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Unique Substrings in Wraparound String in C++
Suppose we have the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so the value s will look like this − "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".
Now we have another string p. Our job is to find out how many unique non-empty substrings of p are present in s. In particular, our input is the string p and we need to output the number of different non-empty substrings of p in the string s.
So if the input is like “zab” the output will be 6. There are 6 substrings “z”, “a”, “b”, “za”, “ab”, “zab” of the string “zab” in the string s
To solve this, we will follow these steps −
Create an array dp of size 26, set x := 0
for I in range 0 to the size of p
if i > 0 and (p[i] – p[i – 1] is 1 or p[i – 1] – p[i] is 25), then increase x by 1, otherwise set x := 1
dp[p[i] – ASCII of ‘a’] := maximum of (x, dp[p[i] – ASCII of ‘a’])
ret := 0
for I in range 0 to 25
ret := ret + dp[i]
return ret
Example (C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int findSubstringInWraproundString(string p) {
vector <int> dp(26);
int x = 0;
for(int i = 0; i < p.size(); i++){
if(i > 0 && (p[i] - p[i - 1] == 1 || p[i - 1] - p[i] == 25)){
x++;
}
else x = 1;
dp[p[i] - 'a'] = max(x, dp[p[i] - 'a']);
}
int ret = 0;
for(int i = 0; i < 26; i++){
ret += dp[i];
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.findSubstringInWraproundString("zab"));
}Input
"zab"
Output
6
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