Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Program to find start indices of all anagrams of a string S in T in Python
Suppose we have two strings S and T, we have to find all the start indices of S's anagrams in T. The strings consist of lowercase letters only and the length of both strings S and T will not be larger than 20 and 100.
So, if the input is like S = "cab" T = "bcabxabc", then the output will be [0, 1, 5, ], as the substrings "bca", "cab" and "abc".
To solve this, we will follow these steps:
Define a map m, n := size of s, set left := 0, right := 0, counter := size of p
define an array ans
store the frequency of characters in p into the map m
for right := 0 to n – 1
if m has s[right] and m[s[right]] is non zero, then decrease m[s[right]] by 1, decrease counter by 1 and if counter = 0, then insert left into ans
otherwise
while left < right,
if s[left] is not present in m, then increase counter by 1, and increase m[s[left]] by 1
increase left by 1
if m has s[right] and m[s[right]] is non zero, then decrease right by 1, and come out from the loop
if m has no s[left], then set left := right + 1
return ans
Let us see the following implementation to get better understanding:
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
map <char, int> m;
int n = s.size();
int left = 0, right = 0;
int counter = p.size();
vector <int> ans;
for(int i = 0; i < p.size(); i++) m[p[i]]++;
for(int right = 0; right < n; right++){
if(m.find(s[right]) != m.end() && m[s[right]]){
m[s[right]]--;
counter--;
if(counter == 0)ans.push_back(left);
} else {
while(left<right){
if(m.find(s[left]) != m.end()) {
counter++;
m[s[left]]++;
}
left++;
if(m.find(s[right]) != m.end() && m[s[right]]){
right--;
break;
}
}
if(m.find(s[left])==m.end())left = right + 1;
}
}
return ans;
}
};
main(){
Solution ob;
print_vector(ob.findAnagrams("bcabxabc", "cab")) ;
}Input
"bcabxabc", "cab"
Output
[0, 1, 5, ]
99 Views
