Check if an array can be divided into pairs whose sum is divisible by k - GeeksforGeeks

02 Mar, 2022
Given an array of integers and a number k, write a function that returns true if the given array can be divided into pairs such that the sum of every pair is divisible by k.
Examples: 
Input: arr[] = {9, 7, 5, 3}, k = 6 Output: True We can divide the array into (9, 3) and (7, 5). Sum of both of these pairs is a multiple of 6.
Input: arr[] = {92, 75, 65, 48, 45, 35}, k = 10 Output: True We can divide the array into (92, 48), (75, 65). and (45, 35). The sum of all these pairs is a multiple of 10.
Input: arr[] = {91, 74, 66, 48}, k = 10 Output: False 
A Simple Solution is to iterate through every element arr[i]. Find if there is another not yet visited element that has a remainder like (k – arr[i]%k). If there is no such element, return false. If a pair is found, then mark both elements as visited. The time complexity of this solution is O(n2 and it requires O(n) extra space.
An Efficient Solution is to use Hashing.
1) If length of given array is odd, return false. 
    An odd length array cannot be divided into pairs.
2) Traverse input array and count occurrences of 
    all remainders (use (arr[i] % k)+k)%k for handling the case of negative integers as well). 
      freq[((arr[i] % k)+k)%k]++
3) Traverse input array again. 
   a) Find the remainder of the current element.
   b) If remainder divides k into two halves, then
      there must be even occurrences of it as it 
      forms pair with itself only.
   c) If the remainder is 0, then there must be 
      even occurrences.
   d) Else, number of occurrences of current 
      the remainder must be equal to a number of 
      occurrences of "k - current remainder".
An efficient approach is to use hashing (unordered_map in C++ and HashMap in Java).
The below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// A C++ program to check if arr[0..n-1] can be divided// in pairs such that every pair is divisible by k.#include <bits/stdc++.h>using namespace std; // Returns true if arr[0..n-1] can be divided into pairs// with sum divisible by k.bool canPairs(int arr[], int n, int k){    // An odd length array cannot be divided into pairs    if (n & 1)        return false;     // Create a frequency array to count occurrences    // of all remainders when divided by k.    unordered_map<int, int> freq;     // Count occurrences of all remainders    for (int i = 0; i < n; i++)        freq[((arr[i] % k) + k) % k]++;     // Traverse input array and use freq[] to decide    // if given array can be divided in pairs    for (int i = 0; i < n; i++) {        // Remainder of current element        int rem = ((arr[i] % k) + k) % k;         // If remainder with current element divides        // k into two halves.        if (2 * rem == k) {            // Then there must be even occurrences of            // such remainder            if (freq[rem] % 2 != 0)                return false;        }         // If remainder is 0, then there must be even        // number of elements with 0 remainder        else if (rem == 0) {            if (freq[rem] & 1)                return false;        }         // Else number of occurrences of remainder        // must be equal to number of occurrences of        // k - remainder        else if (freq[rem] != freq[k - rem])            return false;    }    return true;} // Driver codeint main(){    int arr[] = { 92, 75, 65, 48, 45, 35 };    int k = 10;    int n = sizeof(arr) / sizeof(arr[0]);     // Function call    canPairs(arr, n, k) ? cout << "True" : cout << "False";    return 0;}
// JAVA program to check if arr[0..n-1] can be divided// in pairs such that every pair is divisible by k.import java.util.HashMap;public class Divisiblepair {    // Returns true if arr[0..n-1] can be divided into pairs    // with sum divisible by k.    static boolean canPairs(int ar[], int k)    {        // An odd length array cannot be divided into pairs        if (ar.length % 2 == 1)            return false;         // Create a frequency array to count occurrences        // of all remainders when divided by k.        HashMap<Integer, Integer> hm = new HashMap<>();         // Count occurrences of all remainders        for (int i = 0; i < ar.length; i++) {            int rem = ((ar[i] % k) + k) % k;            if (!hm.containsKey(rem)) {                hm.put(rem, 0);            }            hm.put(rem, hm.get(rem) + 1);        }         // Traverse input array and use freq[] to decide        // if given array can be divided in pairs        for (int i = 0; i < ar.length; i++) {            // Remainder of current element            int rem = ((ar[i] % k) + k) % k;             // If remainder with current element divides            // k into two halves.            if (2 * rem == k) {                // Then there must be even occurrences of                // such remainder                if (hm.get(rem) % 2 == 1)                    return false;            }             // If remainder is 0, then there must be two            // elements with 0 remainder            else if (rem == 0) {                // Then there must be even occurrences of                // such remainder                if (hm.get(rem) % 2 == 1)                    return false;            }             // Else number of occurrences of remainder            // must be equal to number of occurrences of            // k - remainder            else {                if (hm.get(k - rem) != hm.get(rem))                    return false;            }        }        return true;    }     // Driver code    public static void main(String[] args)    {        int arr[] = { 92, 75, 65, 48, 45, 35 };        int k = 10;         // Function call        boolean ans = canPairs(arr, k);        if (ans)            System.out.println("True");        else            System.out.println("False");    }} // This code is contributed by Rishabh Mahrsee
# Python3 program to check if# arr[0..n-1] can be divided# in pairs such that every# pair is divisible by k.from collections import defaultdict # Returns true if arr[0..n-1] can be# divided into pairs with sum# divisible by k.  def canPairs(arr, n, k):     # An odd length array cannot    # be divided into pairs    if (n & 1):        return 0     # Create a frequency array to    # count occurrences of all    # remainders when divided by k.    freq = defaultdict(lambda: 0)     # Count occurrences of all remainders    for i in range(0, n):        freq[((arr[i] % k) + k) % k] += 1     # Traverse input array and use    # freq[] to decide if given array    # can be divided in pairs    for i in range(0, n):         # Remainder of current element        rem = ((arr[i] % k) + k) % k         # If remainder with current element        # divides k into two halves.        if (2 * rem == k):             # Then there must be even occurrences            # of such remainder            if (freq[rem] % 2 != 0):                return 0         # If remainder is 0, then there        # must be two elements with 0 remainder        else if (rem == 0):            if (freq[rem] & 1):                return 0             # Else number of occurrences of            # remainder must be equal to            # number of occurrences of            # k - remainder        else if (freq[rem] != freq[k - rem]):             return 0     return 1  # Driver codearr = [92, 75, 65, 48, 45, 35]k = 10n = len(arr) # Function callif (canPairs(arr, n, k)):    print("True")else:    print("False") # This code is contributed by Stream_Cipher
// C# program to check if arr[0..n-1]// can be divided in pairs such that// every pair is divisible by k.using System.Collections.Generic;using System; class GFG {     // Returns true if arr[0..n-1] can be    // divided into pairs with sum    // divisible by k.    static bool canPairs(int[] ar, int k)    {         // An odd length array cannot        // be divided into pairs        if (ar.Length % 2 == 1)            return false;         // Create a frequency array to count        // occurrences of all remainders when        // divided by k.        Dictionary<Double, int> hm            = new Dictionary<Double, int>();         // Count occurrences of all remainders        for (int i = 0; i < ar.Length; i++) {            int rem = ((ar[i] % k) + k) % k;            if (!hm.ContainsKey(rem)) {                hm[rem] = 0;            }            hm[rem]++;        }         // Traverse input array and use freq[]        // to decide if given array can be        // divided in pairs        for (int i = 0; i < ar.Length; i++) {             // Remainder of current element            int rem = ((ar[i] % k) + k) % k;             // If remainder with current element            // divides k into two halves.            if (2 * rem == k) {                 // Then there must be even occurrences                // of such remainder                if (hm[rem] % 2 == 1)                    return false;            }             // If remainder is 0, then there            // must be two elements with 0            // remainder            else if (rem == 0) {                // Then there must be even occurrences                // of such remainder                if (hm[rem] % 2 == 1)                    return false;            }             // Else number of occurrences of remainder            // must be equal to number of occurrences of            // k - remainder            else {                if (hm[k - rem] != hm[rem])                    return false;            }        }        return true;    }     // Driver code    public static void Main()    {        int[] arr = { 92, 75, 65, 48, 45, 35 };        int k = 10;         // Function call        bool ans = canPairs(arr, k);        if (ans)            Console.WriteLine("True");        else            Console.WriteLine("False");    }} // This code is contributed by Stream_Cipher
<script> // Javascript program to check if arr[0..n-1] can be divided// in pairs such that every pair is divisible by k.     // Returns true if arr[0..n-1] can be divided into pairs    // with sum divisible by k.       function canPairs(ar, k)    {        // An odd length array cannot be divided into pairs        if (ar.length % 2 == 1)            return false;          // Create a frequency array to count occurrences        // of all remainders when divided by k.        let hm = new Map();          // Count occurrences of all remainders        for (let i = 0; i < ar.length; i++) {            let rem = ((ar[i] % k) + k) % k;            if (!hm.has(rem)) {                hm.set(rem, 0);            }            hm.set(rem, hm.get(rem) + 1);        }          // Traverse input array and use freq[] to decide        // if given array can be divided in pairs        for (let i = 0; i < ar.length; i++) {            // Remainder of current element            let rem = ((ar[i] % k) + k) % k;              // If remainder with current element divides            // k into two halves.            if (2 * rem == k) {                // Then there must be even occurrences of                // such remainder                if (hm.get(rem) % 2 == 1)                    return false;            }              // If remainder is 0, then there must be two            // elements with 0 remainder            else if (rem == 0) {                // Then there must be even occurrences of                // such remainder                if (hm.get(rem) % 2 == 1)                    return false;            }              // Else number of occurrences of remainder            // must be equal to number of occurrences of            // k - remainder            else {                if (hm.get(k - rem) != hm.get(rem))                    return false;            }        }        return true;    }   // Driver program       let arr = [ 92, 75, 65, 48, 45, 35 ];      let k = 10;        // Function call      let ans = canPairs(arr, k);      if (ans)          document.write("True");      else          document.write("False");       </script>
True
Time complexity: O(n).
This article is contributed by Priyanka. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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