Given two strings S1 and S2, the task is to check if S2 contains an anagram of S1 as its substring.

Examples: 

Input: S1 = “ab”, S2 = “bbpobac”
Output: Yes
Explanation: String S2 contains anagram “ba” of S1 (“ba”).

Input: S1 = “ab”, S2 = “cbddaoo”
Output: No

Approach: Follow the steps below to solve the problem:

1. Initialize two Hashmaps s1hash and s2hash, to store the frequency of alphabets of the two strings.
2. If the length of S1 is greater than the length of S2, then print “NO”.
3. Iterate over the characters of the string S1 and update s1hash.
4. Iterate over the characters of the string S2 using the Sliding Window technique and update the HashMap accordingly.
5. For any substring of S2 of length equal to the length of S1, if both the Hashmaps are found to be equal, print “YES”.
6. Otherwise, print “NO”.

Below is the implementation of the above approach:

// C++ Program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if string s2
// contains anagram of the string
// s1 as its substring
bool checkAnagram(string s1, string s2)
{
	// Stores frequencies of
	// characters in substrings of s2
	vector s2hash(26, 0);

	// Stores frequencies of
	// characters in s1
	vector s1hash(26, 0);
	int s1len = s1.size();
	int s2len = s2.size();

	// If length of s2 exceeds length of s1
	if (s1len > s2len)
		return false;

	int left = 0, right = 0;

	// Store frequencies of characters in first
	// substring of length s1len in string s2
	while (right < s1len) {

		s1hash[s1[right] - 'a'] += 1;
		s2hash[s2[right] - 'a'] += 1;
		right++;
	}

	right -= 1;

	// Perform Sliding Window technique
	while (right < s2len) {

		// If hashmaps are found to be
		// identical for any substring
		if (s1hash == s2hash)
			return true;

		right++;

		if (right != s2len)
			s2hash[s2[right] - 'a'] += 1;

		s2hash[s2[left] - 'a'] -= 1;
		left++;
	}
	return false;
}

// Driver Code
int main()
{
	string s1 = "ab";
	string s2 = "bbpobac";

	if (checkAnagram(s1, s2))
		cout << "YES\n";
	else
		cout << "No\n";

	return 0;
}

Output: 

YES