Given a string consisting only of two lowercase characters 
and and two numbers and . The task is to print the lexicographically smallest permutation of the given string such that the count of subsequences of is and of is . If no such string exists, print “Impossible” (without quotes).
Examples:
Input: str = "yxxyx", p = 3, q = 3 Output: xyxyx Input: str = "yxxy", p = 3, q = 2 Output: Impossible
Approach:
First of all, by induction it can prove that the product of count of ‘x’ and count of ‘y’ should be equal to the sum of the count of a subsequence of ‘xy’ and ‘yx’ for any given string. If this does not hold then the answer is ‘Impossible’ else answer always exist.
Now, sort the given string so the count of a subsequence of ‘yx’ becomes zero. Let nx be the count of ‘x’ and ny be count of ‘y’. let a and b be the count of subsequence ‘xy’ and ‘yx’ respectively, then a = nx*ny and b = 0. Then, from beginning of the string find the ‘x’ which has next ‘y’ to it and swap both until you reach end of the string. In each swap a is decremented by 1 and b is incremented by 1. Repeat this until the count of a subsequence of ‘yx’ is achieved i:e a becomes p and b becomes q.
Below is the implementation of the above approach:
C++
// CPP program to find lexicographically smallest// string such that count of subsequence 'xy' and// 'yx' is p and q respectively.#include <bits/stdc++.h>using namespace std;// function to check if answer exitsint nx = 0, ny = 0;bool check(string s, int p, int q){ // count total 'x' and 'y' in string for (int i = 0; i < s.length(); ++i) { if (s[i] == 'x') nx++; else ny++; } // condition to check existence of answer if (nx * ny != p + q) return 1; else return 0;}// function to find lexicographically smallest stringstring smallestPermutation(string s, int p, int q){ // check if answer exist or not if (check(s, p, q) == 1) { return "Impossible"; } sort(s.begin(), s.end()); int a = nx * ny, b = 0, i, j; // check if count of 'xy' and 'yx' becomes // equal to p and q respectively. if (a == p && b == q) { return s; } // Repeat until answer is found. while (1) { // Find index of 'x' to swap with 'y'. for (i = 0; i < s.length() - 1; ++i) { if (s[i] == 'x' && s[i + 1] == 'y') break; } for (j = i; j < s.length() - 1; j++) { if (s[j] == 'x' && s[j + 1] == 'y') { swap(s[j], s[j + 1]); a--; // 'xy' decrement by 1 b++; // 'yx' increment by 1 // check if count of 'xy' and 'yx' becomes // equal to p and q respectively. if (a == p && b == q) { return s; } } } }}// Driver codeint main(){ string s = "yxxyx"; int p = 3, q = 3; cout<< smallestPermutation(s, p, q); return 0;} |
Java
// Java program to find lexicographically // smallest string such that count of // subsequence 'xy' and 'yx' is p and // q respectively.import java.util.*;class GFG{static int nx = 0, ny = 0; static boolean check(String s, int p, int q) { // count total 'x' and 'y' in string for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) == 'x') nx++; else ny++; } // condition to check // existence of answer if ((nx * ny) != (p + q)) return true; else return false; } public static String smallestPermutation(String s, int p, int q){ if (check(s, p, q) == true) { return "Impossible"; } char tempArray[] = s.toCharArray(); Arrays.sort(tempArray); String str = new String(tempArray); int a = nx * ny, b = 0, i = 0, j = 0; if (a == p && b == q) { return str; } while (1 > 0) { // Find index of 'x' to swap with 'y'. for (i = 0; i < str.length() - 1; ++i) { if (str.charAt(i) == 'x' && str.charAt(i + 1) == 'y') break; } for (j = i; j < str.length() - 1; j++) { if (str.charAt(j) == 'x' && str.charAt(j + 1) == 'y') { StringBuilder sb = new StringBuilder(str); sb.setCharAt(j, str.charAt(j + 1)); sb.setCharAt(j + 1, str.charAt(j)); str = sb.toString(); /* char ch[] = str.toCharArray(); char temp = ch[j+1]; ch[j+1] = ch[j]; ch[j] = temp;*/ a--; // 'xy' decrement by 1 b++; // 'yx' increment by 1 // check if count of 'xy' and // 'yx' becomes equal to p // and q respectively. if (a == p && b == q) { return str; } } } }}// Driver Codepublic static void main (String[] args) { String s = "yxxyx"; int p = 3, q = 3; System.out.print(smallestPermutation(s, p, q));}}// This code is contributed by Kirti_Mangal |
Python3
# Python3 program to find lexicographically # smallest string such that count of subsequence # 'xy' and 'yx' is p and q respectively. # Function to check if answer exits def check(s, p, q): global nx global ny # count total 'x' and 'y' in string for i in range(0, len(s)): if s[i] == 'x': nx += 1 else: ny += 1 # condition to check existence of answer if nx * ny != p + q: return 1 else: return 0# Function to find lexicographically # smallest string def smallestPermutation(s, p, q): # check if answer exist or not if check(s, p, q) == 1: return "Impossible" s = sorted(s) a, b, i = nx * ny, 0, 0 # check if count of 'xy' and 'yx' becomes # equal to p and q respectively. if a == p and b == q: return '' . join(s) # Repeat until answer is found. while True: # Find index of 'x' to swap with 'y'. for i in range(0, len(s) - 1): if s[i] == 'x' and s[i + 1] == 'y': break for j in range(i, len(s) - 1): if s[j] == 'x' and s[j + 1] == 'y': s[j], s[j + 1] = s[j + 1], s[j] a -= 1 # 'xy' decrement by 1 b += 1 # 'yx' increment by 1 # check if count of 'xy' and 'yx' becomes # equal to p and q respectively. if a == p and b == q: return '' . join(s) # Driver code if __name__ == "__main__": nx, ny = 0, 0 s = "yxxyx" p, q = 3, 3 print(smallestPermutation(s, p, q)) # This code is contributed by Rituraj Jain |
C#
// C# program to find lexicographically // smallest string such that count of // subsequence 'xy' and 'yx' is p and // q respectively.using System; using System.Text; class GFG{ static int nx = 0, ny = 0; static Boolean check(String s, int p, int q) { // count total 'x' and 'y' in string for (int i = 0; i < s.Length; ++i) { if (s[i] == 'x') nx++; else ny++; } // condition to check // existence of answer if ((nx * ny) != (p + q)) return true; else return false; } public static String smallestPermutation(String s, int p, int q){ if (check(s, p, q) == true) { return "Impossible"; } char []tempArray = s.ToCharArray(); Array.Sort(tempArray); String str = new String(tempArray); int a = nx * ny, b = 0, i = 0, j = 0; if (a == p && b == q) { return str; } while (1 > 0) { // Find index of 'x' to swap with 'y'. for (i = 0; i < str.Length - 1; ++i) { if (str[i] == 'x' && str[i + 1] == 'y') break; } for (j = i; j < str.Length - 1; j++) { if (str[j] == 'x' && str[j + 1] == 'y') { StringBuilder sb = new StringBuilder(str); sb.Remove(j,1); sb.Insert(j, str[j + 1]); sb.Remove(j+1,1); sb.Insert(j + 1, str[j]); str = sb.ToString(); /* char ch[] = str.toCharArray(); char temp = ch[j+1]; ch[j+1] = ch[j]; ch[j] = temp;*/ a--; // 'xy' decrement by 1 b++; // 'yx' increment by 1 // check if count of 'xy' and // 'yx' becomes equal to p // and q respectively. if (a == p && b == q) { return str; } } } }}// Driver Codepublic static void Main (String[] args) { String s = "yxxyx"; int p = 3, q = 3; Console.WriteLine(smallestPermutation(s, p, q));}}// This code has been contributed by 29AjayKumar |
Javascript
<script>// Javascript program to find lexicographically // smallest string such that count of // subsequence 'xy' and 'yx' is p and // q respectively.let nx = 0, ny = 0; function check(s, p, q){ // count total 'x' and 'y' in string for (let i = 0; i < s.length; ++i) { if (s[i] == 'x') nx++; else ny++; } // condition to check // existence of answer if ((nx * ny) != (p + q)) return true; else return false; }function smallestPermutation(s,p,q){ if (check(s, p, q) == true) { return "Impossible"; } let tempArray = s.split(""); (tempArray).sort(); let str = (tempArray).join(""); let a = nx * ny, b = 0, i = 0, j = 0; if (a == p && b == q) { return str; } while (1 > 0) { // Find index of 'x' to swap with 'y'. for (i = 0; i < str.length - 1; ++i) { if (str[i] == 'x' && str[i+1] == 'y') break; } for (j = i; j < str.length - 1; j++) { if (str[j] == 'x' && str[j+1] == 'y') { let sb = (str).split(""); sb[j] = str[j+1]; sb[j + 1] = str[j]; str = sb.join(""); /* char ch[] = str.toCharArray(); char temp = ch[j+1]; ch[j+1] = ch[j]; ch[j] = temp;*/ a--; // 'xy' decrement by 1 b++; // 'yx' increment by 1 // check if count of 'xy' and // 'yx' becomes equal to p // and q respectively. if (a == p && b == q) { return str; } } } }}// Driver Codelet s = "yxxyx"; let p = 3;let q = 3;document.write(smallestPermutation(s, p, q));// This code is contributed by patel2127</script> |
Output
xyxyx
Time Complexity: O(N2)
Auxiliary Space: O(1)
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