Check if rearranging Array elements can form a Palindrome or not - GeeksforGeeks

25 Aug, 2021
Given a positive integer array arr of size N, the task is to check if number formed, from any arrangement of array elements, form a palindrome or not.
Examples:
Input: arr = [1, 2, 3, 1, 2]Output: YesExplanation: The elements of a given array can be rearranged as 1, 2, 3, 2, 1. Since 12321 is a palindrome, so output will be “Yes”
Input: arr = [1, 2, 3, 4, 1]Output: NoExplanation: The elements of a given array cannot be rearranged to form a palindrome within all the possible permutations. So the output will be “No”
Approach: Given problem can be solved using map to store the frequency of array elements
Store the frequency of all array elements
Check if frequency of each element is even
For element whose frequency is odd, if there is only one such element, then print Yes. Else print No.
Below is the implementation of the above approach:
C++14
Java
Python3
C#
Javascript
// C++ implementation of the above approach #include <bits/stdc++.h>using namespace std; #define MAX 256 // Function to check whether elements of// an array can form a palindromebool can_form_palindrome(int arr[], int n){    // create an empty string    // to append elements of an array    string str = "";     // append each element to the string str to form    // a string so that we can solve it in easy way    for (int i = 0; i < n; i++) {        str += arr[i];    }     // Create a freq array and initialize all    // values as 0    int freq[MAX] = { 0 };     // For each character in formed string,    // increment freq in the corresponding    // freq array    for (int i = 0; str[i]; i++) {        freq[str[i]]++;    }    int count = 0;     // Count odd occurring characters    for (int i = 0; i < MAX; i++) {        if (freq[i] & 1) {            count++;        }        if (count > 1) {            return false;        }    }     // Return true if odd count is 0 or 1,    return true;}// Drive codeint main(){    int arr[] = { 1, 2, 3, 1, 2 };    int n = sizeof(arr) / sizeof(int);    can_form_palindrome(arr, n)        ? cout << "YES"        : cout << "NO";    return 0;}
// Java implementation of the above approachimport java.io.*;import java.util.Arrays; class GFG{   static int MAX = 256;   // Function to check whether elements of  // an array can form a palindrome  static boolean can_form_palindrome(int []arr, int n)  {     // create an empty string    // to append elements of an array    String str = "";     // append each element to the string str to form    // a string so that we can solve it in easy way    for (int i = 0; i < n; i++) {      str += arr[i];    }     // Create a freq array and initialize all    // values as 0    int freq[] = new int[MAX];    Arrays.fill(freq,0);     // For each character in formed string,    // increment freq in the corresponding    // freq array    for (int i = 0; i<str.length(); i++) {      freq[str.charAt(i)]++;    }    int count = 0;     // Count odd occurring characters    for (int i = 0; i < MAX; i++) {      if ((freq[i] & 1)!=0) {        count++;      }      if (count > 1) {        return false;      }    }     // Return true if odd count is 0 or 1,    return true;  }   // Drive code  public static void main (String[] args)  {    int []arr = { 1, 2, 3, 1, 2 };    int n = arr.length;    if(can_form_palindrome(arr, n))      System.out.println("YES");    else      System.out.println("NO");  }} // This code is contributed by shivanisinghss2110
# python implementation of the above approach # Function to check whether elements of# an array can form a palindromedef can_form_palindrome(arr, n):    MAX = 256    # create an empty string    # to append elements of an array    s = ""     # append each element to the string str to form    # a string so that we can solve it in easy way    for i in range(n) :        s = s + str(arr[i])     # Create a freq array and initialize all    # values as 0    freq = [0]*MAX     # For each character in formed string,    # increment freq in the corresponding    # freq array    for i in range(N) :        freq[arr[i]]=freq[arr[i]]+1         count = 0     # Count odd occurring characters    for i in range(MAX) :        if (freq[i] & 1) :            count=count+1        if (count > 1) :            return False             # Return true if odd count is 0 or 1,    return True   # Driver Codeif __name__ ==  "__main__":    arr = [ 1, 2, 3, 1, 2 ]    N = len(arr)         # Function Call    if(can_form_palindrome(arr, N)):        print("YES")    else:        print("NO")                     # This code is contributed by anudeep23042002
// C# implementation of the above approachusing System;using System.Collections.Generic; class GFG{ static int MAX = 256; // Function to check whether elements of// an array can form a palindromestatic bool can_form_palindrome(int []arr, int n){    // create an empty string    // to append elements of an array    string str = "";     // append each element to the string str to form    // a string so that we can solve it in easy way    for (int i = 0; i < n; i++) {        str += arr[i];    }     // Create a freq array and initialize all    // values as 0    int []freq = new int[MAX];    Array.Clear(freq,0,MAX);     // For each character in formed string,    // increment freq in the corresponding    // freq array    for (int i = 0; i<str.Length; i++) {         freq[str[i]]++;    }    int count = 0;     // Count odd occurring characters    for (int i = 0; i < MAX; i++) {        if ((freq[i] & 1)!=0) {            count++;        }        if (count > 1) {            return false;        }    }     // Return true if odd count is 0 or 1,    return true;}   // Drive codepublic static void Main(){    int []arr = { 1, 2, 3, 1, 2 };    int n = arr.Length;    if(can_form_palindrome(arr, n))       Console.Write("YES");    else        Console.Write("NO");}} // This code is contributed by SURENDRA_GANGWAR.
<script>        // JavaScript Program to implement        // the above approach        let MAX = 256         // Function to check whether elements of        // an array can form a palindrome        function can_form_palindrome(arr, n)        {                     // create an empty string            // to append elements of an array            let str = "";             // append each element to the string str to form            // a string so that we can solve it in easy way            for (let i = 0; i < n; i++) {                str += toString(arr[i]);            }             // Create a freq array and initialize all            // values as 0            let freq = new Array(n).fill(0);             // For each character in formed string,            // increment freq in the corresponding            // freq array            for (let i = 0; i < str.length; i++) {                freq[str.charCodeAt(i)]++;            }            let count = 0;             // Count odd occurring characters            for (let i = 0; i < MAX; i++) {                if (freq[i] & 1) {                    count++;                }                if (count > 1) {                    return false;                }            }             // Return true if odd count is 0 or 1,            return true;        }                 // Drive code        let arr = [1, 2, 3, 1, 2];        let n = arr.length;        can_form_palindrome(arr, n)            ? document.write("YES")            : document.write("NO"); // This code is contributed by Potta Lokesh    </script>
YES
Time Complexity: O(N)Auxiliary Space: O(N)
 
anudeep23042002
lokeshpotta20
SURENDRA_GANGWAR
shivanisinghss2110
palindrome
Arrays
Strings
Arrays
Strings
palindrome
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