str whose start index is up to i``    ``cntPairs ``=` `0``    ` `    ``# Iterate over the range [0, N -1]``    ``for` `i ``in` `range``(``len``(``str``)):``        ` `        ``# If substring str1 whose``        ``# start index is i``        ``if` `str``[i : i ``+` `len``(str1)] ``=``=` `str1:``            ` `            ``# Update cntStr1``            ``cntStr1 ``+``=` `1``            ` `        ``# If substring str2 whose``        ``# start index is i   ``        ``elif` `str``[i : i ``+` `len``(str2)] ``=``=` `str2:``            ` `            ``# Update cntPairs``            ``cntPairs ``+``=` `cntStr1``    ` `    ``return` `cntPairs``    ` `# Driver Code``str` `=` `'geeksforgeeksfor'``str1 ``=` `'geeks'``str2 ``=` `'for'`       `       ` `print``(CountSubstring(``str``, str1, str2))       ` `# This code is contributed by dharanendralv23`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{``  ` `// Function to find the count of pairs of str1``// and str2 in str such that starting index of``// str1 less than or equal to str2 ``static` `int` `CountSubstring(String str,``                          ``String str1,``                          ``String str2)``{``    ` `    ``// Stores count of pairs of str1 and str2``    ``// such that the start index of str1 is``    ``// less than or equal to str2``    ``int` `cntPairs = 0;` `    ``// Stores count of substring str1 in``    ``// str whose start index is up to i``    ``int` `cntStr1 = 0;` `    ``// Stores length of str``    ``int` `N = str.Length;` `    ``// Stores length of str1``    ``int` `s1 = str1.Length;` `    ``// Stores length of str2``    ``int` `s2 = str2.Length;` `    ``// Iterate over the range [0, N -1]``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// If substring str1 whose``        ``// starting index is i``        ``if` `(((i + s1) <= N) &&``            ``(str.Substring(i, s1)).Equals(str1))``        ``{``            ` `            ``// Update cntStr1``            ``cntStr1++;``        ``}` `        ``// If substring str2 whose``        ``// starting index is i``        ``else` `if` `(((i + s2) <= N) &&``                 ``(str.Substring(i, s2)).Equals(str2))``        ``{``            ` `            ``// Update cntPairs``            ``cntPairs += cntStr1;``        ``}``    ``}``    ``return` `cntPairs;``}` `// Driver code``static` `public` `void` `Main()``{``    ``String str = ``"geeksforgeeksfor"``;        ``    ``String str1 = ``"geeks"``;   ``    ``String str2 = ``"for"``;``    ` `    ``Console.Write(CountSubstring(str, str1, str2));        ``}``}` `// This code is contributed by dharanendralv23`

## Javascript

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Output:

`3`

Time Complexity: O(|str| * (|str1| + |str2|))
Auxiliary Space: O(1)

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