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Find all pairs (a, b) in an array such that a % b = k

Given an array with distinct elements, the task is to find the pairs in the array such that a % b = k, where k is a given integer.

Examples :

Input  :  arr[] = {2, 3, 5, 4, 7}   
              k = 3
Output :  (7, 4), (3, 4), (3, 5), (3, 7)
7 % 4 = 3
3 % 4 = 3
3 % 5 = 3
3 % 7 = 3


A Naive Solution is to make all pairs one by one and check their modulo is equal to k or not. If equals to k, then print that pair.

C++

// C++ implementation to find such pairs
#include <iostream>
using namespace std;
  
// Function to find pair such that (a % b = k)
bool printPairs(int arr[], int n, int k)
{
    bool isPairFound = true;
  
    // Consider each and every pair
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            // Print if their modulo equals to k
            if (i != j && arr[i] % arr[j] == k) {
                cout << "(" << arr[i] << ", "
                     << arr[j] << ")"
                     << " ";
                isPairFound = true;
            }
        }
    }
  
    return isPairFound;
}
  
// Driver program
int main()
{
    int arr[] = { 2, 3, 5, 4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
  
    if (printPairs(arr, n, k) == false)
        cout << "No such pair exists";
  
    return 0;
}

Java

// Java implementation to find such pairs
  
class Test {
    // method to find pair such that (a % b = k)
    static boolean printPairs(int arr[], int n, int k)
    {
        boolean isPairFound = true;
  
        // Consider each and every pair
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                // Print if their modulo equals to k
                if (i != j && arr[i] % arr[j] == k) {
                    System.out.print("(" + arr[i] + ", " + arr[j] + ")"
                                     + " ");
                    isPairFound = true;
                }
            }
        }
  
        return isPairFound;
    }
  
    // Driver method
    public static void main(String args[])
    {
        int arr[] = { 2, 3, 5, 4, 7 };
        int k = 3;
  
        if (printPairs(arr, arr.length, k) == false)
            System.out.println("No such pair exists");
    }
}

Python3

# Python3 implementation to find such pairs
  
# Function to find pair such that (a % b = k)
def printPairs(arr, n, k):
  
    isPairFound = True
  
    # Consider each and every pair
    for i in range(0, n):
      
        for j in range(0, n):
          
            # Print if their modulo equals to k
            if (i != j and arr[i] % arr[j] == k):
              
                print("(", arr[i], ", ", arr[j], ")",
                                 sep = "", end = " ")
                isPairFound = True
              
    return isPairFound
  
# Driver Code
arr = [2, 3, 5, 4, 7]
n = len(arr) 
k = 3
if (printPairs(arr, n, k) == False):
    print("No such pair exists")
  
# This article is contributed by Smitha Dinesh Semwal.

C#

// C# implementation to find such pair
using System;
  
public class GFG {
      
    // method to find pair such that (a % b = k)
    static bool printPairs(int[] arr, int n, int k)
    {
        bool isPairFound = true;
  
        // Consider each and every pair
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) 
            {
                // Print if their modulo equals to k
                if (i != j && arr[i] % arr[j] == k)
                {
                    Console.Write("(" + arr[i] + ", " 
                                + arr[j] + ")" + " ");
                    isPairFound = true;
                }
            }
        }
  
        return isPairFound;
    }
  
    // Driver method
    public static void Main()
    {
        int[] arr = { 2, 3, 5, 4, 7 };
        int k = 3;
  
        if (printPairs(arr, arr.Length, k) == false)
            Console.WriteLine("No such pair exists");
    }
}
  
// This code is contributed by Sam007

PHP

<?php
// PHP implementation to
// find such pairs
  
// Function to find pair
// such that (a % b = k)
function printPairs($arr, $n, $k)
{
    $isPairFound = true;
  
    // Consider each and every pair
    for ($i = 0; $i < $n; $i++) 
    {
        for ( $j = 0; $j < $n; $j++) 
        {
            // Print if their modulo
            // equals to k
            if ($i != $j && $arr[$i] % 
                            $arr[$j] == $k
            {
                echo "(" , $arr[$i] , ", ",
                       $arr[$j] , ")", " ";
                $isPairFound = true;
            }
        }
    }
  
    return $isPairFound;
}
  
// Driver Code
$arr = array(2, 3, 5, 4, 7);
$n = sizeof($arr);
$k = 3;
  
if (printPairs($arr, $n, $k) == false)
    echo "No such pair exists";
  
// This code is contributed by ajit
?>


Output :



(3, 5) (3, 4) (3, 7) (7, 4)

Time Complexity : O(n2)

An Efficient solution is based on below observations :

  1. If k itself is present in arr[], then k forms a pair with all elements arr[i] where k < arr[i]. For all such arr[i], we have k % arr[i] = k.
  2. For all elements greater than or equal to arr[i], we use following fact.
       If arr[i] % arr[j] = k, 
       ==> arr[i] = x * arr[j] + k
       ==> (arr[i] - k) = x * arr[j]
      We find all divisors of (arr[i] - k)
      and see if they are present in arr[].

To quickly check if an element is present in array, we use hashing.

C++

// C++ program to to find all pairs such that
// a % b = k.
#include <bits/stdc++.h>
using namespace std;
  
// Utiltity function to find the divisors of
// n and store in vector v[]
vector<int> findDivisors(int n)
{
    vector<int> v;
  
    // Vector is used to store  the divisors
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            // If n is a square number, push
            // only one occurrence
            if (n / i == i)
                v.push_back(i);
            else {
                v.push_back(i);
                v.push_back(n / i);
            }
        }
    }
    return v;
}
  
// Function to find pairs such that (a%b = k)
bool printPairs(int arr[], int n, int k)
{
    // Store all the elements in the map
    // to use map as hash for finding elements
    // in O(1) time.
    unordered_map<int, bool> occ;
    for (int i = 0; i < n; i++)
        occ[arr[i]] = true;
  
    bool isPairFound = false;
    for (int i = 0; i < n; i++) {
        // Print all the pairs with (a, b) as
        // (k, numbers greater than k) as
        // k % (num (> k)) = k i.e. 2%4 = 2
        if (occ[k] && k < arr[i]) {
            cout << "(" << k << ", " << arr[i] << ") ";
            isPairFound = true;
        }
  
        // Now check for the current element as 'a'
        // how many b exists such that a%b = k
        if (arr[i] >= k) {
            // find all the divisors of (arr[i]-k)
            vector<int> v = findDivisors(arr[i] - k);
  
            // Check for each divisor i.e. arr[i] % b = k
            // or not, if yes then print that pair.
            for (int j = 0; j < v.size(); j++) {
                if (arr[i] % v[j] == k && arr[i] != v[j] && occ[v[j]]) {
                    cout << "(" << arr[i] << ", "
                         << v[j] << ") ";
                    isPairFound = true;
                }
            }
  
            // Clear vector
            v.clear();
        }
    }
  
    return isPairFound;
}
  
// Driver program
int main()
{
    int arr[] = { 3, 1, 2, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
  
    if (printPairs(arr, n, k) == false)
        cout << "No such pair exists";
    return 0;
}

Java

// Java program to to find all pairs such that
// a % b = k.
  
import java.util.HashMap;
import java.util.Vector;
  
class Test {
    // Utility method to find the divisors of
    // n and store in vector v[]
    static Vector<Integer> findDivisors(int n)
    {
        Vector<Integer> v = new Vector<>();
  
        // Vector is used to store  the divisors
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                // If n is a square number, push
                // only one occurrence
                if (n / i == i)
                    v.add(i);
                else {
                    v.add(i);
                    v.add(n / i);
                }
            }
        }
        return v;
    }
  
    // method to find pairs such that (a%b = k)
    static boolean printPairs(int arr[], int n, int k)
    {
        // Store all the elements in the map
        // to use map as hash for finding elements
        // in O(1) time.
        HashMap<Integer, Boolean> occ = new HashMap<>();
        for (int i = 0; i < n; i++)
            occ.put(arr[i], true);
  
        boolean isPairFound = false;
        for (int i = 0; i < n; i++) {
            // Print all the pairs with (a, b) as
            // (k, numbers greater than k) as
            // k % (num (> k)) = k i.e. 2%4 = 2
            if (occ.get(k) && k < arr[i]) {
                System.out.print("(" + k + ", " + arr[i] + ") ");
                isPairFound = true;
            }
  
            // Now check for the current element as 'a'
            // how many b exists such that a%b = k
            if (arr[i] >= k) {
                // find all the divisors of (arr[i]-k)
                Vector<Integer> v = findDivisors(arr[i] - k);
  
                // Check for each divisor i.e. arr[i] % b = k
                // or not, if yes then print that pair.
                for (int j = 0; j < v.size(); j++) {
                    if (arr[i] % v.get(j) == k && arr[i] != v.get(j) && occ.get(v.get(j))) {
                        System.out.print("(" + arr[i] + ", "
                                         + v.get(j) + ") ");
                        isPairFound = true;
                    }
                }
  
                // Clear vector
                v.clear();
            }
        }
  
        return isPairFound;
    }
  
    // Driver method
    public static void main(String args[])
    {
        int arr[] = { 3, 1, 2, 5, 4 };
        int k = 2;
  
        if (printPairs(arr, arr.length, k) == false)
            System.out.println("No such pair exists");
    }
}

Python3

# Python3 program to to find all pairs 
# such that a % b = k.
   
# Utiltity function to find the divisors 
# of n and store in vector v[]
import math as mt
  
def findDivisors(n):
  
    v = []
  
    # Vector is used to store the divisors
    for i in range(1, mt.floor(n**(.5)) + 1):
        if (n % i == 0):
              
            # If n is a square number, push
            # only one occurrence
            if (n / i == i):
                v.append(i)
            else:
                v.append(i)
                v.append(n // i)
                  
    return v
  
# Function to find pairs such that (a%b = k)
def printPairs(arr, n, k):
  
    # Store all the elements in the map
    # to use map as hash for finding elements
    # in O(1) time.
    occ = dict()
    for i in range(n):
        occ[arr[i]] = True
  
    isPairFound = False
  
    for i in range(n):
          
        # Prall the pairs with (a, b) as
        # (k, numbers greater than k) as
        # k % (num (> k)) = k i.e. 2%4 = 2
        if (occ[k] and k < arr[i]):
            print("(", k, ",", arr[i], ")", end = " ")
            isPairFound = True
  
        # Now check for the current element as 'a'
        # how many b exists such that a%b = k
        if (arr[i] >= k):
              
            # find all the divisors of (arr[i]-k)
            v = findDivisors(arr[i] - k)
  
            # Check for each divisor i.e. arr[i] % b = k
            # or not, if yes then prthat pair.
            for j in range(len(v)):
                if (arr[i] % v[j] == k and 
                    arr[i] != v[j] and 
                    occ[v[j]]):
                    print("(", arr[i], ",", v[j], 
                                       ")", end = " ")
                    isPairFound = True
  
    return isPairFound
  
# Driver Code
arr = [3, 1, 2, 5, 4]
n = len(arr)
k = 2
  
if (printPairs(arr, n, k) == False):
    print("No such pair exists")
  
# This code is contributed by mohit kumar

C#

// C# program to to find all pairs 
// such that a % b = k. 
using System;
using System.Collections.Generic;
  
class GFG
{
// Utility method to find the divisors 
// of n and store in vector v[] 
public static List<int> findDivisors(int n)
{
    List<int> v = new List<int>();
  
    // Vector is used to store 
    // the divisors 
    for (int i = 1; 
             i <= Math.Sqrt(n); i++)
    {
        if (n % i == 0)
        {
            // If n is a square number, 
            // push only one occurrence 
            if (n / i == i)
            {
                v.Add(i);
            }
            else
            {
                v.Add(i);
                v.Add(n / i);
            }
        }
    }
    return v;
}
  
// method to find pairs such
// that (a%b = k) 
public static bool printPairs(int[] arr,
                              int n, int k)
{
    // Store all the elements in the 
    // map to use map as hash for 
    // finding elements in O(1) time. 
    Dictionary<int
               bool> occ = new Dictionary<int
                                          bool>();
    for (int i = 0; i < n; i++)
    {
        occ[arr[i]] = true;
    }
  
    bool isPairFound = false;
    for (int i = 0; i < n; i++)
    {
        // Print all the pairs with (a, b) as 
        // (k, numbers greater than k) as 
        // k % (num (> k)) = k i.e. 2%4 = 2 
        if (occ[k] && k < arr[i])
        {
            Console.Write("(" + k + ", "
                           arr[i] + ") ");
            isPairFound = true;
        }
  
        // Now check for the current element
        // as 'a' how many b exists such that 
        // a%b = k 
        if (arr[i] >= k)
        {
            // find all the divisors of (arr[i]-k) 
            List<int> v = findDivisors(arr[i] - k);
  
            // Check for each divisor i.e. 
            // arr[i] % b = k or not, if
            // yes then print that pair. 
            for (int j = 0; j < v.Count; j++)
            {
                if (arr[i] % v[j] == k && 
                    arr[i] != v[j] && occ[v[j]])
                {
                    Console.Write("(" + arr[i] +
                                  ", " + v[j] + ") ");
                    isPairFound = true;
                }
            }
  
            // Clear vector 
            v.Clear();
        }
    }
  
    return isPairFound;
}
  
// Driver Code 
public static void Main(string[] args)
{
    int[] arr = new int[] {3, 1, 2, 5, 4};
    int k = 2;
  
    if (printPairs(arr, arr.Length, k) == false)
    {
        Console.WriteLine("No such pair exists");
    }
}
}
  
// This code is contributed by Shrikant13


Output:

(2, 3) (2, 5) (5, 3) (2, 4)

Time Complexity: O(n* sqrt(max)) where max is the maximum element in the array.

Reference:
https://stackoverflow.com/questions/12732939/find-pairs-in-an-array-such-that-ab-k-where-k-is-a-given-integer

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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