Given a string str, the task is to find the minimum number of moves required to reach lexicographically largest and smallest characters. In one move, a jump can be made from leftmost side or the rightmost side of given string.
Examples:
Input: str = AEDCB, N = 5
Output: 2
Explanation: take two steps from leftmost side to reach A and EInput: str = BACDEFHG, N = 8
Output: 4
Explanation: take two steps from leftmost side to reach A and 2 steps from rightmost side to reach H(2+2=4)Input: str = CDBA, N = 4
Output: 3
Explanation: take three steps from rightmost side to reach A and then D
Approach: This problem is implementation-based. Follow the steps below to solve the given problem.
- Find the index of maximum and minimum elements in the string
- Calculate the minimum and maximum of these indexes to find the min_steps and max_steps that can be taken
- There are only three possible ways to reach both elements
- traverse from the start and cover both elements i.e min_steps+1
- traverse from the last and cover both elements i.e n-max_steps
- traverse from both start and end i.e min_steps+1+n-max_steps
- The final answer is a minimum of all possible three ways of traversal.
Below is the implementation of the above approach:
// C++ program for above approach #include <bits/stdc++.h> using namespace std;
// Minimum number of moves required to // reach and largest and smallest ASCII values int min_moves(string s, int n)
{ int maxpos = 0, minpos = 0;
// Finding index of maximum
// and minimum element in string
for ( int i = 0; i < n; i++) {
if (s[i] > s[maxpos])
maxpos = i;
if (s[i] < s[minpos])
minpos = i;
}
// Calculating minimum ans maximum steps
// that can be taken
int min_steps = min(maxpos, minpos);
int max_steps = max(maxpos, minpos);
// Only three possible ways
// to reach both elements
int ans1, ans2, ans3;
ans1 = n - min_steps;
ans2 = max_steps + 1;
ans3 = min_steps + 1 + n - max_steps;
int result;
// Minimum steps in all three ways
result = min(ans1, min(ans2, ans3));
// Return the final result
return result;
} // Driver code int main()
{ string str = "BACDEFHG" ;
int N = str.length();
cout << min_moves(str, N);
return 0;
} |
// Java code for the above approach import java.io.*;
class GFG
{ // Minimum number of moves required to // reach and largest and smallest ASCII values static int min_moves(String s, int n)
{ int maxpos = 0 , minpos = 0 ;
// Finding index of maximum
// and minimum element in string
for ( int i = 0 ; i < n; i++) {
if (s.charAt(i) > s.charAt(maxpos))
maxpos = i;
if (s.charAt(i) < s.charAt(minpos))
minpos = i;
}
// Calculating minimum ans maximum steps
// that can be taken
int min_steps = Math.min(maxpos, minpos);
int max_steps = Math.max(maxpos, minpos);
// Only three possible ways
// to reach both elements
int ans1, ans2, ans3;
ans1 = n - min_steps;
ans2 = max_steps + 1 ;
ans3 = min_steps + 1 + n - max_steps;
int result;
// Minimum steps in all three ways
result = Math.min(ans1, Math.min(ans2, ans3));
// Return the final result
return result;
} // Driver code public static void main (String[] args) {
String str = "BACDEFHG" ;
int N = str.length();
System.out.println(min_moves(str, N));
}
} // This code is contributed by Potta Lokesh |
# Python code for the above approach # Minimum number of moves required to # reach and largest and smallest ASCII values def min_moves(s, n):
maxpos = 0 ;
minpos = 0 ;
# Finding index of maximum
# and minimum element in string
for i in range (n):
if (s[i] > s[maxpos]):
maxpos = i;
if (s[i] < s[minpos]):
minpos = i;
# Calculating minimum ans maximum steps
# that can be taken
min_steps = min (maxpos, minpos);
max_steps = max (maxpos, minpos);
# Only three possible ways
# to reach both elements
ans1, ans2, ans3 = 0 , 0 , 0 ;
ans1 = n - min_steps;
ans2 = max_steps + 1 ;
ans3 = min_steps + 1 + n - max_steps;
result = 0 ;
# Minimum steps in all three ways
result = min (ans1, min (ans2, ans3));
# Return the final result
return result;
# Driver code if __name__ = = '__main__' :
str = "BACDEFHG" ;
N = len ( str );
print (min_moves( str , N));
# This code is contributed by shikhasingrajput |
// C# program for above approach using System;
class GFG{
// Minimum number of moves required to // reach and largest and smallest ASCII values static int min_moves( string s, int n)
{ int maxpos = 0, minpos = 0;
// Finding index of maximum
// and minimum element in string
for ( int i = 0; i < n; i++)
{
if (s[i] > s[maxpos])
maxpos = i;
if (s[i] < s[minpos])
minpos = i;
}
// Calculating minimum ans maximum steps
// that can be taken
int min_steps = Math.Min(maxpos, minpos);
int max_steps = Math.Max(maxpos, minpos);
// Only three possible ways
// to reach both elements
int ans1, ans2, ans3;
ans1 = n - min_steps;
ans2 = max_steps + 1;
ans3 = min_steps + 1 + n - max_steps;
int result;
// Minimum steps in all three ways
result = Math.Min(ans1, Math.Min(ans2, ans3));
// Return the final result
return result;
} // Driver code public static void Main( string [] args)
{ String str = "BACDEFHG" ;
int N = str.Length;
Console.WriteLine(min_moves(str, N));
} } // This code is contributed by ukasp |
<script> // JavaScript program for above approach // Minimum number of moves required to // reach and largest and smallest ASCII values const min_moves = (s, n) => { let maxpos = 0, minpos = 0;
// Finding index of maximum
// and minimum element in string
for (let i = 0; i < n; i++)
{
if (s[i] > s[maxpos])
maxpos = i;
if (s[i] < s[minpos])
minpos = i;
}
// Calculating minimum ans maximum steps
// that can be taken
let min_steps = Math.min(maxpos, minpos);
let max_steps = Math.max(maxpos, minpos);
// Only three possible ways
// to reach both elements
let ans1, ans2, ans3;
ans1 = n - min_steps;
ans2 = max_steps + 1;
ans3 = min_steps + 1 + n - max_steps;
let result;
// Minimum steps in all three ways
result = Math.min(ans1, Math.min(ans2, ans3));
// Return the final result
return result;
} // Driver code let str = "BACDEFHG" ;
let N = str.length; document.write(min_moves(str, N)); // This code is contributed by rakeshsahni </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(1)