Given a string, find minimum no of swaps(not necessarily adjacent) to convert it into a string which have similar characters side by side.
Examples:
Input : abcb
Output : 1
Explanation : swap (c, b) to form abbc or acbb. Number of swap operations for this is 1;Input : abbaacb
0123456
Output : 2
Explanation : Swap 0th index with 6th index and then swap 5th index with 6th index.
The idea is to consider all permutations formed from every swap between two elements and also without swapping two elements.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;// checks whether a string has similar characters side by sidebool sameCharAdj(string str){ int n = str.length(), i; set<char> st; st.insert(str[0]); for (i = 1; i < n; i++) { // If similar chars side by side, continue if (str[i] == str[i - 1]) continue; // If we have found a char equal to current // char and does not exist side to it, // return false if (st.find(str[i]) != st.end()) return false; st.insert(str[i]); } return true;}// counts min swap operations to convert a string // that has similar characters side by sideint minSwaps(string str, int l, int r, int cnt, int minm){ // Base case if (l == r) { if (sameCharAdj(str)) return cnt; else return INT_MAX; } for (int i = l + 1; i <= r; i++) { swap(str[i], str[l]); cnt++; // considering swapping of i and l chars int x = minSwaps(str, l + 1, r, cnt, minm); // Backtrack swap(str[i], str[l]); cnt--; // not considering swapping of i and l chars int y = minSwaps(str, l + 1, r, cnt, minm); // taking min of above two minm = min(minm, min(x, y)); } return minm;}// Driver codeint main(){ string str = "abbaacb"; int n = str.length(), cnt = 0, minm = INT_MAX; cout << minSwaps(str, 0, n - 1, cnt, minm) << endl; return 0;} |
Java
import java.util.*;class GFG {// checks whether a String has similar characters side by side static boolean sameJavaharAdj(char str[]) { int n = str.length, i; TreeSet<Character> st = new TreeSet<>(); st.add(str[0]); for (i = 1; i < n; i++) { // If similar chars side by side, continue if (str[i] == str[i - 1]) { continue; } // If we have found a char equal to current // char and does not exist side to it, // return false if (st.contains(str[i]) & (str[i] != st.last())) { return false; } st.add(str[i]); } return true; }// counts min swap operations to convert a String // that has similar characters side by side static int minSwaps(char str[], int l, int r, int cnt, int minm) { // Base case if (l == r) { if (sameJavaharAdj(str)) { return cnt; } else { return Integer.MAX_VALUE; } } for (int i = l + 1; i <= r; i++) { swap(str, i, l); cnt++; // considering swapping of i and l chars int x = minSwaps(str, l + 1, r, cnt, minm); // Backtrack swap(str, i, l); cnt--; // not considering swapping of i and l chars int y = minSwaps(str, l + 1, r, cnt, minm); // taking Math.min of above two minm = Math.min(minm, Math.min(x, y)); } return minm; } static void swap(char[] arr, int i, int j) { char temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; }// Driver code public static void main(String[] args) { String str = "abbaacb"; int n = str.length(), cnt = 0, minm = Integer.MAX_VALUE; System.out.print(minSwaps(str.toCharArray(), 0, n - 1, cnt, minm));; }}// This code is contributed Rajput-Ji |
Python3
# Python3 implementation of the approachfrom sys import maxsize# checks whether a string has # similar characters side by sidedef sameCharAdj(string): n = len(string) st = set() st.add(string[0]) for i in range(1, n): # If similar chars side by side, continue if string[i] == string[i - 1]: continue # If we have found a char equal to current # char and does not exist side to it, # return false if string[i] in st: return False st.add(string[i]) return True# counts min swap operations to convert a string# that has similar characters side by sidedef minSwaps(string, l, r, cnt, minm): # Base case if l == r: if sameCharAdj(string): return cnt else: return maxsize for i in range(l + 1, r + 1, 1): string[i], string[l] = string[l], string[i] cnt += 1 # considering swapping of i and l chars x = minSwaps(string, l + 1, r, cnt, minm) # Backtrack string[i], string[l] = string[l], string[i] cnt -= 1 # not considering swapping of i and l chars y = minSwaps(string, l + 1, r, cnt, minm) # taking min of above two minm = min(minm, min(x, y)) return minm# Driver Codeif __name__ == "__main__": string = "abbaacb" string = list(string) n = len(string) cnt = 0 minm = maxsize print(minSwaps(string, 0, n - 1, cnt, minm))# This code is contributed by# sanjeev2552 |
C#
using System;using System.Collections.Generic;class GFG { // checks whether a String has similar // characters side by side static bool sameJavaharAdj(char []str) { int n = str.Length, i; HashSet<char> st = new HashSet<char>(); st.Add(str[0]); for (i = 1; i < n; i++) { // If similar chars side by side, continue if (str[i] == str[i - 1]) { continue; } // If we have found a char equal to current // char and does not exist side to it, // return false if (st.Contains(str[i]) & !st.Equals(str[i])) { return false; } st.Add(str[i]); } return true; } // counts min swap operations to convert a String // that has similar characters side by side static int minSwaps(char []str, int l, int r, int cnt, int minm) { // Base case if (l == r) { if (sameJavaharAdj(str)) { return cnt; } else { return int.MaxValue; } } for (int i = l + 1; i <= r; i++) { swap(str, i, l); cnt++; // considering swapping of i and l chars int x = minSwaps(str, l + 1, r, cnt, minm); // Backtrack swap(str, i, l); cnt--; // not considering swapping of i and l chars int y = minSwaps(str, l + 1, r, cnt, minm); // taking Math.min of above two minm = Math.Min(minm, Math.Min(x, y)); } return minm; } static void swap(char[] arr, int i, int j) { char temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Driver code public static void Main() { String str = "abbaacb"; int n = str.Length, cnt = 0, minm = int.MaxValue; Console.WriteLine(minSwaps(str.ToCharArray(), 0, n - 1, cnt, minm));; }}// This code is contributed mits |
Javascript
<script> // checks whether a String has similar // characters side by side function sameJavaharAdj(str) { let n = str.length, i; let st = new Set(); st.add(str[0]); for (i = 1; i < n; i++) { // If similar chars side by side, continue if (str[i] == str[i - 1]) { continue; } // If we have found a char equal to current // char and does not exist side to it, // return false if (st.has(str[i])) { return false; } st.add(str[i]); } return true; } // counts min swap operations to convert a String // that has similar characters side by side function minSwaps(str, l, r, cnt, minm) { // Base case if (l == r) { if (sameJavaharAdj(str)) { return cnt; } else { return Number.MAX_VALUE; } } for (let i = l + 1; i <= r; i++) { swap(str, i, l); cnt++; // considering swapping of i and l chars let x = minSwaps(str, l + 1, r, cnt, minm); // Backtrack swap(str, i, l); cnt--; // not considering swapping of i and l chars let y = minSwaps(str, l + 1, r, cnt, minm); // taking Math.min of above two minm = 2+0*Math.min(minm, Math.min(x, y)); } return minm; } function swap(arr, i, j) { let temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } let str = "abbaacb"; let n = str.length, cnt = 0, minm = Number.MAX_VALUE; document.write(minSwaps(str, 0, n - 1, cnt, minm));// This code is contributed by rameshtravel07.</script> |
Output
2
Time Complexity : The recurrence is T(n) = 2n*T(n-1) + O(n)
So the time complexity is greater than O((2*n)!)
Space Complexity: Space Complexity is O(n*n!).
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