CODEWITHSUNDEEP
stringalgorithmdata-structures
princess of persia
princess of persia

Reputation: 2232

Sub sequence occurrence in a string

Given 2 strings like bangalore and blr, return whether one appears as a subsequence of the other. The above case returns true whereas bangalore and brl returns false.

Upvotes: 6

Views: 7858

Answers (4)

leo
leo

Reputation: 437

O(N) solution, where N is the length of the substring.

bool subsequence( string s1, string s2 ){
    int n1 = s1.length();
    int n2 = s2.length();

    if( n1 > n2 ){
        return false;
    }

    int i = 0;
    int j = 0;
    while( i < n1 && j < n2 ){
        if( s1[i] == s2[j] ){
            i++;
        } 
        j++;
    }

    return i == n1;
}

Upvotes: 0

Pani Dhakshnamurthy
Pani Dhakshnamurthy

Reputation: 233

// Solution 1
public static boolean isSubSequence(String str1, String str2) {
     int i = 0;
        int j = 0;
        while (i < str1.length() && j < str2.length()) {
            if (str1.charAt(i) == str2.charAt(j)) {
                i++;
                j++;
            } else {
                i++;
            }
        }
    return j == str2.length();
}

// Solution 2 using String indexOf method
public static boolean isSubSequenceUsingIndexOf(String mainStr, String subStr) {
    int i = 0;
    int index = 0;
    while(i<subStr.length()) {
        char c = subStr.charAt(i);
        if((index = mainStr.indexOf(c, index)) == -1) {
            return false;
        }
        i++;
    }

    return true;
}

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727077

Greedy strategy should work for this problem.

  • Find the first letter of the suspected substring (blr) in the big string (*b*angalore)
  • Find the second letter starting at the index of the first letter plus one (anga*l*ore)
  • Find the third letter starting at the index of the second letter plus one (o*r*e)
  • Continue until you can no longer find the next letter of blr in the string (no match), or you run out of letters in the subsequence (you have a match).

Here is a sample code in C++:

#include <iostream>
#include <string>
using namespace std;

int main() {
    string txt = "quick brown fox jumps over the lazy dog";
    string s = "brownfoxzdog";
    int pos = -1;
    bool ok = true;
    for (int i = 0 ; ok && i != s.size() ; i++) {
        ok = (pos = txt.find(s[i], pos+1)) != string::npos;
    }
    cerr << (ok ? "Found" : "Not found") << endl;
    return 0;
}

Upvotes: 17

MBo
MBo

Reputation: 80325

Find the length of the longest common subsequence. If it is equal to the length of small string, return true

Upvotes: -1

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