Find pairs with given sum such that elements of pair are in different rows

• Difficulty Level : Medium
• Last Updated : 25 Nov, 2021

Given a matrix of distinct values and a sum. The task is to find all the pairs in a given matrix whose summation is equal to the given sum. Each element of a pair must be from different rows i.e; the pair must not lie in the same row.

Examples:

Input : mat = {{1, 3, 2, 4},
{5, 8, 7, 6},
{9, 10, 13, 11},
{12, 0, 14, 15}}
sum = 11
Output: (1, 10), (3, 8), (2, 9), (4, 7), (11, 0)

Method 1 (Simple)
A simple solution for this problem is to, one by one, take each element of all rows and find pairs starting from the next immediate row in the matrix. The time complexity for this approach will be O(n4).

Method 2 (Use Sorting) within

• Sort all the rows in ascending order. The time complexity for this preprocessing will be O(n2 logn).
• Now we will select each row one by one and find pair elements in the remaining rows after the current row.
• Take two iterators, left and right. left iterator points left corner of the current i’th row and right iterator points right corner of the next j’th row in which we are going to find a pair of elements.
• If mat[i][left] + mat[j][right] < sum then left++ i.e; move in i’th row towards the right corner, otherwise right++ i.e; move in j’th row towards the left corner.

C++

 // C++ program to find a pair with given sum such that // every element of pair is in different rows. #include using namespace std;   const int MAX = 100;   // Function to find pair for given sum in matrix // mat[][] --> given matrix // n --> order of matrix // sum --> given sum for which we need to find pair void pairSum(int mat[][MAX], int n, int sum) {     // First sort all the rows in ascending order     for (int i=0; i=0)             {                 if ((mat[i][left] + mat[j][right]) == sum)                 {                     cout << "(" << mat[i][left]                          << ", "<< mat[j][right] << "), ";                     left++;                     right--;                 }                 else                 {                     if ((mat[i][left] + mat[j][right]) < sum)                         left++;                     else                         right--;                 }             }         }     } }   // Driver program to run the case int main() {     int n = 4, sum = 11;     int mat[][MAX] = {{1, 3, 2, 4},                       {5, 8, 7, 6},                       {9, 10, 13, 11},                       {12, 0, 14, 15}};     pairSum(mat, n, sum);     return 0; }

Java

 // Java program to find a pair with // given sum such that every element // of pair is in different rows. import java.util.Arrays; class GFG { static final int MAX = 100;   // Function to find pair for given sum in // matrix mat[][] --> given matrix // n --> order of matrix // sum --> given sum for which we need to find pair static void pairSum(int mat[][], int n, int sum) {           // First sort all the rows in ascending order     for (int i = 0; i < n; i++)     Arrays.sort(mat[i]);       // Select i'th row and find pair for element in i'th     // row in j'th row whose summation is equal to given sum     for (int i = 0; i < n - 1; i++) {         for (int j = i + 1; j < n; j++) {             int left = 0, right = n - 1;             while (left < n && right >= 0) {                 if ((mat[i][left] + mat[j][right]) == sum) {                 System.out.print("(" + mat[i][left] + ", " +                                      mat[j][right] + "), ");                 left++;                 right--;                 }                 else {                     if ((mat[i][left] + mat[j][right]) < sum)                         left++;                     else                         right--;                 }             }         }     } }   // Driver code public static void main(String[] args) {     int n = 4, sum = 11;     int mat[]         [] = {{1 ,  3,  2,  4},               {5 ,  8,  7,  6},               {9 , 10, 13, 11},               {12,  0, 14, 15}};     pairSum(mat, n, sum); } } // This code is contributed by Anant Agarwal.

Python 3

 # Python 3 program to find a pair with # given sum such that every element of # pair is in different rows. MAX = 100   # Function to find pair for given # sum in matrix mat[][] --> given matrix # n --> order of matrix # sum --> given sum for which we # need to find pair def pairSum(mat, n, sum):       # First sort all the rows     # in ascending order     for i in range(n):         mat[i].sort()       # Select i'th row and find pair for     # element in i'th row in j'th row     # whose summation is equal to given sum     for i in range(n - 1):         for j in range(i + 1, n):             left = 0             right = n - 1             while (left < n and right >= 0):                 if ((mat[i][left] + mat[j][right]) == sum):                     print( "(", mat[i][left],                            ", ", mat[j][right], "), ",                                             end = " ")                     left += 1                     right -= 1                                   else:                     if ((mat[i][left] +                          mat[j][right]) < sum):                         left += 1                     else:                         right -= 1   # Driver Code if __name__ == "__main__":     n = 4     sum = 11     mat = [[1, 3, 2, 4],            [5, 8, 7, 6],            [9, 10, 13, 11],            [12, 0, 14, 15]]     pairSum(mat, n, sum)   # This code is contributed # by ChitraNayal

C#

 // C# program to find a pair with // given sum such that every element // of pair is in different rows. using System; using System.Collections.Generic;   public class GFG {   // Function to find pair for given sum in // matrix mat[][] --> given matrix // n --> order of matrix // sum --> given sum for which we need to find pair static void pairSum(int [,]mat, int n, int sum) {           // First sort all the rows in ascending order     for (int i = 0; i < n; i++)     {         List l = new List();         for(int j = 0; j= 0) {                 if ((mat[i,left] + mat[j,right]) == sum) {                 Console.Write("(" + mat[i,left] + ", " +                                      mat[j,right] + "), ");                 left++;                 right--;                 }                 else {                     if ((mat[i,left] + mat[j,right]) < sum)                         left++;                     else                         right--;                 }             }         }     } }   // Driver code public static void Main(string[] args) {     int n = 4, sum = 11;     int [,]mat = {{1 ,  3,  2,  4},               {5 ,  8,  7,  6},               {9 , 10, 13, 11},               {12,  0, 14, 15}};     pairSum(mat, n, sum); } }   // This code is contributed by rutvik_56.

Javascript



Output:

(1, 10), (3, 8), (2, 9), (4, 7), (11, 0)

Time complexity : O(n2logn + n^3)
Auxiliary space : O(1)

Method 3 (Hashing)

1. Create an empty hash table and store all elements of the matrix in the hash as keys and their locations as values.
2. Traverse the matrix again to check for every element whether its pair exists in the hash table or not. If it exists, then compare row numbers. If row numbers are not the same, then print the pair.

CPP

 // C++ program to find pairs with given sum such // the two elements of pairs are from different rows #include using namespace std;   const int MAX = 100;   // Function to find pair for given sum in matrix // mat[][] --> given matrix // n --> order of matrix // sum --> given sum for which we need to find pair void pairSum(int mat[][MAX], int n, int sum) {     // Create a hash and store all elements of matrix     // as keys, and row as values     unordered_map hm;       // Traverse the matrix to check for every     // element whether its pair exists or not.     for (int i=0; i

Java

 // Java program to find pairs with given sum such // the two elements of pairs are from different rows import java.io.*; import java.util.*; class GFG {     static int MAX = 100;     // Function to find pair for given sum in matrix   // mat[][] --> given matrix   // n --> order of matrix   // sum --> given sum for which we need to find pair   static void pairSum(int mat[][], int n, int sum)   {       // Create a hash and store all elements of matrix     // as keys, and row and column numbers as values     Map> hm = new HashMap>();     for (int i = 0; i < n; i++)     {       for (int j = 0; j < n; j++)       {         hm.put(mat[i][j], new ArrayList(Arrays.asList(i, j)) );       }       }       // Traverse the matrix again to check for every     // element whether its pair exists or not.     for (int i = 0; i < n; i++)     {       for (int j = 0; j < n; j++)       {           // Look for remaining sum in hash         int remSum = sum - mat[i][j];           // If an element with value equal to remaining sum exists         if(hm.containsKey(remSum))         {             // Find row and column numbers of element with           // value equal to remaining sum.           ArrayList p = hm.get(remSum);           int row = p.get(0), col = p.get(1);             // If row number of pair is not same as current           // row, then print it as part of result.           // Second condition is there to make sure that a            // pair is printed only once.             if (row != i && row > i)           {             System.out.print("(" + mat[i][j] + "," + mat[row][col] + "), ");           }         }       }     }   }     // Driver code   public static void main (String[] args) {     int n = 4, sum = 11;     int[][] mat = {{1, 3, 2, 4},                    {5, 8, 7, 6},                    {9, 10, 13, 11},                    {12, 0, 14, 15}};     pairSum(mat, n, sum);   } }   // This code is contributed by avanitrachhadiya2155

Python3

 # JavaScript program to find pairs with given sum such # the two elements of pairs are from different rows MAX = 100   # Function to find pair for given sum in matrix   # mat[][] --> given matrix   # n --> order of matrix   # sum --> given sum for which we need to find pair def pairSum(mat, n, sum):           # Create a hash and store all elements of matrix     # as keys, and row and column numbers as values     hm = {}           for i in range(n):         for j in range(n):             hm[mat[i][j]] = [i, j]           # Traverse the matrix again to check for every     # element whether its pair exists or not.     for i in range(n):         for j in range(n):                         # Look for remaining sum in hash             remSum = sum - mat[i][j]                           # If an element with value equal to remaining sum exists             if remSum in hm:                                 # Find row and column numbers of element with                 # value equal to remaining sum.                 p=hm[remSum]                 row = p                 col = p                                   # If row number of pair is not same as current                 # row, then print it as part of result.                 # Second condition is there to make sure that a                 # pair is printed only once.                 if (row != i and row > i):                     print("(" , mat[i][j] , "," , mat[row][col] , "), ", end="")                       # Driver code n = 4 sum = 11 mat = [[1, 3, 2, 4],                    [5, 8, 7, 6],                    [9, 10, 13, 11],                    [12, 0, 14, 15]] pairSum(mat, n, sum)   # This code is contributed by patel2127

C#

 // C# program to find pairs with given sum such // the two elements of pairs are from different rows using System; using System.Collections.Generic; public class GFG {        // Function to find pair for given sum in matrix   // mat[][] --> given matrix   // n --> order of matrix   // sum --> given sum for which we need to find pair   static void pairSum(int[,] mat, int n, int sum)   {       // Create a hash and store all elements of matrix     // as keys, and row and column numbers as values     Dictionary> hm = new Dictionary>();     for (int i = 0; i < n; i++)     {       for (int j = 0; j < n; j++)       {         hm.Add(mat[i,j],new List(){i,j});       }     }       // Traverse the matrix again to check for every     // element whether its pair exists or not.     for (int i = 0; i < n; i++)     {       for (int j = 0; j < n; j++)       {           // Look for remaining sum in hash         int remSum = sum - mat[i,j];           // If an element with value equal to remaining sum exists         if(hm.ContainsKey(remSum))         {             // Find row and column numbers of element with           // value equal to remaining sum.           List p = hm[remSum];           int row = p, col = p;             // If row number of pair is not same as current           // row, then print it as part of result.           // Second condition is there to make sure that a            // pair is printed only once.             if (row != i && row > i)           {             Console.Write("(" + mat[i, j] + "," + mat[row, col] + "), ");           }         }       }     }   }     // Driver code   static public void Main (){     int n = 4, sum = 11;     int[,] mat = {{1, 3, 2, 4},                   {5, 8, 7, 6},                   {9, 10, 13, 11},                   {12, 0, 14, 15}};     pairSum(mat, n, sum);   } }   // This code is contributed by rag2127

Javascript



Output:

(1, 10), (3, 8), (2, 9), (4, 7), (11, 0),

One important thing is, when we traverse a matrix, a pair may be printed twice. To make sure that a pair is printed only once, we check if the row number of other elements picked from the hash table is more than the row number of the current element.
Time Complexity: O(n2) under the assumption that hash table inserts and search operations take O(1) time.

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