Given string str of length L and an integer N, the task is to form a total of (L / N) contiguous sub-segments of the string which contain distinct subsequent characters.
Note: that the integer N will be a factor of the length of the string i.e L.
Examples:
Input: str = “neveropengfg”, N = 4
Output:
gek
sfor
gek
sgf
The length of “neveropengfg” is 16, therefore there will be 4 subsegments.
The first subsegment will contain the characters g, e, e and k but the alphabet ‘e’ is repeated,
So we discard one ‘e’. Therefore, the final subsegment will be “gek”.
Similarly, the other three subsegments will be “sfor”, “gek” and “sgf”.
Each subsegment should have subsequent distinct characters.Input: str = “aabdekfgf”, N = 3
Output:
ab
dek
fg
Approach: An array is created every time the iteration is started over the new sub segment. Those elements are stored in a sequence in that array. If any element already exists in the array, then it is not pushed into the array.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include<bits/stdc++.h>using namespace std;// Function that prints the segments void sub_segments(string str, int n){ int l = str.length(); for (int x = 0; x < l; x += n) { string newlist = str.substr(x, n); // New array for every iteration list<char> arr; // Iterator for new array list<char>::iterator it; for(auto y:newlist) { it = find(arr.begin(), arr.end(), y); // Check if iterator points to end or not if(it == arr.end()) arr.push_back(y); } for(auto y:arr) cout << y; cout << endl; } } // Driver code int main() { string str = "neveropengfg"; int n = 4; sub_segments(str, n); } // This code is contributed by Sanjit_Prasad |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function that prints the segmentsstatic void sub_segments(String str, int n) { int l = str.length(); for (int x = 0; x < l; x += n) { String newlist = str.substring(x, x + n); // New array for every iteration List<Character> arr = new ArrayList<Character>(); for (char y : newlist.toCharArray()) { // Check if the character is in the array if (!arr.contains(y)) arr.add(y); } for (char y : arr) System.out.print(y); System.out.println(); }}// Driver codepublic static void main(String[] args){ String str = "neveropengfg"; int n = 4; sub_segments(str, n);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach# Function that prints the segmentsdef sub_segments (string, n): l = len (string) for x in range (0, l, n): newlist = string[x : x + n] # New array for every iteration arr = [] for y in newlist: # Check if the character is in the array if y not in arr: arr.append (y) print (''.join (arr))# Driver codestring = "neveropengfg"n = 4sub_segments (string, n) |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {// Function that prints the segmentsstatic void sub_segments(String str, int n) { int l = str.Length; for (int x = 0; x < l; x += n) { String newlist = str.Substring(x, n); // New array for every iteration List<char> arr = new List<char>(); foreach (char y in newlist.ToCharArray()) { // Check if the character is in the array if (!arr.Contains(y)) arr.Add(y); } foreach (char y in arr) Console.Write(y); Console.WriteLine(); }}// Driver codepublic static void Main(String[] args){ String str = "neveropengfg"; int n = 4; sub_segments(str, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function that prints the segmentsfunction sub_segments(str, n) { let l = str.length; for (let x = 0; x < l; x += n) { let newlist = str.substr(x, n); // New array for every iteration let arr = []; for (let y of newlist) { // Check if the character is in the array if (!arr.includes(y)) arr.push(y); } for (let y of arr) document.write(y); document.write("<br>"); }}// Driver codelet str = "neveropengfg";let n = 4;sub_segments(str, n);// This code is contributed by gfgking</script> |
gek sfor gek sgf
Time Complexity: O(L), Here L is the length of the string.
Auxiliary Space: O(n), The extra space is used to store the substrings.
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