Given two strings S and T, find a string of the same length which is lexicographically greater than S and smaller than T. Print “-1” if no such string is formed.(S > T)
Note: string S = s1s2… sn is said to be lexicographically smaller than string T = t1t2… tn, if there exists an i, such that s1 = t1, s2 = t2, … si – 1 = ti – 1, si < ti.
Examples:
Input : S = "aaa", T = "ccc"
Output : aab
Explanation:
Here, 'b' is greater than any
letter in S[]('a') and smaller
than any letter in T[]('c').
Input : S = "abcde", T = "abcdf"
Output : -1
Explanation:
There is no other string between
S and T.
Approach: Find a string which is lexicographically greater than string S and check if it is smaller than string T, if yes print the string next else print “-1”.
To find string, iterate the string S in the reverse order, if the last letter is not ‘z’, increase the letter by one (to move to next letter). If it is ‘z’, change it to ‘a’ and move to the second last character.
Compare the resultant string with string T, if both strings are equal print ‘-1’, else print the resultant string.
Below is the implementation of above approach:
C++
// CPP program to find the string // in lexicographic order which is // in between given two strings #include <bits/stdc++.h> using namespace std; // Function to find the lexicographically // next string string lexNext(string s, int n) { // Iterate from last character for (int i = n - 1; i >= 0; i--) { // If not 'z', increase by one if (s[i] != 'z') { s[i]++; return s; } // if 'z', change it to 'a' s[i] = 'a'; } } // Driver Code int main() { string S = "abcdeg", T = "abcfgh"; int n = S.length(); string res = lexNext(S, n); // If not equal, print the // resultant string if (res != T) cout << res << endl; else cout << "-1" << endl; return 0; } |
Java
//Java program to find the string // in lexicographic order which is // in between given two strings class GFG { // Function to find the lexicographically // next string static String lexNext(String str, int n) { char[] s = str.toCharArray(); // Iterate from last character for (int i = n - 1; i >= 0; i--) { // If not 'z', increase by one if (s[i] != 'z') { s[i]++; return String.valueOf(s); } // if 'z', change it to 'a' s[i] = 'a'; } return null; } // Driver Code static public void main(String[] args) { String S = "abcdeg", T = "abcfgh"; int n = S.length(); String res = lexNext(S, n); // If not equal, print the // resultant string if (res != T) { System.out.println(res); } else { System.out.println("-1"); } } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to find the string # in lexicographic order which is # in between given two strings # Function to find the lexicographically # next string def lexNext(s, n): # Iterate from last character for i in range(n - 1, -1, -1): # If not 'z', increase by one if s[i] != 'z': k = ord(s[i]) s[i] = chr(k + 1) return ''.join(s) # if 'z', change it to 'a' s[i] = 'a' # Driver Code if __name__ == "__main__": S = "abcdeg" T = "abcfgh" n = len(S) S = list(S) res = lexNext(S, n) # If not equal, print the # resultant string if res != T: print(res) else: print(-1) # This code is conributed by # sanjeev2552 |
C#
//C# program to find the string // in lexicographic order which is // in between given two strings using System; public class GFG { // Function to find the lexicographically // next string static String lexNext(String str, int n) { char[] s = str.ToCharArray(); // Iterate from last character for (int i = n - 1; i >= 0; i--) { // If not 'z', increase by one if (s[i] != 'z') { s[i]++; return new String(s); } // if 'z', change it to 'a' s[i] = 'a'; } return null; } // Driver Code static public void Main() { String S = "abcdeg", T = "abcfgh"; int n = S.Length; String res = lexNext(S, n); // If not equal, print the // resultant string if (res != T) { Console.Write(res); } else { Console.Write("-1"); } } } // This code is contributed by 29AjayKumar |
Output:
abcdeh
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