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I have two arrays with letters. I want to know if array A is contained in array B as follows: The letters in A must appear adjacent to each other in array B but do not have to appear in the same order as they were in array A. Example that would be accepted

A = abcd B = hbadcg

A = aabc B = abcag

Examples that would not be accepted

A = aabcd B = adcbga

A = abcd B = abbcdg

What I could do is for every variation of A check if its a sub string in B. but I'm looking for a better way

Mark Rotteveel
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megoan
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    You should show what you did so far and wherr exactly the problem is. Otherwise the question is too broad. – Laurenz Albe Oct 07 '18 at 10:07
  • what i wrote at the end is the only way I could think of. that's why i wanted to know if anyone knows of a better way – megoan Oct 07 '18 at 10:42

1 Answers1

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Consider using a two-pointer approach for the given problem.

  • Store the count of each character in A in a hash map - HashMapA
  • Maintain two pointers, start and end as we iterate over the array B.
  • Maintain yet another hash map to store the count of characters in the range [start, end] appearing in array B - HashMapB

Sharing ideone link for the same: https://ideone.com/vLmaxL

for(char c : A) HashMapA[c]++;

start = 0
for(int end = 0; end < B.length(); end++) {
  char c = B[end];
  HashMapB[c]++;
  while(HashMapB[c] > HashMapA[c] && start <= end) {
    HashMapB[ B[start] ]--;
    start++;  
  }
  if(end - start + 1 == A.length())
    return true;
} 

return false;
Nilesh
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  • thanx @nellex for the answer. it was the fastest i was able to find. i would just add at the beginning that if B contains A then return true right away. but other then that looks good! – megoan Oct 07 '18 at 14:23