Given two arrays of equal size N, form maximum number of pairs by using their elements, one from the first array and second from the second array, such that an element from each array is used at-most-once and the absolute difference between the selected elements used for forming a pair is less than or equal to a given element K.
Examples:Â
Input : a[] = {3, 4, 5, 2, 1}
b[] = {6, 5, 4, 7, 15}
k = 3
Output : 4
The maximum number of pairs that can be formed
using the above 2 arrays is 4 and the corresponding
pairs are [1, 4], [2, 5], [3, 6], [4, 7], we can't
pair the remaining elements.
Other way of pairing under given constraint is
[2, 5], [3, 6], [4, 4], but count of pairs here
is 3 which is less than the result 4.
Simple Approach: By taking few examples, we can observe that if we sort both arrays. Then one by picking the closest feasible element for every element, we get the optimal answer.
In this approach we first sort both the arrays and then compare each element of the first array with each element of the second array for the possible pair, if it’s possible to form a pair, we form the pair and move to check for the next possible pair for the next element of the first array.
Implementation:
C++
// C++ implementation of above approach #include <bits/stdc++.h> #define ll long long int using namespace std;   // Returns count of maximum pairs that can // be formed from a[] and b[] under given // constraints. ll findMaxPairs(ll a[], ll b[], ll n, ll k) {     sort(a, a+n); // Sorting the first array.     sort(b, b+n); // Sorting the second array.       // To keep track of visited elements of b[]     bool flag[n];     memset(flag, false, sizeof(flag));       // For every element of a[], find a pair     // for it and break as soon as a pair is     // found.     int result = 0;     for (int i=0; i<n; i++)     {         for (int j=0; j<n; j++)         {             if (abs(a[i]-b[j])<=k && flag[j]==false)             {                 // Increasing the count if a pair is formed.                 result++;                   /* Making the corresponding flag array                    element as 1 indicating the element                    in the second array element has                    been used. */                flag[j] = true;                   // We break the loop to make sure an                 // element of a[] is used only once.                 break;             }         }     }     return result; }   // Driver code int main() {     ll a[] = {10, 15, 20}, b[] = {17, 12, 24};     int n = sizeof(a)/sizeof(a[0]);     int k = 3;     cout << findMaxPairs(a, b, n, k);     return 0; } |
Java
// Java implementation of above approach import java.util.*;   class solution {   // Returns count of maximum pairs that can // be formed from a[] and b[] under given // constraints. static int findMaxPairs(int a[], int b[], int n, int k) {     Arrays.sort(a); // Sorting the first array.     Arrays.sort(b); // Sorting the second array.       // To keep track of visited elements of b[]     boolean []flag = new boolean[n];     Arrays.fill(flag,false);       // For every element of a[], find a pair     // for it and break as soon as a pair is     // found.     int result = 0;     for (int i=0; i<n; i++)     {         for (int j=0; j<n; j++)         {             if (Math.abs(a[i]-b[j])<=k && flag[j]==false)             {                 // Increasing the count if a pair is formed.                 result++;                   /* Making the corresponding flag array                 element as 1 indicating the element                 in the second array element has                 been used. */                flag[j] = true;                   // We break the loop to make sure an                 // element of a[] is used only once.                 break;             }         }     }     return result; }   // Driver code public static void main(String args[]) {     int[] a = {10, 15, 20};     int[] b = {17, 12, 24};     int n = a.length;     int k = 3;     System.out.println(findMaxPairs(a, b, n, k));   } }   // This code is contributed by // Shashank_Sharma |
Python 3
# Returns count of maximum pairs # that can be formed from a[] and # b[] under given constraints. def findMaxPairs(a, b, n, k):       a.sort() # Sorting the first array.     b.sort() # Sorting the second array.       # To keep track of visited     # elements of b[]     flag = [False] * n       # For every element of a[], find     # a pair for it and break as soon     # as a pair is found.     result = 0    for i in range(n):         for j in range(n):             if (abs(a[i] - b[j]) <= k and                          flag[j] == False):                       # Increasing the count if                 # a pair is formed.                 result += 1                  ''' Making the corresponding flag array                 element as 1 indicating the element                 in the second array element has                 been used. '''                flag[j] = True                  # We break the loop to make sure an                 # element of a[] is used only once.                 break    return result   # Driver code if __name__ == "__main__":     a = [10, 15, 20]     b = [17, 12, 24]     n = len(a)     k = 3    print(findMaxPairs(a, b, n, k))   # This code is contributed # by ChitraNayal |
C#
// C# implementation of above approach using System;   class GFG {       // Returns count of maximum pairs that can     // be formed from a[] and b[] under given     // constraints.     static int findMaxPairs(int []a, int []b,                                 int n, int k)     {         Array.Sort(a); // Sorting the first array.         Array.Sort(b); // Sorting the second array.           // To keep track of visited elements of b[]         bool []flag = new bool[n];               //Arrays.fill(flag,false);           // For every element of a[], find a pair         // for it and break as soon as a pair is         // found.         int result = 0;         for (int i = 0; i < n; i++)         {             for (int j = 0; j < n; j++)             {                 if (Math.Abs(a[i] - b[j]) <= k && flag[j] == false)                 {                     // Increasing the count if a pair is formed.                     result++;                       /* Making the corresponding flag array                     element as 1 indicating the element                     in the second array element has                     been used. */                    flag[j] = true;                       // We break the loop to make sure an                     // element of a[] is used only once.                     break;                 }             }         }         return result;     }       // Driver code     public static void Main()     {         int[] a = {10, 15, 20};         int[] b = {17, 12, 24};         int n = a.Length;         int k = 3;         Console.WriteLine(findMaxPairs(a, b, n, k));     } }   // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of above approach           // Returns count of maximum pairs that can     // be formed from a[] and b[] under given     // constraints.     function findMaxPairs(a,b,n,k)     {         a.sort(function(c,d){return c-d;}); // Sorting the first array.         b.sort(function(c,d){return c-d;}) // Sorting the second array.        // To keep track of visited elements of b[]     let flag = new Array(n);     for(let i=0;i<flag.length;i++)     {         flag[i]=false;     }        // For every element of a[], find a pair     // for it and break as soon as a pair is     // found.     let result = 0;     for (let i=0; i<n; i++)     {         for (let j=0; j<n; j++)         {             if (Math.abs(a[i]-b[j])<=k && flag[j]==false)             {                 // Increasing the count if a pair is formed.                 result++;                    /* Making the corresponding flag array                 element as 1 indicating the element                 in the second array element has                 been used. */                flag[j] = true;                    // We break the loop to make sure an                 // element of a[] is used only once.                 break;             }         }     }     return result;     }           // Driver code     let a=[10, 15, 20];     let b=[17, 12, 24];     let n = a.length;     let k = 3;     document.write(findMaxPairs(a, b, n, k));                   // This code is contributed by rag2127 </script> |
Output
2
Time complexity : O(n2)Â
Auxiliary Space : O(n)Â
Efficient Approach: In this approach, rather than checking all the possible combinations of pairs, we optimize our code by checking only the feasible combination of pairs using the 2 pointer approach.
Implementation:
C++
#include <bits/stdc++.h> #define ll long long int using namespace std;   // Returns count of maximum pairs that can // be formed from a[] and b[] under given // constraints. ll findMaxPairs(ll a[], ll b[], ll n, ll k) {     sort(a, a+n); // Sorting the first array.     sort(b, b+n); // Sorting the second array.       int result = 0;     for (int i=0, j=0; i<n && j<n;)     {         if (abs(a[i] - b[j]) <= k)         {             result++;               // Increasing array pointer of             // both the first and the second array.             i++;             j++;         }           // Increasing array pointer of the second array.         else if(a[i] > b[j])             j++;           // Increasing array pointer of the first array.         else            i++;     }     return result; }   // Driver code int main() {     ll a[] = {10, 15, 20};     ll b[] = {17, 12, 24};     int n = sizeof(a)/sizeof(a[0]);     int k = 3;     cout << findMaxPairs(a, b, n, k);     return 0; } |
Java
// Java program for Maximizing Unique Pairs // from two arrays import java.util.*;   class GFG {   // Returns count of maximum pairs that can // be formed from a[] and b[] under given // constraints. static int findMaxPairs(int a[], int b[],                         int n, int k) {     Arrays.sort(a); // Sorting the first array.     Arrays.sort(b); // Sorting the second array.       int result = 0;     for (int i = 0, j = 0; i < n && j < n;)     {         if (Math.abs(a[i] - b[j]) <= k)         {             result++;               // Increasing array pointer of             // both the first and the second array.             i++;             j++;         }           // Increasing array pointer         // of the second array.         else if(a[i] > b[j])             j++;           // Increasing array pointer         // of the first array.         else            i++;     }     return result; }   // Driver code public static void main(String args[]) {     int a[] = {10, 15, 20};     int b[] = {17, 12, 24};     int n = a.length;     int k = 3;     System.out.println(findMaxPairs(a, b, n, k)); } }   // This code is contributed by // Sanjit_Prasad |
Python3
# Python3 program for # Maximizing Unique Pairs # from two arrays   # Returns count of maximum pairs that can # be formed from a[] and b[] under given # constraints. def findMaxPairs(a,b,n,k):           # Sorting the first array.     a.sort()           # Sorting the second array.     b.sort()           result =0    j=0    for i in range(n):         if j<n:             if abs(a[i]-b[j])<=k:                 result+=1              # Increasing array pointer of             # both the first and the second array.                 j +=1                              # Increasing array pointer of             # the second array.             elif a[i]>b[j]:                 j+=1    return result   # Driver code if __name__=='__main__':     a = [10,15,20]     b = [17,12,24]     n = len(a)     k =3    print(findMaxPairs(a,b,n,k))   # This code is contributed by # Shrikant13 |
C#
// C# program for Maximizing Unique Pairs // from two arrays using System;   class GFG {   // Returns count of maximum pairs that can // be formed from a[] and b[] under given // constraints. static int findMaxPairs(int []a, int []b,                         int n, int k) {     Array.Sort(a); // Sorting the first array.     Array.Sort(b); // Sorting the second array.       int result = 0;     for (int i = 0, j = 0; i < n && j < n;)     {         if (Math.Abs(a[i] - b[j]) <= k)         {             result++;               // Increasing array pointer of             // both the first and the second array.             i++;             j++;         }           // Increasing array pointer         // of the second array.         else if(a[i] > b[j])             j++;           // Increasing array pointer         // of the first array.         else            i++;     }     return result; }   // Driver code public static void Main(String []args) {     int []a = {10, 15, 20};     int []b = {17, 12, 24};     int n = a.Length;     int k = 3;     Console.WriteLine(findMaxPairs(a, b, n, k)); } }   // This code has been contributed by 29AjayKumar |
Javascript
<script>       // Javascript program for Maximizing     // Unique Pairs from two arrays           // Returns count of maximum pairs that can     // be formed from a[] and b[] under given     // constraints.     function findMaxPairs(a, b, n, k)     {         // Sorting the first array.         a.sort(function(a, b){return a - b});         // Sorting the second array.         b.sort(function(a, b){return a - b});           let result = 0;         for (let i = 0, j = 0; i < n && j < n;)         {             if (Math.abs(a[i] - b[j]) <= k)             {                 result++;                   // Increasing array pointer of                 // both the first and the second array.                 i++;                 j++;             }               // Increasing array pointer             // of the second array.             else if(a[i] > b[j])                 j++;               // Increasing array pointer             // of the first array.             else                i++;         }         return result;     }           let a = [10, 15, 20];     let b = [17, 12, 24];     let n = a.length;     let k = 3;     document.write(findMaxPairs(a, b, n, k));           </script> |
Output
2
Time complexity : O(n Log n)Â
Auxiliary Space : O(1)Â
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