Check if words in given sentence occur based on the given pattern

Given a pattern ‘pattern‘ and a sentence ‘s‘, the task is to check if the words in the given sentence occur based on the pattern represented in the ‘pattern‘.
Examples:

Input: pattern = “abba”, s = “geeks for for geeks”
Output: true
Explanation: In the pattern, ‘a’ denotes ‘geeks’ and ‘b’ denotes ‘for’. Therefore, sentence ‘s’ follows the pattern ‘pattern’

Input: pattern = “abc”, s = “geeks for geeks”
Output: false

Approach: The given problem can be solved by generalizing the pattern formed by words in a given sentence and the characters in the given pattern. Then just check if both the generalized pattern and the given pattern are the same or not. Follow the below steps to solve the problem:

• Create a map to store each word and assign a value to each unique word based on its occurrence.
• Example: for sentence “geeks for for geeks”, the map will be [{“geeks”, 0}, {“for”, 1}]
• Similarly, map the occurrence of each character in the pattern
• Then match the pattern index by index in both maps and print the result

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the words in given
// sentence follows the given pattern
bool wordPattern(string pattern, string s)
{
    // Stores the occurrence of each word
    // of a sentence
    map<string, int> mp;
    int ch = -1;
    string temp = "";
    string str = "", p = "";
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == ' ') {

            if (!temp.empty()
                && mp.find(temp) == mp.end()) {

                mp.insert({ temp, ++ch });
            }
            if (mp.find(temp) != mp.end())
                str += ((char)((mp.find(temp))->second
                               + 'a'));
            temp = "";
        }
        else
            temp += s[i];
    }
    if (!temp.empty()
        && mp.find(temp) == mp.end())
        mp.insert({ temp, ++ch });

    if (mp.find(temp) != mp.end())
        str += ((char)((mp.find(temp))->second + 'a'));

    map<char, int> m;
    ch = -1;
    for (int i = 0; i < pattern.length(); i++) {
        if (m.find(pattern[i]) == m.end())
            m.insert({ pattern[i], ++ch });
        if (m.find(pattern[i]) != m.end())
            p += ((char)((m.find(pattern[i]))->second
                         + 'a'));
    }

    return p == str;
}

// Driver Code
int main()
{

    string pattern = "abba",
           s = "geeks for for geeks";
    cout << (wordPattern(pattern, s)
                 ? "true"
                 : "false");
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if the words in given
// sentence follows the given pattern
static boolean wordPattern(String pattern, String s)
{
    // Stores the occurrence of each word
    // of a sentence
    HashMap<String, Integer> mp = new HashMap<>();
    int ch = -1;
    String temp = "";
    String str = "", p = "";
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == ' ') {

            if (!temp.isEmpty()
                && mp.containsKey(temp)) {

                mp.put( temp, ++ch );
            }
            if (mp.containsKey(temp))
                str += mp.get(temp)
                               + 'a';
            temp = "";
        }
        else
            temp += s.charAt(i);
    }
    if (!temp.isEmpty()
        && mp.containsKey(temp) )
        mp.put( temp, ++ch );

    if (mp.containsKey(temp))
        str += mp.get(temp) + 'a';

    HashMap<Character,Integer> m = new HashMap<>();
    ch = -1;
    for (int i = 0; i < pattern.length(); i++) {
        if (m.containsKey(pattern.charAt(i)))
            m.put( pattern.charAt(i), ++ch );
        if (m.containsKey(pattern.charAt(i)))
            p +=m.get(pattern.charAt(i)) + 'a';
    }

    return p == str;
}

// Driver Code
public static void main(String[] args)
{

    String pattern = "abba",
           s = "geeks for for geeks";
    System.out.print((wordPattern(pattern, s)
                 ? "true"
                 : "false"));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python program for the above approach
# Function to check if the words in given
# sentence follows the given pattern
def wordPattern(pattern, s):

    # Stores the occurrence of each word
    # of a sentence
    mp = {}
    ch = -1
    temp = ""
    st = ""
    p = ""

    for i in range(0, len(s)):
        if (s[i] == ' '):
            if ((len(temp)) and (not mp.__contains__(temp))):
                ch += 1
                mp[temp] = ch
            if (mp.__contains__(temp)):
                st += ((chr)((mp[temp])
                             + 97))
            temp = ""
        else:
            temp += s[i]

    if ((len(temp)) and (not mp.__contains__(temp))):
        ch += 1
        mp[temp] = ch

    if (mp.__contains__(temp)):
        st += ((chr)((mp[temp]) + 97))

    m = {}
    ch = -1

    for i in range(0, len(pattern)):
        if (not m.__contains__(pattern[i])):
            ch += 1
            m[pattern[i]] = ch

        if (m.__contains__(pattern[i])):
            p += ((chr)(m[pattern[i]] + 97))

    return p == st

# Driver Code
pattern = "abba"
s = "geeks for for geeks"
ans = wordPattern(pattern, s)
print("true" if ans else "false")

# This code is contributed by ninja_hattori.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG {

  // Function to check if the words in given
  // sentence follows the given pattern
  static bool wordPattern(String pattern, String s)
  {

    // Stores the occurrence of each word
    // of a sentence
    Dictionary<String, int> mp = new Dictionary<String, int>();
    int ch = -1;
    String temp = "";
    String str = "", p = "";
    for (int i = 0; i < s.Length; i++) {
      if (s[i] == ' ') {

        if (temp.Length != 0 && mp.ContainsKey(temp)) {

          mp.Add(temp, ++ch);
        }
        if (mp.ContainsKey(temp))
          str += mp[temp] + 'a';
        temp = "";
      } else
        temp += s[i];
    }
    if (temp.Length != 0 && mp.ContainsKey(temp))
      mp.Add(temp, ++ch);

    if (mp.ContainsKey(temp))
      str += mp[temp] + 'a';

    Dictionary<char, int> m = new Dictionary<char,int>();
    ch = -1;
    for (int i = 0; i < pattern.Length; i++) {
      if (m.ContainsKey(pattern[i]))
        m.Add(pattern[i], ++ch);
      if (m.ContainsKey(pattern[i]))
        p += m[pattern[i]] + 'a';
    }

    return p == str;
  }

  // Driver Code
  public static void Main(String[] args) {

    String pattern = "abba", s = "geeks for for geeks";
    Console.Write((wordPattern(pattern, s) ? "true" : "false"));
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
       // JavaScript code for the above approach
       // Function to check if the words in given
       // sentence follows the given pattern
       function wordPattern(pattern, s)
       {

           // Stores the occurrence of each word
           // of a sentence
           let mp = new Map();
           let ch = -1;
           let temp = "";
           let str = "";
           let p = "";
           for (let i = 0; i < s.length; i++) {
               if (s[i] == ' ') {

                   if (temp.length != 0
                       && !mp.has(temp)) {
                       ch++;
                       mp.set(temp, ch);

                   }
                   if (mp.has(temp))
                       str += String.fromCharCode(mp.get(temp)
                           + 'a'.charCodeAt(0))
                   temp = "";
               }
               else
                   temp += s[i];
           }
           if (temp.length != 0
               && !mp.has(temp)) {
               ch++;
               mp.set(temp, ch);

           }
           if (mp.has(temp))
               str += String.fromCharCode(mp.get(temp)
                   + 'a'.charCodeAt(0))

           let m = new Map()
           ch = -1;
           for (let i = 0; i < pattern.length; i++) {
               if (!m.has(pattern[i])) {
                   ch++;
                   m.set((pattern[i], ch));

               }

               if (m.has(pattern[i]))
                   p += String.fromCharCode(m.get(pattern[i])
                       + 'a'.charCodeAt(0));
           }

           return p != str;
       }

       // Driver Code
       let pattern = "abba",
           s = "geeks for for geeks";
       document.write((wordPattern(pattern, s)
           ? "true"
           : "false"));

      // This code is contributed by Potta Lokesh
   </script>

Output

true

Time Complexity: O(NlogN), as we are inserting elements into map and finding values from the map inside the loop. 
Auxiliary Space: O(N), as we are using extra space for map.