Check if two strings can be made equal by swapping pairs of adjacent characters
Last Updated : 28 Oct, 2023
Given two strings A and B of length N and M respectively and an array arr[] consisting of K integers, the task is to check if the string B can be obtained from the string A by swapping any pair of adjacent characters of the string A any number of times such that the swapped indices are not present in the array arr[]. If it is possible to convert string A to string B, print "Yes". Otherwise, print "No".
Examples:
Input: A = "abcabka", B = "acbakba", arr[] = {0, 3, 6}
Output: Yes
Explanation:
Swap A1 and A2. Now the string becomes A = "acbabka".
Swap A4 and A5. Now the string becomes A = "acbakba", which is the same as string B.
Input: A = "aaa", B = "bbb", arr[] = {0}
Output : No
Approach: Follow the below steps to solve the problem:
- If the length of both the strings are not equal, then string A can't be transformed to string B. Therefore, print "No".
- Traverse the given array arr[] and check if characters at A[arr[i]] and B[arr[i]] are same or not. If found to be true, then print "No". Otherwise, perform the following steps:
- If the first element of arr[i] is not 0:
- Similarly, if the last element of arr[i] is not equal to the element at index (N - 1), then:
- Iterate a loop from 1 to N and initialize two pointers, L = arr[i - 1] and R = arr[i]:
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count frequency of the // characters of two given strings bool isEqual(string& A, string& B) { // Stores the frequencies of // characters of strings A and B map<char, int> mp, mp1; // Traverse the string A for (auto it : A) { // Update frequency of // each character mp[it]++; } // Traverse the string B for (auto it : B) { // Update frequency of // each character mp1[it]++; } // Traverse the Map for (auto it : mp) { // If the frequency a character // is not the same in both the strings if (it.second != mp1[it.first]) { return false; } } return true; } // Function to check if it is possible // to convert string A to string B void IsPossible(string& A, string& B, int arr[], int N) { // Base Case if (A == B) { cout << "Yes" << endl; return; } // If length of both the // strings are not equal if (A.length() != B.length()) { cout << "No" << endl; return; } // Traverse the array and check // if the blocked indices // contains the same character for (int i = 0; i < N; i++) { int idx = arr[i]; // If there is a different // character, return No if (A[idx] != B[idx]) { cout << "No" << endl; return; } } // If the first index is not present if (arr[0]) { string X = ""; string Y = ""; // Extract characters from // string A and B for (int i = 0; i < arr[0]; i++) { X += A[i]; Y += B[i]; } // If frequency is not same if (!isEqual(A, B)) { cout << "No" << endl; return; } } // If last index is not present if (arr[N - 1] != (A.length() - 1)) { string X = ""; string Y = ""; // Extract characters from // string A and B for (int i = arr[N - 1] + 1; i < A.length(); i++) { X += A[i]; Y += B[i]; } // If frequency is not same if (!isEqual(A, B)) { cout << "No" << endl; return; } } // Traverse the array arr[] for (int i = 1; i < N; i++) { int L = arr[i - 1]; int R = arr[i]; string X = ""; string Y = ""; // Extract characters from strings A and B for (int j = L + 1; j < R; j++) { X += A[j]; Y += B[j]; } // If frequency is not same if (!isEqual(A, B)) { cout << "No" << endl; return; } } // If all conditions are satisfied cout << "Yes" << endl; } // Driver Code int main() { string A = "abcabka"; string B = "acbakba"; int arr[] = { 0, 3, 6 }; int N = sizeof(arr) / sizeof(arr[0]); IsPossible(A, B, arr, N); return 0; } // Java program for the above approach // importing input-output and utility classes import java.io.*; import java.util.*; class GFG { // method to count frequency of the // characters of two given strings static boolean isEqual(String A, String B) { // storing the frequencies of characters HashMap<Character, Integer> map = new HashMap<>(); HashMap<Character, Integer> map2 = new HashMap<>(); // Traversing the String A for (int i = 0; i < A.length(); i++) { if (map.containsKey(A.charAt(i))) map.put(A.charAt(i), map.get(A.charAt(i)) + 1); else map.put(A.charAt(i), 1); } // Traversing the String B for (int i = 0; i < B.length(); i++) { if (map.containsKey(B.charAt(i))) map.put(B.charAt(i), map.get(B.charAt(i)) + 1); else map.put(B.charAt(i), 1); } // checking if both map have same character // frequency if (!map.equals(map2)) return false; return true; } // method to check possibility to convert A into B static void isPossible(String A, String B, int arr[], int N) { // Base Case if (A.equals(B)) { System.out.println("Yes"); return; } // If length is not equal if (A.length() != B.length()) { System.out.println("No"); return; } for (int i = 0; i < N; i++) { int idx = arr[i]; if (A.charAt(idx) != B.charAt(idx)) { System.out.println("No"); return; } } // If first index is not present if (arr[0] == 0) { String X = ""; String Y = ""; // Extracting Characters from A and B for (int i = 0; i < arr[0]; i++) { X += A.charAt(i); Y += B.charAt(i); } // If frequency is not same if (!isEqual(A, B)) { System.out.println("No"); return; } } // If last index is not present if (arr[N - 1] != (A.length() - 1)) { String X = ""; String Y = ""; for (int i = arr[N - 1] + 1; i < A.length(); i++) { X += A.charAt(i); Y += B.charAt(i); } if (!isEqual(A, B)) { System.out.println("No"); return; } } // Traversing the Array for (int i = 1; i < N; i++) { int L = arr[i - 1]; int R = arr[i - 1]; String X = ""; String Y = ""; // Extract Characters from Strings A and B for (int j = L + 1; j < R; j++) { X += A.charAt(j); Y += B.charAt(j); } // if frequency is not same if (!isEqual(A, B)) { System.out.println("No"); return; } } // if all condition satisfied System.out.println("Yes"); } // main method public static void main(String[] args) { String A = "abcabka"; String B = "abcabka"; int arr[] = { 0, 3, 6 }; int N = arr.length; // calling method isPossible(A, B, arr, N); } } # Python3 program for the above approach from collections import defaultdict # Function to count frequency of the # characters of two given strings def isEqual(A, B): # Stores the frequencies of # characters of strings A and B mp = defaultdict(int) mp1 = defaultdict(int) # Traverse the string A for it in A: # Update frequency of # each character mp[it] += 1 # Traverse the string B for it in B: # Update frequency of # each character mp1[it] += 1 # Traverse the Map for it in mp: # If the frequency a character # is not the same in both the strings if (mp[it] != mp1[it]): return False return True # Function to check if it is possible # to convert string A to string B def IsPossible(A, B, arr, N): # Base Case if (A == B): print("Yes") return # If length of both the # strings are not equal if (len(A) != len(B)): print("No") return # Traverse the array and check # if the blocked indices # contains the same character for i in range(N): idx = arr[i] # If there is a different # character, return No if (A[idx] != B[idx]): print("No") return # If the first index is not present if (arr[0]): X = "" Y = "" # Extract characters from # string A and B for i in range(arr[0]): X += A[i] Y += B[i] # If frequency is not same if (not isEqual(A, B)): print("No") return # If last index is not present if (arr[N - 1] != (len(A) - 1)): X = "" Y = "" # Extract characters from # string A and B for i in range(arr[N - 1] + 1, len(A)): X += A[i] Y += B[i] # If frequency is not same if (not isEqual(A, B)): print("No") return # Traverse the array arr[] for i in range(1, N): L = arr[i - 1] R = arr[i] X = "" Y = "" # Extract characters from strings A and B for j in range(L + 1, R): X += A[j] Y += B[j] # If frequency is not same if (not isEqual(A, B)): print("No") return # If all conditions are satisfied print("Yes") # Driver Code if __name__ == "__main__": A = "abcabka" B = "acbakba" arr = [ 0, 3, 6 ] N = len(arr) IsPossible(A, B, arr, N) # This code is contributed by ukasp // C# program for the above approach // importing input-output and utility classes using System; using System.Collections.Generic; class GFG { // method to count frequency of the // characters of two given strings static bool isEqual(string A, string B) { // storing the frequencies of characters Dictionary<char, int> map = new Dictionary<char, int>(); Dictionary<char, int> map2 = new Dictionary<char, int>(); // Traversing the String A for (int i = 0; i < A.Length; i++) { if (map.ContainsKey(A[i])) map[A[i]] = map[A[i]] + 1; else map.Add(A[i], 1); } // Traversing the String B for (int i = 0; i < B.Length; i++) { if (map2.ContainsKey(B[i])) map2[B[i]] = map2[B[i]] + 1; else map2.Add(B[i], 1); } // checking if both map have same character // frequency if (!map.Equals(map2)) return false; return true; } // method to check possibility to convert A into B static void isPossible(string A, string B, int[] arr, int N) { // Base Case if (A.Equals(B)) { Console.WriteLine("Yes"); return; } // If length is not equal if (A.Length != B.Length) { Console.WriteLine("No"); return; } for (int i = 0; i < N; i++) { int idx = arr[i]; if (A[idx] != B[idx]) { Console.WriteLine("No"); return; } } // If first index is not present if (arr[0] == 0) { String X = ""; String Y = ""; // Extracting Characters from A and B for (int i = 0; i < arr[0]; i++) { X += A[i]; Y += B[i]; } // If frequency is not same if (!isEqual(A, B)) { Console.WriteLine("No"); return; } } // If last index is not present if (arr[N - 1] != (A.Length - 1)) { string X = ""; string Y = ""; for (int i = arr[N - 1] + 1; i < A.Length; i++) { X += A[i]; Y += B[i]; } if (!isEqual(A, B)) { Console.WriteLine("No"); return; } } // Traversing the Array for (int i = 1; i < N; i++) { int L = arr[i - 1]; int R = arr[i - 1]; string X = ""; string Y = ""; // Extract Characters from Strings A and B for (int j = L + 1; j < R; j++) { X += A[j]; Y += B[j]; } // if frequency is not same if (!isEqual(A, B)) { Console.WriteLine("No"); return; } } // if all condition satisfied Console.WriteLine("Yes"); } // main method public static void Main() { string A = "abcabka"; string B = "abcabka"; int[] arr = { 0, 3, 6 }; int N = arr.Length; // calling method isPossible(A, B, arr, N); } } // This code is contributed by Samim Hossain Mondal. <script> // Javascript program for the above approach // Function to count frequency of the // characters of two given strings function isEqual(A, B) { // Stores the frequencies of // characters of strings A and B let mp = new Map(); let mp1 = new Map(); // Traverse the string A for(let it of A) { // Update frequency of // each character if (mp.has(it)) { mp.set(it, mp.get(it) + 1) } else { mp.set(it, 1) } } // Traverse the string B for(let it of B) { // Update frequency of // each character if (mp1.has(it)) { mp1.set(it, mp1.get(it) + 1) } else { mp1.set(it, 1) } } // Traverse the Map for(let it of mp) { // If the frequency a character // is not the same in both the strings if (it[1] != mp1.get(it[0])) { return false; } } return true; } // Function to check if it is possible // to convert string A to string B function IsPossible(A, B, arr, N) { // Base Case if (A == B) { document.write("Yes" + "<br>"); return; } // If length of both the // strings are not equal if (A.length != B.length) { document.write("No" + "<br>"); return; } // Traverse the array and check // if the blocked indices // contains the same character for(let i = 0; i < N; i++) { let idx = arr[i]; // If there is a different // character, return No if (A[idx] != B[idx]) { document.write("No" + "<br>"); return; } } // If the first index is not present if (arr[0]) { let X = ""; let Y = ""; // Extract characters from // string A and B for(let i = 0; i < arr[0]; i++) { X += A[i]; Y += B[i]; } // If frequency is not same if (!isEqual(A, B)) { document.write("No" + "<br>"); return; } } // If last index is not present if (arr[N - 1] != (A.length - 1)) { let X = ""; let Y = ""; // Extract characters from // string A and B for(let i = arr[N - 1] + 1; i < A.length; i++) { X += A[i]; Y += B[i]; } // If frequency is not same if (!isEqual(A, B)) { document.write("No" + "<br>"); return; } } // Traverse the array arr[] for(let i = 1; i < N; i++) { let L = arr[i - 1]; let R = arr[i]; let X = ""; let Y = ""; // Extract characters from strings A and B for(let j = L + 1; j < R; j++) { X += A[j]; Y += B[j]; } // If frequency is not same if (!isEqual(A, B)) { document.write("No" + "<br>"); return; } } // If all conditions are satisfied document.write("Yes" + "<br>"); } // Driver Code let A = "abcabka"; let B = "acbakba"; let arr = [ 0, 3, 6 ]; let N = arr.length IsPossible(A, B, arr, N); // This code is contributed by gfgking </script> Time Complexity: O(N)
Auxiliary Space: O(N)
Another Approach:
- Check if the length of strings A and B are equal, if not, return false.
- Create a 2D boolean dp table of size N x M, where N and M are the length of strings A and B respectively.
- Initialize dp[0][0] as true if A[0] == B[0].
- Initialize dp[0][j] for j > 0, using the constraints given in the arr vector.
- Initialize dp[i][0] for i > 0, using the constraints given in the arr vector.
- Fill the remaining cells of the dp table using the given constraints, and the following conditions:
If A[i] == B[j], then dp[i][j] = dp[i-1][j-1]
If i > 1 and j > 0 and A[i] == B[j-1] and A[i-1] == B[j] and the constraints allow, then dp[i][j] = dp[i-2][j-2]
If j > 1 and A[i] == B[j-2] and A[i] == B[j-1] and the constraints allow, then dp[i][j] = dp[i][j-2]
If i > 1 and A[i] == B[j] and A[i-1] == B[j-1] and the constraints allow, then dp[i][j] = dp[i-2][j-1] - Return the value of dp[N-1][M-1].
- In the main function, call the isPossible function with the given input strings A and B, and the array arr.
- If isPossible returns true, print "Yes", else print "No".
Below is the implementation of the above approach:
C++ #include <iostream> #include <vector> #include <cstring> using namespace std; bool isPossible(string A, string B, vector<int> arr) { int N = A.length(); int M = B.length(); if (N != M) { return false; } // Create a 2D boolean array to store if it is possible to // transform A[0...i] to B[0...j] with the given constraints bool dp[N][M]; memset(dp, false, sizeof(dp)); // Initialize dp[0][0] as true if A[0] == B[0] dp[0][0] = (A[0] == B[0]); // Initialize dp[0][j] for j > 0 for (int j = 1; j < M; j++) { dp[0][j] = (A[0] == B[j]); dp[0][j] = dp[0][j] && (arr[0] != j); dp[0][j] = dp[0][j] && dp[0][j - 1]; } // Initialize dp[i][0] for i > 0 for (int i = 1; i < N; i++) { dp[i][0] = (A[i] == B[0]); dp[i][0] = dp[i][0] && (arr[0] != i); dp[i][0] = dp[i][0] && dp[i - 1][0]; } // Fill the remaining cells of the dp table for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { dp[i][j] = false; if (A[i] == B[j]) { dp[i][j] = dp[i][j] || dp[i - 1][j - 1]; } if (i > 1 && j > 0 && A[i] == B[j - 1] && A[i - 1] == B[j] && arr[j - 1] != j && arr[j] != j - 1) { dp[i][j] = dp[i][j] || dp[i - 2][j - 2]; } if (j > 1) { dp[i][j] = dp[i][j] || (A[i] == B[j - 2] && A[i] == B[j - 1] && arr[j - 2] != j && arr[j - 1] != j && dp[i][j - 2]); } if (i > 1) { dp[i][j] = dp[i][j] || (A[i] == B[j] && A[i - 1] == B[j - 1] && arr[j] != j && arr[j - 1] != j - 1 && dp[i - 2][j - 1]); } } } return dp[N - 1][M - 1]; } int main() { string A = "abcabka"; string B = "acbakba"; vector<int> arr = {0, 3, 6}; if (isPossible(A, B, arr)) { cout << "Yes\n"; } else { cout << "No\n"; } } //This code is contributed by rudra1807raj public class Main { static boolean isPossible(String A, String B, int[] arr) { int N = A.length(); int M = B.length(); // Create a 2D boolean array to store if it is possible to // transform A[0...i] to B[0...j] with the given constraints boolean[][] dp = new boolean[N][M]; // Initialize dp[0][0] as True if A[0] == B[0] dp[0][0] = A.charAt(0) == B.charAt(0); // Initialize dp[0][j] for j > 0 for (int j = 1; j < M; j++) { dp[0][j] = A.charAt(0) == B.charAt(j) && !contains(arr, j) && dp[0][j - 1]; } // Initialize dp[i][0] for i > 0 for (int i = 1; i < N; i++) { dp[i][0] = A.charAt(i) == B.charAt(0) && !contains(arr, i) && dp[i - 1][0]; } // Fill the remaining cells of the dp table for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { dp[i][j] = false; if (A.charAt(i) == B.charAt(j)) { dp[i][j] = dp[i][j] || dp[i - 1][j - 1]; } if (i > 1 && j > 0 && A.charAt(i) == B.charAt(j - 1) && A.charAt(i - 1) == B.charAt(j) && !contains(arr, j) && !contains(arr, j - 1)) { dp[i][j] = dp[i][j] || dp[i - 2][j - 2]; } if (j > 1) { dp[i][j] = dp[i][j] || (A.charAt(i) == B.charAt(j - 2) && A.charAt(i) == B.charAt(j - 1) && !contains(arr, j - 2) && !contains(arr, j - 1) && dp[i][j - 2]); } if (i > 1) { dp[i][j] = dp[i][j] || (A.charAt(i) == B.charAt(j) && A.charAt(i - 1) == B.charAt(j - 1) && !contains(arr, j) && !contains(arr, j - 1) && dp[i - 2][j - 1]); } } } return dp[N - 1][M - 1]; } // Helper method to check if an array contains a given value static boolean contains(int[] arr, int value) { for (int num : arr) { if (num == value) { return true; } } return false; } public static void main(String[] args) { String A = "abcabka"; String B = "acbakba"; int[] arr = {0, 3, 6}; if (isPossible(A, B, arr)) { System.out.println("Yes"); } else { System.out.println("No"); } } } def isPossible(A, B, arr): N = len(A) M = len(B) # Create a 2D boolean array to store if it is possible to # transform A[0...i] to B[0...j] with the given constraints dp = [[False] * M for _ in range(N)] # Initialize dp[0][0] as True if A[0] == B[0] dp[0][0] = A[0] == B[0] # Initialize dp[0][j] for j > 0 for j in range(1, M): dp[0][j] = A[0] == B[j] and (not j in arr) and dp[0][j - 1] # Initialize dp[i][0] for i > 0 for i in range(1, N): dp[i][0] = A[i] == B[0] and (not i in arr) and dp[i - 1][0] # Fill the remaining cells of the dp table for i in range(1, N): for j in range(1, M): dp[i][j] = False if A[i] == B[j]: dp[i][j] = dp[i][j] or dp[i - 1][j - 1] if i > 1 and j > 0 and A[i] == B[j - 1] and A[i - 1] == B[j] and (not j in arr) and (not (j - 1) in arr): dp[i][j] = dp[i][j] or dp[i - 2][j - 2] if j > 1: dp[i][j] = dp[i][j] or (A[i] == B[j - 2] and A[i] == B[j - 1] and (not (j - 2) in arr) and (not (j - 1) in arr) and dp[i][j - 2]) if i > 1: dp[i][j] = dp[i][j] or (A[i] == B[j] and A[i - 1] == B[j - 1] and (not j in arr) and (not (j - 1) in arr) and dp[i - 2][j - 1]) return dp[N - 1][M - 1] # Example usage A = "abcabka" B = "acbakba" arr = [0, 3, 6] if isPossible(A, B, arr): print("Yes") else: print("No") using System; using System.Collections.Generic; class Program { static bool IsPossible(string A, string B, List<int> arr) { int N = A.Length; int M = B.Length; if (N != M) { return false; } // Create a 2D boolean array to store if it is possible to // transform A[0...i] to B[0...j] with the given constraints bool[,] dp = new bool[N, M]; // Initialize dp[0][0] as true if A[0] == B[0] dp[0, 0] = (A[0] == B[0]); // Initialize dp[0][j] for j > 0 for (int j = 1; j < M; j++) { dp[0, j] = (A[0] == B[j]); dp[0, j] = dp[0, j] && !arr.Contains(j); dp[0, j] = dp[0, j] && dp[0, j - 1]; } // Initialize dp[i][0] for i > 0 for (int i = 1; i < N; i++) { dp[i, 0] = (A[i] == B[0]); dp[i, 0] = dp[i, 0] && !arr.Contains(i); dp[i, 0] = dp[i, 0] && dp[i - 1, 0]; } // Fill the remaining cells of the dp table for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { dp[i, j] = false; if (A[i] == B[j]) { dp[i, j] = dp[i, j] || dp[i - 1, j - 1]; } if (i > 1 && j > 0 && A[i] == B[j - 1] && A[i - 1] == B[j] && !arr.Contains(j - 1) && !arr.Contains(j)) { dp[i, j] = dp[i, j] || dp[i - 2, j - 2]; } if (j > 1) { dp[i, j] = dp[i, j] || (A[i] == B[j - 2] && A[i] == B[j - 1] && !arr.Contains(j - 2) && !arr.Contains(j - 1) && dp[i, j - 2]); } if (i > 1) { dp[i, j] = dp[i, j] || (A[i] == B[j] && A[i - 1] == B[j - 1] && !arr.Contains(j) && !arr.Contains(j - 1) && dp[i - 2, j - 1]); } } } return dp[N - 1, M - 1]; } static void Main() { string A = "abcabka"; string B = "acbakba"; List<int> arr = new List<int> { 0, 3, 6 }; if (IsPossible(A, B, arr)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } } function isPossible(A, B, arr) { const N = A.length; const M = B.length; // Create a 2D boolean array to store if it is possible to // transform A[0...i] to B[0...j] with the given constraints const dp = new Array(N).fill(0).map(() => new Array(M).fill(false)); // Initialize dp[0][0] as true if A[0] == B[0] dp[0][0] = A[0] === B[0]; // Initialize dp[0][j] for j > 0 for (let j = 1; j < M; j++) { dp[0][j] = A[0] === B[j] && arr[0] !== j && dp[0][j - 1]; } // Initialize dp[i][0] for i > 0 for (let i = 1; i < N; i++) { dp[i][0] = A[i] === B[0] && arr[0] !== i && dp[i - 1][0]; } // Fill the remaining cells of the dp table for (let i = 1; i < N; i++) { for (let j = 1; j < M; j++) { dp[i][j] = false; if (A[i] === B[j]) { dp[i][j] = dp[i][j] || dp[i - 1][j - 1]; } if (i > 1 && j > 0 && A[i] === B[j - 1] && A[i - 1] === B[j] && arr[j - 1] !== j && arr[j] !== j - 1) { dp[i][j] = dp[i][j] || dp[i - 2][j - 2]; } if (j > 1) { dp[i][j] = dp[i][j] || (A[i] === B[j - 2] && A[i] === B[j - 1] && arr[j - 2] !== j && arr[j - 1] !== j && dp[i][j - 2]); } if (i > 1) { dp[i][j] = dp[i][j] || (A[i] === B[j] && A[i - 1] === B[j - 1] && arr[j] !== j && arr[j - 1] !== j - 1 && dp[i - 2][j - 1]); } } } return dp[N - 1][M - 1]; } // Example usage const A = "abcabka"; const B = "acbakba"; const arr = [0, 3, 6]; if (isPossible(A, B, arr)) { console.log("Yes"); } else { console.log("No"); } Output
Yes
Time complexity: O(NM), where N is the length of string A and M is the length of string B.
Auxiliary Space: O(NM), to store the 2D boolean dp array.
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Given two strings A and B of length N, the task is to check whether the two strings can be made equal by swapping any character of A with any other character of B only once.Examples: Input: A = "SEEKSFORGEEKS", B = "GEEKSFORGEEKG" Output: Yes "SEEKSFORGEEKS" and "GEEKSFORGEEKG" After removing the el
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Check if a string can be made equal to another string by swapping or replacement of characters
Given two strings S1 and S2, the task is to check if string S1 can be made equal to string S2 by swapping any pair of characters replacing any character in the string S1. If it is possible to make the string S1 equal to S2, then print "Yes". Otherwise, print "No". Examples: Input: S1 = “abcâ€, S2 = “
8 min read
Check if String T can be made Substring of S by replacing given characters
Given two strings S and T and a 2D array replace[][], where replace[i] = {oldChar, newChar} represents that the character oldChar of T is replaced with newChar. The task is to find if it is possible to make string T a substring of S by replacing characters according to the replace array. Note: Each
9 min read
Check if given strings can be made same by swapping two characters of same or different strings
Given an array of equal-length strings, arr[] of size N, the task is to check if all the strings can be made equal by repeatedly swapping any pair of characters of same or different strings from the given array. If found to be true, then print "YES". Otherwise, print "NO". Examples: Input: arr[] = {
11 min read
Check whether two strings can be made equal by copying their characters with the adjacent ones
Given two strings str1 and str2, the task is to check whether both of the string can be made equal by copying any character of the string with its adjacent character. Note that this operation can be performed any number of times.Examples: Input: str1 = "abc", str2 = "def" Output: No As all the chara
5 min read
Check if a string can be converted to another by swapping of adjacent characters of given type
Given two strings str1 and str2 of size N consisting of only three characters A, B, and C, the task is to check whether the string str1 can be changed to str2 using the below operations: Replacing one occurrence of “BC†with “CB†i.e swap adjacent ‘B’ and ‘C’.Replacing one occurrence of “CA†with “A
7 min read
Check if a two character string can be made using given words
Given a string of two characters and n distinct words of two characters. The task is to find if it is possible to arrange given words in such a way that the concatenated string has the given two character string as a substring. We can append a word multiple times. Examples: Input : str = "ya" words[
6 min read
Check if two binary strings can be made equal by swapping 1s occurring before 0s
Given two binary strings str1 and str2 having same length, the task is to find if it is possible to make the two binary strings str1 and str2 equal by swapping all 1s occurring at indices less than that of 0s index in the binary string str1. Examples: Input: str1 = "0110", str2 = "0011" Output: Poss
15+ min read
Minimum flips or swapping of adjacent characters required to make a string equal to another
Given two binary strings A and B of length N, the task is to convert the string A to B by either flipping any character of A or swapping adjacent characters of A minimum number of times. If it is not possible to make both the strings equal, print -1. Examples: Input: A = "10010010", B = "00001000" O
6 min read