Solution 3 for Scaler Topics Fortnightly Contest - 2

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DSA Problem Solving for Interviews using Java
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DSA Problem Solving for Interviews using Java
DSA Problem Solving for Interviews using Java
by Jitender Punia
1000
4.9
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This article is part of the Scaler Topics Fortnightly Contest - 2.

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Solution Approach

  • Let n be length of A, and m be the length of B.
  • Let's say for an ith position in string A, L[i] is the largest integer such that string B0...L[i] is present as a subsequence in A0..i .
  • Similarly, for ith position in string A, R[i] is the smallest integer such that string BR[i]...m-1 is present as a subsequence is Ai..n-1.
  • Then for each i, if L[i] >= R[i] is true, then your answer is "Yes". Otherwise "No". So now, How to find L[i] and R[i]?
  • For L[i] -
  • Make a pos array of length 26 and initialize it to 0. Initialize integer j = 0, with which we will be iterating B.
  • Where pos[i] will store the largest index where ith character of lowercase english letters exists. e.g 0th position is for character 'a'.
  • Now start traversing string A. Whenever you find a match i.e. A[i] == B[j], do the following :
  • update pos[A[i]-'a'] = j, increment j.
  • For each i do the following L[i] = pos[A[i]-'a'].
  • Similiarily you can do for R[i], but from back.

C++ Implementation

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Java Implementation

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Python Implementation