Count unique substrings of a string S present in a wraparound string
Last Updated :
23 Jul, 2025
Given a string S which is an infinite wraparound string of the string "abcdefghijklmnopqrstuvwxyz", the task is to count the number of unique non-empty substrings of a string p are present in s.
Examples:
Input: S = "zab"
Output: 6
Explanation: All possible substrings are "z", "a", "b", "za", "ab", "zab".
Input: S = "cac"
Output: 2
Explanation: All possible substrings are "a" and "c" only.
Approach: Follow the steps below to solve the problem
- Iterate over each character of the string
- Initialize an auxiliary array arr[] of size 26, to store the current length of substring that is present in string S starting from each character of string P.
- Initialize a variable, say curLen, which stores the length of substring present in P including the current character if the current character is not a part of the previous substring.
- Initialize a variable, say ans, to store the unique count of non-empty substrings of p present in S.
- Iterate over the characters of the string and check for the following two cases:
- Check if the current character can be added with previous substring to form the required substring or not.
- Add the difference of curLen and arr[curr] to ans if (curLen + 1) is greater than arr[curr] to avoid repetition of substrings.
- Print the value of ans.
Below is the implementation of the above approach:
C++
// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of
// non-empty substrings of p present in s
int findSubstringInWraproundString(string p)
{
// Stores the required answer
int ans = 0;
// Stores the length of
// substring present in p
int curLen = 0;
// Stores the current length
// of substring that is
// present in string s starting
// from each character of p
int arr[26] = { 0 };
// Iterate over the characters of the string
for (int i = 0; i < (int)p.length(); i++) {
int curr = p[i] - 'a';
// Check if the current character
// can be added with previous substring
// to form the required substring
if (i > 0
&& (p[i - 1]
!= ((curr + 26 - 1) % 26 + 'a'))) {
curLen = 0;
}
// Increment current length
curLen++;
if (curLen > arr[curr]) {
// To avoid repetition
ans += (curLen - arr[curr]);
// Update arr[cur]
arr[curr] = curLen;
}
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
string p = "zab";
// Function call to find the
// count of non-empty substrings
// of p present in s
findSubstringInWraproundString(p);
return 0;
}
import java.util.*;
class GFG
{
// Function to find the count of
// non-empty substrings of p present in s
static void findSubstringInWraproundString(String p)
{
// Stores the required answer
int ans = 0;
// Stores the length of
// substring present in p
int curLen = 0;
// Stores the current length
// of substring that is
// present in string s starting
// from each character of p
int arr[] = new int[26];
// Iterate over the characters of the string
for (int i = 0; i < p.length(); i++)
{
int curr = p.charAt(i) - 'a';
// Check if the current character
// can be added with previous substring
// to form the required substring
if (i > 0
&& (p.charAt(i - 1)
!= ((curr + 26 - 1) % 26 + 'a')))
{
curLen = 0;
}
// Increment current length
curLen++;
if (curLen > arr[curr])
{
// To avoid repetition
ans += (curLen - arr[curr]);
// Update arr[cur]
arr[curr] = curLen;
}
}
// Print the answer
System.out.println(ans);
}
// Driver Code
public static void main(String args[])
{
String p = "zab";
// Function call to find the
// count of non-empty substrings
// of p present in s
findSubstringInWraproundString(p);
}
}
// This code is contributed by hemanth gadarla
# Python3 program for
# the above approach
# Function to find the count of
# non-empty substrings of p present in s
def findSubstringInWraproundString(p) :
# Stores the required answer
ans = 0
# Stores the length of
# substring present in p
curLen = 0
# Stores the current length
# of substring that is
# present in string s starting
# from each character of p
arr = [0]*26
# Iterate over the characters of the string
for i in range(0, len(p)) :
curr = ord(p[i]) - ord('a')
# Check if the current character
# can be added with previous substring
# to form the required substring
if (i > 0 and (ord(p[i - 1]) != ((curr + 26 - 1) % 26 + ord('a')))) :
curLen = 0
# Increment current length
curLen += 1
if (curLen > arr[curr]) :
# To avoid repetition
ans += (curLen - arr[curr])
# Update arr[cur]
arr[curr] = curLen
# Print the answer
print(ans)
p = "zab"
# Function call to find the
# count of non-empty substrings
# of p present in s
findSubstringInWraproundString(p)
# This code is contributed by divyeshrabadiya07.
// C# program for
// the above approach
using System;
class GFG
{
// Function to find the count of
// non-empty substrings of p present in s
static void findSubstringInWraproundString(string p)
{
// Stores the required answer
int ans = 0;
// Stores the length of
// substring present in p
int curLen = 0;
// Stores the current length
// of substring that is
// present in string s starting
// from each character of p
int[] arr = new int[26];
// Iterate over the characters of the string
for (int i = 0; i < (int)p.Length; i++)
{
int curr = p[i] - 'a';
// Check if the current character
// can be added with previous substring
// to form the required substring
if (i > 0 && (p[i - 1] != ((curr + 26 - 1) % 26 + 'a')))
{
curLen = 0;
}
// Increment current length
curLen++;
if (curLen > arr[curr])
{
// To avoid repetition
ans += (curLen - arr[curr]);
// Update arr[cur]
arr[curr] = curLen;
}
}
// Print the answer
Console.Write(ans);
}
// Driver code
static void Main()
{
string p = "zab";
// Function call to find the
// count of non-empty substrings
// of p present in s
findSubstringInWraproundString(p);
}
}
// This code is contributed by divyesh072019.
<script>
// Javascript program for the above approach
// Function to find the count of
// non-empty substrings of p present in s
function findSubstringInWraproundString(p)
{
// Stores the required answer
let ans = 0;
// Stores the length of
// substring present in p
let curLen = 0;
// Stores the current length
// of substring that is
// present in string s starting
// from each character of p
let arr = new Array(26);
arr.fill(0);
// Iterate over the characters of the string
for (let i = 0; i < p.length; i++)
{
let curr = p[i].charCodeAt() - 'a'.charCodeAt();
// Check if the current character
// can be added with previous substring
// to form the required substring
if (i > 0 && (p[i - 1].charCodeAt() != ((curr + 26 - 1) % 26 + 'a'.charCodeAt())))
{
curLen = 0;
}
// Increment current length
curLen++;
if (curLen > arr[curr])
{
// To avoid repetition
ans += (curLen - arr[curr]);
// Update arr[cur]
arr[curr] = curLen;
}
}
// Print the answer
document.write(ans);
}
let p = "zab";
// Function call to find the
// count of non-empty substrings
// of p present in s
findSubstringInWraproundString(p);
// This code is contributed by surehs07.
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
Method 2:
Approach Steps:
- Initialize a dictionary to store the count of distinct substrings starting with each letter of the English alphabet.
- Initialize the length of the longest increasing substring ending at each position in the given string 'p' to 0.
- Iterate over the characters of 'p' and update the length of the longest increasing substring ending at each position using dynamic programming.
- Update the count of distinct substrings starting with each letter of 'p' based on the length of the longest increasing substring ending at each position.
- Return the sum of counts for all letters.
C++
#include <iostream>
#include <cstring>
using namespace std;
int countSubstringsInWraparoundString(string p) {
// Initialize an array to store the count
// of distinct substrings starting with each letter
int count[26];
memset(count, 0, sizeof(count));
// Initialize the length of the longest increasing
// substring ending at each position to 0
int len_inc_substring[p.length()];
memset(len_inc_substring, 0, sizeof(len_inc_substring));
// Iterate over the characters of the string
for (int i = 0; i < p.length(); i++) {
// Update the length of the longest increasing
// substring ending at the current position
if (i > 0 && (p[i] - p[i-1] + 26) % 26 == 1) {
len_inc_substring[i] = len_inc_substring[i-1] + 1;
}
else {
len_inc_substring[i] = 1;
}
// Update the count of distinct substrings
// starting with the current letter
count[p[i]-'a'] = max(count[p[i]-'a'], len_inc_substring[i]);
}
// Return the sum of counts for all letters
int total_count = 0;
for (int i = 0; i < 26; i++) {
total_count += count[i];
}
return total_count;
}
int main() {
string p = "zab";
cout << countSubstringsInWraparoundString(p) << endl; // Output: 6
return 0;
}
// Java code to count the number of distinct substrings in a wraparound string
import java.util.Arrays;
public class Main {
public static int countSubstringsInWraparoundString(String p) {
// Initialize an array to store the count
// of distinct substrings starting with each letter
int[] count = new int[26];
Arrays.fill(count, 0);
// Initialize the length of the longest increasing
// substring ending at each position to 0
int[] len_inc_substring = new int[p.length()];
Arrays.fill(len_inc_substring, 0);
// Iterate over the characters of the string
for (int i = 0; i < p.length(); i++) {
// Update the length of the longest increasing
// substring ending at the current position
if (i > 0 && (p.charAt(i) - p.charAt(i-1) + 26) % 26 == 1) {
len_inc_substring[i] = len_inc_substring[i-1] + 1;
}
else {
len_inc_substring[i] = 1;
}
// Update the count of distinct substrings
// starting with the current letter
count[p.charAt(i)-'a'] = Math.max(count[p.charAt(i)-'a'], len_inc_substring[i]);
}
// Return the sum of counts for all letters
int total_count = 0;
for (int i = 0; i < 26; i++) {
total_count += count[i];
}
return total_count;
}
public static void main(String[] args) {
String p = "zab";
System.out.println(countSubstringsInWraparoundString(p)); // Output: 6
}
}
def countSubstringsInWraparoundString(p):
# Initialize a dictionary to store the count
# of distinct substrings starting with each letter
count = {chr(i): 0 for i in range(ord('a'), ord('z')+1)}
# Initialize the length of the longest increasing
# substring ending at each position to 0
len_inc_substring = [0] * len(p)
# Iterate over the characters of the string
for i in range(len(p)):
# Update the length of the longest increasing
# substring ending at the current position
if i > 0 and (ord(p[i]) - ord(p[i-1])) % 26 == 1:
len_inc_substring[i] = len_inc_substring[i-1] + 1
else:
len_inc_substring[i] = 1
# Update the count of distinct substrings
# starting with the current letter
count[p[i]] = max(count[p[i]], len_inc_substring[i])
# Return the sum of counts for all letters
return sum(count.values())
# Test the function with a sample input
p = "zab"
print(countSubstringsInWraparoundString(p)) # Output: 6
using System;
public class MainClass {
public static int
CountSubstringsInWraparoundString(string p)
{
// Initialize an array to store the count
// of distinct substrings starting with each letter
int[] count = new int[26];
Array.Fill(count, 0);
// Initialize the length of the longest increasing
// substring ending at each position to 0
int[] len_inc_substring = new int[p.Length];
Array.Fill(len_inc_substring, 0);
// Iterate over the characters of the string
for (int i = 0; i < p.Length; i++) {
// Update the length of the longest increasing
// substring ending at the current position
if (i > 0 && (p[i] - p[i - 1] + 26) % 26 == 1) {
len_inc_substring[i]
= len_inc_substring[i - 1] + 1;
}
else {
len_inc_substring[i] = 1;
}
// Update the count of distinct substrings
// starting with the current letter
count[p[i] - 'a'] = Math.Max(
count[p[i] - 'a'], len_inc_substring[i]);
}
// Return the sum of counts for all letters
int total_count = 0;
for (int i = 0; i < 26; i++) {
total_count += count[i];
}
return total_count;
}
public static void Main()
{
string p = "zab";
Console.WriteLine(CountSubstringsInWraparoundString(
p)); // Output: 6
}
}
function countSubstringsInWraparoundString(p) {
// Initialize a dictionary to store the count
// of distinct substrings starting with each letter
let count = {};
for (let i = 'a'.charCodeAt(0); i <= 'z'.charCodeAt(0); i++) {
count[String.fromCharCode(i)] = 0;
}
// Initialize the length of the longest increasing
// substring ending at each position to 0
let len_inc_substring = new Array(p.length).fill(0);
// Iterate over the characters of the string
for (let i = 0; i < p.length; i++) {
// Update the length of the longest increasing
// substring ending at the current position
if (i > 0 && (p.charCodeAt(i) - p.charCodeAt(i-1) + 26) % 26 == 1) {
len_inc_substring[i] = len_inc_substring[i-1] + 1;
} else {
len_inc_substring[i] = 1;
}
// Update the count of distinct substrings starting with the current letter
count[p[i]] = Math.max(count[p[i]], len_inc_substring[i]);
}
// Return the sum of counts for all letters
return Object.values(count).reduce((a,b) => a+b);
}
// Test the function with a sample input
let p = "zab";
console.log(countSubstringsInWraparoundString(p)); // Output: 6
Time Complexity:
The time complexity of this approach is O(n), where n is the length of the input string 'p'. This is because we iterate over the characters of 'p' only once.
Auxiliary Space:
The auxiliary space of this approach is O(26), which is constant. This is because we use a dictionary to store the count of distinct substrings starting with each letter of the English alphabet, and the size of the dictionary is fixed at 26. We also use a list of size n to store the length of the longest increasing substring ending at each position in 'p'. Therefore, the total auxiliary space used by the algorithm is O(26 + n), which is equivalent to O(n).
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