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

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

Examples:

Input : mat[4][4] = {{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)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

Method 2 (Use Sorting)

• Sort all the rows in ascending order. Time complexity for this preprocessing will be O(n2 logn).
• Now we will select each row one by one and find pair element in remaining rows after current row.
• Take two iterators left and right. left iterator points left corner of current i’th row and right iterator points right corner of next j’th row in which we are going to find pair element.
• If mat[i][left] + mat[j][right] < sum then left++ i.e; move in i’th row towards right corner, otherwise right++ i.e; move in j’th row towards 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 [tabbyending] 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 matrix in hash as key and their locations as values.
2. Traverse the matrix again to check for every element whether its pair exists in hash table or not. If exists, then compare row numbers. If row numbers are not 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 and column numbers as values     unordered_map > hm;     for (int i=0; i p = it->second;                 int row = p.first, col = p.second;                    // 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)                    cout << "(" << mat[i][j] << ", "                         << mat[row][col] << "), ";             }         }     } }    // Driver program  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; }

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 row number of other element picked from hash table is more than row number of current element.

Time Complexity : O(n2) under the assumption that hash table insert and search operations take O(1) time.

## tags:

Hash Matrix Hash Matrix