Rearrange string such that no pair of adjacent characters are of the same type

21 Jun, 2021
Given alphanumeric string str, the task is to rearrange the string such that no two adjacent characters are of the same type, i.e., no two adjacent characters can be alphabets or digits. If no such arrangement is possible, print -1.
Examples:
Input: str = “geeks2020”Output: g2e0e2k0s
Input: str = “IPL20”Output: I2P0L
Naive Approach: The simplest approach is to generate all possible permutation of the given string and for every permutation, check if it satisfies the given conditions or not. If fount to be true for any permutation, print that permutation. If no such permutation exists, then print -1.
Time Complexity: O(2N)Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to store all the alphabets and the digits separately and rearrange them by placing them alternatively in the resultant string.  If the count of the alphabets and the digits differ by more than 1, print -1 as no desired arrangement is possible.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach #include <bits/stdc++.h>using namespace std; // Function to rearrange given// alphanumeric string such that// no two adjacent characters// are of the same typestring rearrange(string s){    // Stores alphabets and digits    string s1 = "", s2 = "";     // Store the alphabets and digits    // separately in the strings    for (char x : s) {        isalpha(x) ? s1.push_back(x)                   : s2.push_back(x);    }     // Stores the count of    // alphabets and digits    int n = s1.size();    int m = s2.size();     // If respective counts    // differ by 1    if (abs(n - m) > 1)         // Desired arrangement        // not possible        return "-1";     // Stores the indexes    int i = 0, j = 0, k = 0;     // Check if first character    // should be alphabet or digit    int flag = (n >= m) ? 1 : 0;     // Place alphabets and digits    // alternatively    while (i < n and j < m) {         // If current character        // needs to be alphabet        if (flag)            s[k++] = s1[i++];         // If current character        // needs to be a digit        else            s[k++] = s2[j++];         // Flip flag for alternate        // arrangement        flag = !flag;    }     // Return resultant string    return s;} // Driver Codeint main(){    // Given String    string str = "geeks2020";     // Function Call    cout << rearrange(str) << endl;     return 0;}
// Java program to implement// the above approachclass GFG{ // Function to rearrange given// alphanumeric String such that// no two adjacent characters// are of the same typestatic String rearrange(String s){  // Stores alphabets and digits  String s1 = "", s2 = "", ans = "";  char []s3 = s.toCharArray();   // Store the alphabets and digits  // separately in the Strings  for (char x : s3)  {    if(x >= 'a' && x <= 'z')      s1 += x ;    else      s2 += x;  }   // Stores the count of  // alphabets and digits  int n = s1.length();  int m = s2.length();   // If respective counts  // differ by 1  if (Math.abs(n - m) > 1)     // Desired arrangement    // not possible    return "-1";   // Stores the indexes  int i = 0, j = 0, k = 0;   // Check if first character  // should be alphabet or digit  int flag = (n >= m) ? 1 : 0;   // Place alphabets and digits  // alternatively  while (i < n && j < m)  {    // If current character    // needs to be alphabet    if (flag != 0)      ans += s1.charAt(i++);     // If current character    // needs to be a digit    else      ans += s2.charAt(j++);     // Flip flag for alternate    // arrangement    if(flag == 1)      flag = 0;    else      flag = 1;  }   // Return resultant String  return ans;} // Driver Codepublic static void main(String[] args){  // Given String  String str = "geeks2020";   // Function Call  System.out.print(rearrange(str) + "\n");}} // This code is contributed by gauravrajput1
# Python3 program to implement# the above approach # Function to rearrange given# alphanumeric such that no# two adjacent characters# are of the same typedef rearrange(s):         # Stores alphabets and digits    s1 = []    s2 = []     # Store the alphabets and digits    # separately in the strings    for x in s:        if x.isalpha():            s1.append(x)        else:            s2.append(x)     # Stores the count of    # alphabets and digits    n = len(s1)    m = len(s2)     # If respective counts    # differ by 1    if (abs(n - m) > 1):         # Desired arrangement        # not possible        return "-1"     # Stores the indexes    i = 0    j = 0    k = 0     # Check if first character    # should be alphabet or digit    flag = 0    if (n >= m):        flag = 1    else:        flag = 0     # Place alphabets and digits    # alternatively    while (i < n and j < m):         # If current character        # needs to be alphabet        if (flag):            s[k] = s1[i]            k += 1            i += 1         # If current character        # needs to be a digit        else:            s[k] = s2[j]            k += 1            j += 1         # Flip flag for alternate        # arrangement        flag = not flag     # Return resultant string    return "".join(s) # Driver Codeif __name__ == '__main__':         # Given String    str = "geeks2020"     str1 = [i for i in str]     # Function call    print(rearrange(str1)) # This code is contributed by mohit kumar 29
// C# program to implement// the above approachusing System;class GFG{ // Function to rearrange given// alphanumeric String such that// no two adjacent characters// are of the same typestatic String rearrange(String s){  // Stores alphabets and digits  String s1 = "", s2 = "", ans = "";  char []s3 = s.ToCharArray();   // Store the alphabets and digits  // separately in the Strings  foreach (char x in s3)  {    if(x >= 'a' && x <= 'z')      s1 += x ;    else      s2 += x;  }   // Stores the count of  // alphabets and digits  int n = s1.Length;  int m = s2.Length;   // If respective counts  // differ by 1  if (Math.Abs(n - m) > 1)     // Desired arrangement    // not possible    return "-1";   // Stores the indexes  int i = 0, j = 0, k = 0;   // Check if first character  // should be alphabet or digit  int flag = (n >= m) ? 1 : 0;   // Place alphabets and digits  // alternatively  while (i < n && j < m)  {    // If current character    // needs to be alphabet    if (flag != 0)      ans += s1[i++];     // If current character    // needs to be a digit    else      ans += s2[j++];     // Flip flag for alternate    // arrangement    if(flag == 1)      flag = 0;    else      flag = 1;  }   // Return resultant String  return ans;} // Driver Codepublic static void Main(String[] args){  // Given String  String str = "geeks2020";   // Function Call  Console.Write(rearrange(str) + "\n");}} // This code is contributed by 29AjayKumar
<script>// Javascript program to implement// the above approach // Function to rearrange given// alphanumeric String such that// no two adjacent characters// are of the same typefunction rearrange(s){     // Stores alphabets and digits  let s1 = "", s2 = "", ans = "";  let s3 = s.split("");    // Store the alphabets and digits  // separately in the Strings  for (let x = 0; x < s3.length; x++)  {    if(s3[x] >= 'a' && s3[x] <= 'z')      s1 += s3[x] ;    else      s2 += s3[x];  }    // Stores the count of  // alphabets and digits  let n = s1.length;  let m = s2.length;    // If respective counts  // differ by 1  if (Math.abs(n - m) > 1)      // Desired arrangement    // not possible    return "-1";    // Stores the indexes  let i = 0, j = 0, k = 0;    // Check if first character  // should be alphabet or digit  let flag = (n >= m) ? 1 : 0;    // Place alphabets and digits  // alternatively  while (i < n && j < m)  {    // If current character    // needs to be alphabet    if (flag != 0)      ans += s1[i++];      // If current character    // needs to be a digit    else      ans += s2[j++];      // Flip flag for alternate    // arrangement    if(flag == 1)      flag = 0;    else      flag = 1;  }    // Return resultant String  return ans;} // Driver Code// Given Stringlet str = "geeks2020"; // Function Calldocument.write(rearrange(str) + "<br>"); // This code is contributed by patel2127</script>
g2e0e2k00
 
Time Complexity: O(N)Auxiliary Space: O(N)
mohit kumar 29
GauravRajput1
29AjayKumar
patel2127
frequency-counting
permutation
Hash
Strings
Hash
Strings
permutation
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