Given a string S consisting of N lowercase characters, the task is to find the starting and the ending indices ( 0-based indexing ) of the substring of the given string S that needed to be reversed to make the string S sorted. If it is not possible to sort the given string S by reversing any substring, then print “-1”.
Examples:
Input: S = “abcyxuz”
Output: 3 5
Explanation: Reversing the substring from indices [3, 5] modifies the string to “abcuxyz”, which is sorted.
Therefore, print 3 and 5.Input: S = “GFG”
Output: 0 1
Naive Approach: The simplest approach to solve the given problem is to generate all possible substring of the given string S and if there exists any substring such reversing it makes the string sorted, then print the indices of that substrings. Otherwise, print “-1”.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized based on the observation that to sort the string by only reversing one substring, the original string must be in one of the following formats:
- Decreasing string
- Increasing substring + Decreasing substring
- Decreasing substring + Increasing substring
- Increasing substring + Decreasing substring + Increasing substring
Follow the steps below to solve the problem:
- Initialize two variables, say start and end as -1, that stores the starting and ending indices of the substring to be reversed respectively.
- Initialize a variable, say flag as 1, that stores if it is possible to sort the string or not.
-
Iterate over the range [1, N] and perform the following operations:
- If the characters S[i] is less than characters S[i – 1] then find the index of the right boundary of the decreasing substring starting from the index (i – 1) and store it in end.
- Check if reversing the substring S[i – 1, end] makes the string sorted or not. If found to be false then print “-1” and return. Otherwise, mark the flag as false.
- After completing the above steps update the value of i with the right boundary of the substring.
- If the characters S[i] is less than characters S[i – 1] and the flag is false, then print “-1” and return.
- If the start is equal to -1, then update the value of start and end as 1.
- After completing the above steps, print the value of start and end as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the substring // in S required to be reversed bool adjust(string& S, int & i,
int & start, int & end)
{ // Stores the size of the string
int N = S.length();
// Stores the starting point
// of the substring
start = i - 1;
// Iterate over the string S
// while i < N
while (i < N && S[i] < S[i - 1]) {
// Increment the value of i
i++;
}
// Stores the ending index of
// the substring
end = i - 1;
// If start <= 0 or i >= N
if (start <= 0 && i >= N)
return true ;
// If start >= 1 and i <= N
if (start >= 1 && i <= N) {
// Return the boolean value
return (S[end] >= S[start - 1]
&& S[start] <= S[i]);
}
// If start >= 1
if (start >= 1) {
// Return the boolean value
return S[end] >= S[start - 1];
}
// If i < N
if (i < N) {
// Return true if S[start]
// is less than or equal to
// S[i]
return S[start] <= S[i];
}
// Otherwise
return false ;
} // Function to check if it is possible // to sort the string or not void isPossible(string& S, int N)
{ // Stores the starting and the
// ending index of substring
int start = -1, end = -1;
// Stores whether it is possible
// to sort the substring
bool flag = true ;
// Traverse the range [1, N]
for ( int i = 1; i < N; i++) {
// If S[i] is less than S[i-1]
if (S[i] < S[i - 1]) {
// If flag stores true
if (flag) {
// If adjust(S, i, start,
// end) return false
if (adjust(S, i, start, end)
== false ) {
// Print -1
cout << -1 << endl;
return ;
}
// Unset the flag
flag = false ;
}
// Otherwise
else {
// Print -1
cout << -1 << endl;
return ;
}
}
}
// If start is equal to -1
if (start == -1) {
// Update start and end
start = end = 1;
}
// Print the value of start
// and end
cout << start << " "
<< end << "\n" ;
} // Driver Code int main()
{ string S = "abcyxuz" ;
int N = S.length();
isPossible(S, N);
return 0;
} |
// Java program for the above approach import java.util.*;
import java.lang.*;
class GFG{
static int i, start, end;
// Function to find the substring // in S required to be reversed static boolean adjust(String S)
{ // Stores the size of the string
int N = S.length();
// Stores the starting point
// of the substring
start = i - 1 ;
// Iterate over the string S
// while i < N
while (i < N && S.charAt(i) <
S.charAt(i - 1 ))
{
// Increment the value of i
i++;
}
// Stores the ending index of
// the substring
end = i - 1 ;
// If start <= 0 or i >= N
if (start <= 0 && i >= N)
return true ;
// If start >= 1 and i <= N
if (start >= 1 && i <= N)
{
// Return the boolean value
return (S.charAt(end) >= S.charAt(start - 1 ) &&
S.charAt(start) <= S.charAt(i));
}
// If start >= 1
if (start >= 1 )
{
// Return the boolean value
return S.charAt(end) >= S.charAt(start - 1 );
}
// If i < N
if (i < N)
{
// Return true if S[start]
// is less than or equal to
// S[i]
return S.charAt(start) <= S.charAt(i);
}
// Otherwise
return false ;
} // Function to check if it is possible // to sort the string or not static void isPossible(String S, int N)
{ // Stores the starting and the
// ending index of substring
start = - 1 ; end = - 1 ;
// Stores whether it is possible
// to sort the substring
boolean flag = true ;
// Traverse the range [1, N]
for (i = 1 ; i < N; i++)
{
// If S[i] is less than S[i-1]
if (S.charAt(i) < S.charAt(i - 1 ))
{
// If flag stores true
if (flag)
{
// If adjust(S, i, start,
// end) return false
if (adjust(S) == false )
{
// Print -1
System.out.println(- 1 );
return ;
}
// Unset the flag
flag = false ;
}
// Otherwise
else
{
// Print -1
System.out.println(- 1 );
return ;
}
}
}
// If start is equal to -1
if (start == - 1 )
{
// Update start and end
start = end = 1 ;
}
// Print the value of start
System.out.println(start + " " + end);
} // Driver code public static void main(String[] args)
{ String S = "abcyxuz" ;
int N = S.length();
isPossible(S, N);
} } // This code is contributed by offbeat |
# Python3 program for the above approach # Function to find the substring # in S required to be reversed def adjust(S, i, start, end):
# Stores the size of the string
N = len (S)
# Stores the starting point
# of the substring
start = i - 1
# Iterate over the string S
# while i < N
while (i < N and S[i] < S[i - 1 ]):
# Increment the value of i
i + = 1
# Stores the ending index of
# the substring
end = i - 1
# If start <= 0 or i >= N
if (start < = 0 and i > = N):
return True ,start,i,end
# If start >= 1 and i <= N
if (start > = 1 and i < = N):
# Return the boolean value
return (S[end] > = S[start - 1 ] and S[start] < = S[i]),start,i,end
# If start >= 1
if (start > = 1 ):
# Return the boolean value
return (S[end] > = S[start - 1 ]),start,i,end
# If i < N
if (i < N):
# Return true if S[start]
# is less than or equal to
# S[i]
return (S[start] < = S[i]),start,i,end
# Otherwise
return False ,start,i,end
# Function to check if it is possible # to sort the string or not def isPossible(S, N):
# global start,end,i
# Stores the starting and the
# ending index of substring
start, end = - 1 , - 1
# Stores whether it is possible
# to sort the substring
flag = True
# Traverse the range [1, N]
i = 1
while i < N:
# If S[i] is less than S[i-1]
if (S[i] < S[i - 1 ]):
# If flag stores true
if (flag):
# If adjust(S, i, start,
# end) return false
f, start, i, end = adjust(S, i, start, end)
if (f = = False ):
# Pr-1
print ( - 1 )
return
# Unset the flag
flag = False
# Otherwise
else :
# Pr-1
print ( - 1 )
return
i + = 1 # If start is equal to -1
if (start = = - 1 ):
# Update start and end
start, end = 1 , 1
# Print the value of start
# and end
print (start, end)
# Driver Code if __name__ = = '__main__' :
S = "abcyxuz"
N = len (S)
isPossible(S, N)
# This code is contributed by mohit kumar 29.
|
// C# program for the above approach using System;
class GFG{
static int i, start, end;
// Function to find the substring // in S required to be reversed static bool adjust( string S)
{ // Stores the size of the string
int N = S.Length;
// Stores the starting point
// of the substring
start = i - 1;
// Iterate over the string S
// while i < N
while (i < N && S[i] < S[i - 1])
{
// Increment the value of i
i++;
}
// Stores the ending index of
// the substring
end = i - 1;
// If start <= 0 or i >= N
if (start <= 0 && i >= N)
return true ;
// If start >= 1 and i <= N
if (start >= 1 && i <= N)
{
// Return the boolean value
return (S[end] >= S[start - 1] &&
S[start] <= S[i]);
}
// If start >= 1
if (start >= 1)
{
// Return the boolean value
return S[end] >= S[start - 1];
}
// If i < N
if (i < N)
{
// Return true if S[start]
// is less than or equal to
// S[i]
return S[start] <= S[i];
}
// Otherwise
return false ;
} // Function to check if it is possible // to sort the string or not static void isPossible( string S, int N)
{ // Stores the starting and the
// ending index of substring
start = -1; end = -1;
// Stores whether it is possible
// to sort the substring
bool flag = true ;
// Traverse the range [1, N]
for (i = 1; i < N; i++)
{
// If S[i] is less than S[i-1]
if (S[i] < S[i - 1])
{
// If flag stores true
if (flag)
{
// If adjust(S, i, start,
// end) return false
if (adjust(S) == false )
{
// Print -1
Console.WriteLine(-1);
return ;
}
// Unset the flag
flag = false ;
}
// Otherwise
else
{
// Print -1
Console.WriteLine(-1);
return ;
}
}
}
// If start is equal to -1
if (start == -1)
{
// Update start and end
start = end = 1;
}
// Print the value of start
Console.WriteLine(start + " " + end);
} // Driver code static void Main()
{ string S = "abcyxuz" ;
int N = S.Length;
isPossible(S, N);
} } // This code is contributed by SoumikMondal |
<script> // JavaScript program for the above approach
var i, start, end;
// Function to find the substring
// in S required to be reversed
function adjust(S) {
// Stores the size of the string
var N = S.length;
// Stores the starting point
// of the substring
start = i - 1;
// Iterate over the string S
// while i < N
while (i < N && S[i] < S[i - 1]) {
// Increment the value of i
i++;
}
// Stores the ending index of
// the substring
end = i - 1;
// If start <= 0 or i >= N
if (start <= 0 && i >= N) return true ;
// If start >= 1 and i <= N
if (start >= 1 && i <= N) {
// Return the boolean value
return S[end] >= S[start - 1] && S[start] <= S[i];
}
// If start >= 1
if (start >= 1) {
// Return the boolean value
return S[end] >= S[start - 1];
}
// If i < N
if (i < N) {
// Return true if S[start]
// is less than or equal to
// S[i]
return S[start] <= S[i];
}
// Otherwise
return false ;
}
// Function to check if it is possible
// to sort the string or not
function isPossible(S, N) {
// Stores the starting and the
// ending index of substring
start = -1;
end = -1;
// Stores whether it is possible
// to sort the substring
var flag = true ;
// Traverse the range [1, N]
for (i = 1; i < N; i++) {
// If S[i] is less than S[i-1]
if (S[i] < S[i - 1]) {
// If flag stores true
if (flag) {
// If adjust(S, i, start,
// end) return false
if (adjust(S) === false ) {
// Print -1
document.write(-1);
return ;
}
// Unset the flag
flag = false ;
}
// Otherwise
else {
// Print -1
document.write(-1);
return ;
}
}
}
// If start is equal to -1
if (start === -1) {
// Update start and end
start = end = 1;
}
// Print the value of start
document.write(start + " " + end + "<br>" );
}
// Driver code
var S = "abcyxuz" ;
var N = S.length;
isPossible(S, N);
</script> |
3 5
Time Complexity: O(N)
Auxiliary Space: O(1)