Given two strings, write a method to decide if one is a permutation of the other

Problem

Given two strings, write a method to decide if one is a permutation of the other.

Example) 

Given two strings, "mom", "grandmom", your algorithm should return True because "mom" is the permutation of "grandmom".

String comparison using the HashTable, $O(N)$

The brute force approach can be checking each string is the other permutation by generating the permutations of one string and checking the other word exists in the permutation.

This takes $O(N \times N!)$ for each case, which is terribly inefficient.

Can we do better? Of course, we can !!

Say we are given two words, "mom" and "grandmom". We know instantly that grandmom can't be the permutation of "mom" since the former is short than the latter.

How do we know "mom" is the permutation of "grandmom"?

As long as all the characters of "mom" exist in "grandmom" we can say that there is one permutation that is the same as "mom".

Using Hash Table, we can record the occurrence of the characters consisting of each word. The hash table for the short word, $H_s$ and the one for the longer word, $H_l$.

Once this is done, we can check if all the characters of the shorter word exist in the longer word as follows:

      Check the occurrence count of chars in the short words in $H_s$ $\le$ the occurrence count of chars in $H_l$. In expression,

     { $\forall c$  $\in$ $S_1$, $H_s$[c] <= $H_l$[c] }

The overall running time of this algorithm will be $O(N)$, where N is the length of the longer string.

C++ solution

	#include<iostream>
	#include<string>
	using namespace std;
	bool is_permutation(string str1, string str2)
	{
	int map_short[256];
	int map_long[256];
	string shorter, longer;
	int i, j;
	memset(map_short, 0, sizeof(int)*256);
	memset(map_long, 0, sizeof(int)*256);
	if (str1.length() > str2.length())
	{
	longer = str1;
	shorter = str2;
	} else {
	longer = str2;
	shorter = str1;
	}
	for(i = 0; i<shorter.length(); i++)
	{
	map_short[shorter[i]]++;
	}
	for(i = 0; i<longer.length(); i++)
	{
	map_long[longer[i]]++;
	}
	for(i = 0; i<shorter.length(); i++)
	{
	if(map_short[shorter[i]] > map_long[shorter[i]])
	break;
	}
	if (i == shorter.length())
	{
	cout<<shorter <<" is the permuation of "<<longer<<endl;
	}
	return i == shorter.length();
	}
	int main()
	{
	string str1 ="mon";
	string str2 = "mathonstic";
	cout<<" str1, str2 are in permutation: "<< is_permutation(str1, str2)<<endl;
	return 1;
	}

view raw check_if_one_is_permuate_of_the_other.cpp hosted with ❤ by GitHub

Python solution

	def is_permutation(str1, str2):
	len_str1 = len(str1)
	len_str2 = len(str2)
	if (len_str1 >= len_str2):
	longer = str1
	shorter = str2
	else:
	longer = str2
	shorter = str1
	map_longer = {char: 0 for char in longer}
	map_shorter = {char: 0 for char in shorter}
	# create a map of chars in longer with occurrence
	for char in longer:
	map_longer[char] += 1
	# create a map of chars in shorter with occurrence
	for char in shorter:
	map_shorter[char] += 1
	#check if all of chars in shorter exists in longer char
	for char in shorter:
	if map_shorter[char] > map_longer[char]:
	return False
	return True
	str1 = 'mom'
	str2 = 'grandmom'
	print('%s, %s are permutation: %r'% (str1, str2, is_permutation(str1, str2)))
	str3 = "momendum"
	print('%s, %s are permutation: %r'% (str1, str3, is_permutation(str1, str3)))
	str4 = "monday"
	print('%s, %s are permutation: %r'% (str1, str4, is_permutation(str1, str4)))

view raw check_two_in_permutation2.py hosted with ❤ by GitHub

Practice statistics

20:00 mins: to write up the code and test the code

UPDATE(2022-06-13): It turned out that we don't need to create the hashtable for the shorter one. We can just iterate over the shorter word and check if the character exists in the longer one's hashtable and decrement for each occurrence, Code is way shorter.

	#code
	def check_for_permutation(word1, word2):
	longer = word1 if len(word1) > len(word2) else word2
	shorter = word1 if word1 != longer else word2
	# create the alphabet dictionary
	dic = {}
	for c in longer:
	if c in dic:
	dic[c] = dic[c] + 1
	else:
	dic[c] = 1
	# check if all the char in the shorter word exists in the longer one
	for char in shorter:
	if char in dic and dic[char] > 0:
	dic[char] = dic[char] - 1
	else:
	return False
	return True
	# test
	word1 = "mom"
	word2 = "grandmom"
	result = check_for_permutation(word1, word2)
	print ("are {}, {} in permutation? {}".format(word1, word2, result))
	word1 = "mad"
	word2 = "grandmom"
	result = check_for_permutation(word1, word2)
	print ("are {}, {} in permutation? {}".format(word1, word2, result))
	word1 = "gone"
	word2 = "grandmom"
	result = check_for_permutation(word1, word2)
	print ("are {}, {} in permutation? {}".format(word1, word2, result))

view raw check_one_is_permutation_of_other.py hosted with ❤ by GitHub