Minimum number of adjacent swaps to reverse a String
Last Updated :
30 Nov, 2022
Given a string s. The task is to minimize the number of adjacent swaps required to reverse the string.
Examples:
Input: s = “abc”
Output: 3
Explanation: Follow the operations below to solve the given problem.
swap(1, 2)-> “bac”
swap(2, 3)-> “bca”
swap(1, 2)-> “cba”
Input: s = “ba”
Output: 1
Approach: This problem can be solved by comparing the given string with its reverse and counting the number of swaps required to form the reversed string. Follow the steps below to solve the problem:
- Initialize a string s2 as the copy of the original string s.
- reverse string s2
- Initialize result = 0, to store the number of adjacent swaps required to reverse the string.
- Iterate using two pointers i and j for both the strings and find each occurrence of s in s2 through two pointers i and j
- Each time set result = result + j – i.
- Return result as the final answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int min_swaps(string s)
{
string s2 = s;
reverse(s2.begin(), s2.end());
int i = 0, j = 0;
int result = 0;
int n = s.length();
while (i < n) {
j = i;
while (s[j] != s2[i]) {
j += 1;
}
while (i < j) {
char temp = s[j];
s[j] = s[j - 1];
s[j - 1] = temp;
j -= 1;
result += 1;
}
i += 1;
}
return result;
}
int main()
{
string s = "abc";
cout << min_swaps(s);
return 0;
}
|
Java
public class GFG {
static int min_swaps(String s)
{
String s2 = "";
char[] cArray = s.toCharArray();
for (int k = cArray.length - 1; k > -1; k--) {
s2 += cArray[k];
}
int i = 0, j = 0;
int result = 0;
int n = s.length();
while (i < n) {
j = i;
while (s.charAt(j) != s2.charAt(i)) {
j += 1;
}
while (i < j) {
char temp = s.charAt(j);
char[] ch = s.toCharArray();
ch[j] = ch[j - 1];
ch[j - 1] = temp;
s = new String(ch);
j -= 1;
result += 1;
}
i += 1;
}
return result;
}
static public void main(String []args)
{
String s = "abc";
System.out.println(min_swaps(s));
}
}
|
Python3
def min_swaps(s):
s2 = s.copy()
s2.reverse()
i = 0
j = 0
result = 0
n = len(s)
while (i < n):
j = i
while (s[j] != s2[i]):
j += 1
while (i < j):
temp = s[j]
s[j] = s[j - 1]
s[j - 1] = temp
j -= 1
result += 1
i += 1
return result
if __name__ == "__main__":
s = "abc"
s = list(s)
print(min_swaps(s))
|
C#
using System;
public class GFG {
static int min_swaps(string s)
{
string s2 = String.Empty;
char[] cArray = s.ToCharArray();
for (int k = cArray.Length - 1; k > -1; k--) {
s2 += cArray[k];
}
int i = 0, j = 0;
int result = 0;
int n = s.Length;
while (i < n) {
j = i;
while (s[j] != s2[i]) {
j += 1;
}
while (i < j) {
char temp = s[j];
char[] ch = s.ToCharArray();
ch[j] = ch[j - 1];
ch[j - 1] = temp;
s = new string(ch);
j -= 1;
result += 1;
}
i += 1;
}
return result;
}
static public void Main()
{
string s = "abc";
Console.WriteLine(min_swaps(s));
}
}
|
Javascript
<script>
function min_swaps(s)
{
var s2 = JSON.parse(JSON.stringify(s));
s = s.split('');
s2 = s2.split('');
s2.reverse();
var i = 0, j = 0;
var result = 0;
var n = s.length;
while (i < n) {
j = i;
while (s[j] != s2[i]) {
j += 1;
}
while (i < j) {
var temp = s[j];
s[j] = s[j - 1];
s[j - 1] = temp;
j -= 1;
result += 1;
}
i += 1;
}
return result;
}
var s = "abc";
document.write(min_swaps(s));
</script>
|
Time Complexity: O(N2), Where N is the size of the given string.
Auxiliary Space: O(N)
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