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

Last Updated : 15 Jul, 2025
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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 found 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++ 
// 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 type
string 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 Code
int main()
{
    // Given String
    string str = "geeks2020";

    // Function Call
    cout << rearrange(str) << endl;

    return 0;
}
Java 
// Java program to implement
// the above approach
class GFG{

// Function to rearrange given
// alphanumeric String such that
// no two adjacent characters
// are of the same type
static 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 Code
public 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 
# Python3 program to implement
# the above approach

# Function to rearrange given
# alphanumeric such that no 
# two adjacent characters
# are of the same type
def 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 Code
if __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# 
// C# program to implement
// the above approach
using System;
class GFG{

// Function to rearrange given
// alphanumeric String such that
// no two adjacent characters
// are of the same type
static 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 Code
public static void Main(String[] args)
{
  // Given String
  String str = "geeks2020";

  // Function Call
  Console.Write(rearrange(str) + "\n");
}
}

// This code is contributed by 29AjayKumar
JavaScript 
<script>
// Javascript program to implement
// the above approach

// Function to rearrange given
// alphanumeric String such that
// no two adjacent characters
// are of the same type
function 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 String
let str = "geeks2020";

// Function Call
document.write(rearrange(str) + "<br>");

// This code is contributed by patel2127
</script>

Output: 
g2e0e2k00

 

Time Complexity: O(N)
Auxiliary Space: O(N)


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