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Print list items containing all characters of a given word
  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021

There is a list of items. Given a specific word, e.g., “sun”, print out all the items in list which contain all the characters of “sun”. 
For example if the given word is “sun” and the items are “sunday”, “geeksforgeeks”, “utensils”, “”just” and “sss”, then the program should print “sunday” and “utensils”.

Algorithm: Thanks to geek4u for suggesting this algorithm. 

1) Initialize a binary map:
        map[256] = {0, 0, ..}
2) Set values in map[] for the given word "sun"
        map['s'] = 1,  map['u'] = 1,  map['n'] = 1
3) Store length of the word "sun":
        len = 3 for "sun"
4) Pick words (or items)one by one from the list
    a) set count = 0;
    b) For each character ch of the picked word
         if(map['ch'] is set)
            increment count and unset map['ch'] 
    c) If count becomes equal to len (3 for "sun"),
           print the currently picked word.
    d) Set values in map[] for next list item
          map['s'] = 1,  map['u'] = 1,  map['n'] = 1

C++




// C++ program to print all strings that contain all
// characters of a word
#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
using namespace std;
# define NO_OF_CHARS 256
 
/* prints list items having all caharacters of word */
void print(char list[][50], char *word, int list_size)
{
    /*Since calloc is used, map[] is initialized as 0 */
    int *map = new int[(sizeof(int)*NO_OF_CHARS)];
    int i, j, count, word_size;
 
    /*Set the values in map */
    for (i = 0; *(word+i); i++)
        map[*(word + i)] = 1;
 
    /* Get the length of given word */
    word_size = strlen(word);
 
    /* Check each item of list if has all characters
    of word*/
    for (i = 0; i < list_size; i++)
    {
        for (j = 0, count = 0; *(list[i] + j); j++)
        {
            if (map[*(list[i] + j)])
            {
                count++;
 
                /* unset the bit so that strings like
                sss not printed*/
                map[*(list[i] + j)] = 0;
            }
        }
        if (count == word_size)
            cout << list[i] << endl;
 
        /*Set the values in map for next item*/
        for (j = 0; *(word + j); j++)
            map[*(word + j)] = 1;
    }
}
 
/* Driver code*/
int main()
{
    char str[] = "sun";
    char list[][50] = {"geeksforgeeks", "unsorted", "sunday",
                    "just", "sss" };
    print(list, str, 5);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C




// C program to print all strings that contain all
// characters of a word
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# define NO_OF_CHARS 256
 
/* prints list items having all caharacters of word */
void print(char *list[], char *word, int list_size)
{
    /*Since calloc is used, map[] is initialized as 0 */
    int *map = (int *)calloc(sizeof(int), NO_OF_CHARS);
    int i, j, count, word_size;
 
    /*Set the values in map */
    for (i = 0; *(word+i); i++)
        map[*(word + i)] = 1;
 
    /* Get the length of given word */
    word_size = strlen(word);
 
    /* Check each item of list if has all characters
     of word*/
    for (i = 0; i < list_size; i++)
    {
        for (j = 0, count = 0; *(list[i] + j); j++)
        {
            if (map[*(list[i] + j)])
            {
                count++;
 
                /* unset the bit so that strings like
                   sss not printed*/
                map[*(list[i] + j)] = 0;
            }
        }
        if (count == word_size)
            printf("\n %s", list[i]);
 
        /*Set the values in map for next item*/
        for (j = 0; *(word+j); j++)
            map[*(word + j)] = 1;
    }
}
 
/* Driver program to test to pront printDups*/
int main()
{
    char str[] = "sun";
    char *list[] = {"geeksforgeeks", "unsorted", "sunday",
                    "just", "sss" };
    print(list, str, 5);
    getchar();
    return 0;
}

Java




// Java program to print all strings that contain all
// characters of a word
class GFG
{
  static final int NO_OF_CHARS = 256;
 
  /* prints list items having all caharacters of word */
  static void print(String[] list, String word, int list_size)
  {
    /*
         * Since calloc is used, map[] is initialized as 0
         */
    int[] map = new int[NO_OF_CHARS];
    int i, j, count, word_size;
 
    /* Set the values in map */
    for (i = 0; i < word.length(); i++)
      map[word.charAt(i)] = 1;
 
    /* Get the length() of given word */
    word_size = word.length();
 
    /*
         * Check each item of list if has all characters of word
         */
    for (i = 0; i < list_size; i++)
    {
      for (j = 0, count = 0; j < list[i].length(); j++)
      {
        if (map[list[i].charAt(j)] > 0)
        {
          count++;
 
          /*
            * unset the bit so that strings like sss not printed
             */
          map[list[i].charAt(j)] = 0;
        }
      }
      if (count == word_size)
        System.out.println(list[i]);
 
      /* Set the values in map for next item */
      for (j = 0; j < word.length(); j++)
        map[word.charAt(j)] = 1;
    }
  }
 
  /* Driver code */
  public static void main(String[] args)
  {
 
    String str = "sun";
    String[] list = { "geeksforgeeks", "unsorted",
                     "sunday", "just", "sss" };
    print(list, str, 5);
  }
}
 
// This code is contributed by sanjeev2552

Python




# Python program to print the list items containing all
# characters of a given word
NO_OF_CHARS = 256
 
# Prints list items having all characters of word
def printList(list, word, list_size):
    map = [0] * NO_OF_CHARS
 
    # Set the values in map
    for i in word:
        map[ord(i)] = 1
 
    # Get the length of given word
    word_size = len(word)
 
    # Check each item of list if has all characters
    # of words
    for i in list:
        count = 0
        for j in i:
            if map[ord(j)]:
                count+=1
 
                # unset the bit so that strings like sss
                # not printed
                map[ord(j)] = 0
        if count==word_size:
            print i
 
        # Set the values in map for next item
        for j in xrange(len(word)):
            map[ord(word[j])] = 1
 
# Driver program to test the above function
string = "sun"
list = ["geeksforgeeks", "unsorted", "sunday", "just", "sss"]
printList(list, string, 5)
 
# This code is contributed by Bhavya Jain

C#




// C# program to print all strings that contain
// all characters of a word
using System;
 
class GFG{
     
static int NO_OF_CHARS = 256;
 
// Prints list items having all caharacters of word
static void print(string[] list, string word,
                  int list_size)
{
    // Since calloc is used, map[] is
    // initialized as 0
    int[] map = new int[NO_OF_CHARS];
    int i, j, count, word_size;
 
    // Set the values in map
    for (i = 0; i < word.Length; i++)
        map[word[i]] = 1;
 
    // Get the length() of given word
    word_size = word.Length;
 
    // Check each item of list if has all
    // characters of word
    for(i = 0; i < list_size; i++)
    {
        for(j = 0, count = 0; j < list[i].Length; j++)
        {
            if (map[list[i][j]] > 0)
            {
                count++;
 
                // unset the bit so that strings like
                // sss not printed
                map[list[i][j]] = 0;
            }
        }
        if (count == word_size)
            Console.WriteLine(list[i]);
 
        // Set the values in map for next item
        for(j = 0; j < word.Length; j++)
            map[word[j]] = 1;
    }
}
 
// Driver code
public static void Main(string[] args)
{
    string str = "sun";
    string[] list = { "geeksforgeeks", "unsorted",
                      "sunday", "just", "sss" };
    print(list, str, 5);
}
}
     
// This code is contributed by ukasp

Output: 

unsorted
 sunday

Time Complexity: O(n + m) where n is total number of characters in the list of items. And m = (number of items in list) * (number of characters in the given word)

Please write comments if you find any bug in above code/algorithm, or find other ways to solve the same problem
 

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