Find largest word in dictionary by deleting some characters of given string

Giving a dictionary and a string ‘str’, find the longest string in dictionary which can be formed by deleting some characters of the given ‘str’.

Examples:

Input : dict = {"ale", "apple", "monkey", "plea"}   
        str = "abpcplea"  
Output : apple 

Input  : dict = {"pintu", "geeksfor", "geeksgeeks", 
                                        " forgeek"} 
         str = "geeksforgeeks"
Output : geeksgeeks

Asked In : Google Interview

This problem reduces to finding if a string is subsequence of another string or not. We traverse all dictionary words and for every word, we check if it is subsequence of given string and is largest of all such words. We finally return the longest word with given string as subsequence.

Below is the implementation of above idea

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find largest word in Dictionary
// by deleting some characters of given string
#include <bits/stdc++.h>
using namespace std;
  
// Returns true if str1[] is a subsequence of str2[].
// m is length of str1 and n is length of str2
bool isSubSequence(string str1, string str2)
{
    int m = str1.length(), n = str2.length();
  
    int j = 0; // For index of str1 (or subsequence
  
    // Traverse str2 and str1, and compare current
    // character of str2 with first unmatched char
    // of str1, if matched then move ahead in str1
    for (int i=0; i<n&&j<m; i++)
        if (str1[j] == str2[i])
            j++;
  
    // If all characters of str1 were found in str2
    return (j==m);
}
  
// Returns the longest string in dictionary which is a
// subsequence of str.
string findLongestString(vector <string > dict, string str)
{
    string result = "";
    int length = 0;
  
    // Traverse through all words of dictionary
    for (string word : dict)
    {
        // If current word is subsequence of str and is largest
        // such word so far.
        if (length < word.length() && isSubSequence(word, str))
        {
            result = word;
            length = word.length();
        }
    }
  
    // Return longest string
    return result;
}
  
// Driver program to test above function
int main()
{
    vector <string > dict = {"ale", "apple", "monkey", "plea"};
    string str = "abpcplea" ;
    cout << findLongestString(dict, str) << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find largest
// word in Dictionary by deleting
// some characters of given String
  
import java.util.*;
  
class GFG
{
  
    // Returns true if str1[] is a 
    // subsequence of str2[]. m is 
    // length of str1 and n is length of str2 
    static boolean isSubSequence(String str1,
                                String str2) 
    {
        int m = str1.length(), n = str2.length();
  
        int j = 0; // For index of str1 (or subsequence) 
  
        // Traverse str2 and str1, and compare current 
        // character of str2 with first unmatched char 
        // of str1, if matched then move ahead in str1 
        for (int i = 0; i < n && j < m; i++)
        {
            if (str1.charAt(j) == str2.charAt(i)) 
            {
                j++;
            }
        }
  
        // If all characters of str1 
        // were found in str2 
        return (j == m);
    }
  
// Returns the longest String 
// in dictionary which is a 
// subsequence of str. 
    static String findLongestString(Vector<String> dict, 
                                            String str)
    {
        String result = "";
        int length = 0;
  
        // Traverse through all words of dictionary 
        for (String word : dict)
        {
              
            // If current word is subsequence of str 
            // and is largest such word so far. 
            if (length < word.length() &&
                isSubSequence(word, str))
            {
                result = word;
                length = word.length();
            }
        }
  
        // Return longest String 
        return result;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        String[] arr = {"ale", "apple", "monkey", "plea"};
        Vector dict = new Vector(Arrays.asList(arr));
        String str = "abpcplea";
        System.out.println(findLongestString(dict, str));
    }
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find largest word in Dictionary
# by deleting some characters of given string
  
# Returns true if str1[] is a subsequence of str2[].
# m is length of str1 and n is length of str2
def isSubSequence(str1, str2):
  
    m = len(str1);
    n = len(str2);
  
    j = 0; # For index of str1 (or subsequence
  
    # Traverse str2 and str1, and compare current
    # character of str2 with first unmatched char
    # of str1, if matched then move ahead in str1
    i = 0;
    while (i < n and j < m):
        if (str1[j] == str2[i]):
            j += 1;
        i += 1;
  
    # If all characters of str1 were found in str2
    return (j == m);
  
# Returns the longest string in dictionary which is a
# subsequence of str.
def findLongestString(dict1, str1):
    result = "";
    length = 0;
  
    # Traverse through all words of dictionary
    for word in dict1:
          
        # If current word is subsequence of str and is largest
        # such word so far.
        if (length < len(word) and isSubSequence(word, str1)):
            result = word;
            length = len(word);
  
    # Return longest string
    return result;
  
# Driver program to test above function
  
dict1 = ["ale", "apple", "monkey", "plea"];
str1 = "abpcplea" ;
print(findLongestString(dict1, str1));
      
# This code is conribued by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find largest
// word in Dictionary by deleting
// some characters of given String
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Returns true if str1[] is a 
    // subsequence of str2[]. m is 
    // length of str1 and n is length of str2 
    static bool isSubSequence(String str1,
                                String str2) 
    {
        int m = str1.Length, n = str2.Length;
  
        int j = 0; // For index of str1 (or subsequence) 
  
        // Traverse str2 and str1, and compare current 
        // character of str2 with first unmatched char 
        // of str1, if matched then move ahead in str1 
        for (int i = 0; i < n && j < m; i++)
        {
            if (str1[j] == str2[i]) 
            {
                j++;
            }
        }
  
        // If all characters of str1 
        // were found in str2 
        return (j == m);
    }
  
    // Returns the longest String 
    // in dictionary which is a 
    // subsequence of str. 
    static String findLongestString(List<String> dict, 
                                            String str)
    {
        String result = "";
        int length = 0;
  
        // Traverse through all words of dictionary 
        foreach (String word in dict)
        {
              
            // If current word is subsequence of str 
            // and is largest such word so far. 
            if (length < word.Length &&
                isSubSequence(word, str))
            {
                result = word;
                length = word.Length;
            }
        }
  
        // Return longest String 
        return result;
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        String[] arr = {"ale", "apple", "monkey", "plea"};
        List<String> dict = new List<String>(arr);
        String str = "abpcplea";
        Console.WriteLine(findLongestString(dict, str));
    }
  
// This code contributed by Rajput-Ji

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find largest word in Dictionary
// by deleting some characters of given string
  
// Returns true if str1[] is a subsequence of str2[].
// m is length of str1 and n is length of str2
function isSubSequence($str1, $str2)
{
    $m = strlen($str1);
    $n = strlen($str2);
  
    $j = 0; // For index of str1 (or subsequence
  
    // Traverse str2 and str1, and compare current
    // character of str2 with first unmatched char
    // of str1, if matched then move ahead in str1
    for ($i = 0; $i < $n && $j < $m; $i++)
        if ($str1[$j] == $str2[$i])
            $j++;
  
    // If all characters of str1 were found in str2
    return ($j == $m);
}
  
// Returns the longest string in dictionary which is a
// subsequence of str.
function findLongestString($dict, $str)
{
    $result = "";
    $length = 0;
  
    // Traverse through all words of dictionary
    foreach ($dict as $word)
    {
        // If current word is subsequence of str and is largest
        // such word so far.
        if ($length < strlen($word) && isSubSequence($word, $str))
        {
            $result = $word;
            $length = strlen($word);
        }
    }
  
    // Return longest string
    return $result;
}
  
// Driver code
$dict = array("ale", "apple", "monkey", "plea");
$str = "abpcplea" ;
echo findLongestString($dict, $str);
      
// This code is conribued by mits
?>

chevron_right


Output:

apple

Time Complexity : O(N*K*n) Here N is the length of dictionary and n is the length of given string ‘str’ and K – maximum length of words in the dictionary.

Auxiliary Space : O(1)

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks 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.



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.