anagram detecting using linked lists in c - c

Question

I'm trying to write a code that'll check 2 words if they are anagrams,using linked lists.To do that,I guess it should receive 2 words from the user and pass every letter they contain to linked list's nodes,and compare the nodes if they have the same letter,if so,remove the same letter from the second word.When the process is done,if the second list is empty,then they are anagrams.Even if 1 letter is not matching,it should return 0,but I don't know how to determine the length of these words,here is what I wrote so far
#include <stdio.h>
#include <stdlib.h>
struct node
{
struct node *prev;
struct node *next;
char data;
};
typedef struct node *NODE,NOD;
NODE last(NODE list)
{
if(list!=NULL)
while(list->next!=NULL)
list=list->next;
NODE lst;
lst=list;
return lst;
}
void insert( char letter, NODE list)
{
NODE nod;
nod=(NODE)malloc(sizeof(NOD));
nod->next=NULL;
nod->prev=NULL;
nod->data=letter;
if(list==NULL)
{
list=nod;
}
else
{
nod->prev=last(list);
last(list)->next=nod;
}
}

Answer 1

Just check that each word has the same number of each letter in it.
int anagrams(const char *a, const char *b) {
int counts[256] = {0};
while (*a) counts[*a++]++;
while (*b) counts[*b++]--;
for (int i = 0; i < 256; i++) {
if (counts[i]) return 0;
}
return 1;
}

Answer 2

Why use linked lists for such an easy problem?
O(N) solution:
Calculate frequencies of every letter for each word and then compare these 2 histograms. If they're equal, then one word can be obtained from another.
If you want to use your linked-list-based solution, then, the length of the word is, indeed:
Length of each input word (they must have the same length) - it can be calculated with a single traversal from the linked list head to the tail.
Amount of removed symbols

Answer 3

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
typedef struct node {
char data;
struct node *prev;
struct node *next;
} NODE;
NODE *newNode(char ch){
NODE *p = malloc(sizeof(NODE));
if(p){
p->data = ch;
p->prev = p->next = NULL;
}
return p;
}
void insert(NODE **list, NODE *node){
if(*list == NULL){
*list = node;
return ;
}
NODE *curr = *list;
while(curr){
if(curr->data >= node->data){//insert sort
if(curr->prev == NULL)
*list = node;
else
curr->prev->next = node;
node->prev = curr->prev;
curr->prev = node;
node->next = curr;
break;
}
if(curr->next == NULL){
curr->next = node;
node->prev = curr;
break;
} else
curr = curr->next;
}
}
NODE *input_word(){
NODE *word=NULL;
int ch;
while(EOF!=(ch=getchar()) && ch != '\n'){
insert(&word, newNode(ch));
}
return word;
}
bool isAnagram(NODE *word1, NODE *word2){
while(word1 && word2){
if(word1->data != word2-> data)
return false;
word1 = word1->next;
word2 = word2->next;
}
return word1 == NULL && word2 == NULL;
}
int main(){
NODE *word1, *word2;
printf("input word : ");
word1 = input_word();
printf("input other word : ");
word2 = input_word();
if(isAnagram(word1, word2))
printf("YES\n");
else
printf("NO\n");
//drop(word1);drop(word2);
return 0;
}