Classic interview question: Hash table

Simple

Given two strings: ransomNote and magazine , determine ransomNote whether they can magazine be composed of the characters inside.

Return if possible true ; otherwise return false .

magazine Each character in can only ransomNote be used once in .

Example 1:

Input: ransomNote = "a", magazine = "b"
 Output: false

Example 2:

Input: ransomNote = "aa", magazine = "ab"
 Output: false

Example 3:

Input: ransomNote = "aa", magazine = "aab"
 Output: true

hint:

1 <= ransomNote.length, magazine.length <= 105
ransomNote and magazine consists of lowercase English letters

bool canConstruct(char* ransomNote, char* magazine) {
    int rr=strlen(ransomNote);
    int mm=strlen(magazine);
    int count=0;
    for(int i=0;i<rr;i++){
        for(int j=0;j<mm;j++){
            if(ransomNote[i]==magazine[j]){
               count++;
               magazine[j]=' ';
               break;
            }
        }
    }
    if(count==rr){
        return true;
    }
     return false;
}

205. Isomorphic strings

Simple

Given two strings s and t , determine whether they are isomorphic.

If s the characters in can be replaced according to a certain mapping relationship t , then the two strings are isomorphic.

Each occurrence of a character should be mapped to another character without changing the order of the characters. Different characters cannot be mapped to the same character, the same character can only be mapped to the same character, and characters can be mapped to themselves.

Example 1:
Input: s = "egg", t = "add"
 Output: true
Example 2:
Input: s = "foo", t = "bar"
 Output: false
Example 3:
Input: s = "paper", t = "title"
 Output: true
hint:

1 <= s.length <= 5 * 104

t.length == s.length

s and t consists of any valid ASCII characters
bool isIsomorphic(char* s, char* t) {
    int a[200],f[200]={0};
    for(int i=0;i<200;i++){
        a[i]=-1;
    }
    int ss=strlen(s);
    int tt=strlen(t);
    if(ss!=tt){
        return false;
    }
    for(int i=0;i<ss;i++){
        int p=s[i]-' ';
        if(a[p]==-1&&f[t[i]-' ']==0){
            a[p]=t[i]-' ';
            f[(t[i]-' ')]=1;
         }else if(a[p]==-1&&f[t[i]-' ']!=0){
                  return false;
         }else{
            if(a[p]!=t[i]-' '){
                   return false;
            }
       }

    }

    return true;
}

290. Word patterns

Simple

Given a pattern pattern and a string s , determine s whether it follows the same pattern.

Follow here refers to a complete match. For example, there is a two-way connection between pattern each letter in and each non-empty word in the string .s

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"
 Output: true

Example 2:

Input: pattern = "abba", s = "dog cat cat fish"
 Output: false

Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"
 Output: false

hint:

1 <= pattern.length <= 300
pattern Contains only lowercase English letters
1 <= s.length <= 3000
s Contains only lowercase English letters and ' '
s Does not contain any leading or trailing pairs of spaces
s Each word in is separated by a single space

bool wordPattern(char* pattern, char* s) {
    char *hashset[26];
    for(int i=0;i<26;i++){
        hashset[i]=NULL;
    }
    int pp=strlen(pattern);
    const char ff[2]=" ";
    char *token=NULL;
    for(int i=0;i<pp;i++){
      
        if(token == NULL){
            token = strtok(s, ff);
        }
        else{
             token = strtok(NULL, ff);
             if(token == NULL) {
                 return false;
             }
        }
        if(hashset[pattern[i]-'a']==NULL){
            
                 hashset[pattern[i]-'a']=token;
       }else{
           if(strcmp(hashset[pattern[i]-'a'],token)!=0){
                   return false;
            }
        }
    }
    token = strtok(NULL, ff);
    if(token != NULL)
    {
        return false;
    }
    for(int i =0; i < 26; i++)
    {
        if(hashset[i]!= NULL)
        {
            for(int j = i+ 1; j < 26; j++)
            {
                if(hashset[j]!= NULL)
                {
                    if(strcmp(hashset[i], hashset[j]) == 0)
                    {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}

242. Valid anagrams

Simple

Given two strings s and t , write a function to determine t whether is s an anagram of .

Note: If each character in s and t occurs the same number of times, then s and are said t to be anagrams of each other.

Example 1:

Input:  s = "anagram", t = "nagaram"
 Output: true

Example 2:

Input:  s = "rat", t = "car"
 Output: false

hint:

1 <= s.length, t.length <= 5 * 104
s and t contain only lowercase letters

bool isAnagram(char* s, char* t) {
    int snum[26]={0};
    int tnum[26]={0};
    int ss=strlen(s);
    int tt=strlen(t);
    for(int i=0;i<ss;i++){
        snum[s[i]-'a']++;
    }
    for(int i=0;i<tt;i++){
        tnum[t[i]-'a']++;
    }
    for(int i=0;i<26;i++){
        if(snum[i]!=tnum[i]){
            return false;
        }
    }
    return true;
}

1. The sum of two numbers

Simple

Given an integer array nums and an integer target value , please find the twotarget integers in the array whose sum is the target value and return their array subscripts.target

You can assume that each input will correspond to only one answer. However, the same element in the array cannot appear repeatedly in the answer.

You can return answers in any order.

Example 1:
Input: nums = [2,7,11,15], target = 9
 Output: [0,1]
 Explanation: Because nums[0] + nums[1] == 9, [0, 1] is returned.
Example 2:
Input: nums = [3,2,4], target = 6
 Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
 Output: [0,1]
hint:

2 <= nums.length <= 104

-109 <= nums[i] <= 109

-109 <= target <= 109

There will only be one valid answer
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
    int *a=(int*)malloc(sizeof(int)*2);
    for(int i=0;i<numsSize;i++){
        for(int j=i+1;j<numsSize;j++){
            if(nums[i]+nums[j]==target){
               a[0]=i;
               a[1]=j;
               (*returnSize)=2;
                break;
            }
        }
    }
     return a;
}

Classic interview question: Hash table

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