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)



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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find a pair with given sum such that
// every element of pair is in different rows.
#include<bits/stdc++.h>
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<n; i++)
        sort(mat[i], mat[i]+n);
  
    // 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)
                {
                    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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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[]
        [] = {{1 324},
              {5 876},
              {9 , 10, 13, 11},
              {120, 14, 15}};
    pairSum(mat, n, sum);
}
}
// This code is contributed by Anant Agarwal.

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right



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

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find pairs with given sum such
// the two elements of pairs are from different rows
#include<bits/stdc++.h>
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<int, pair<int, int> > hm;
    for (int i=0; i<n; i++)
        for (int j=0; j<n; j++)
            hm[mat[i][j]] = make_pair(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];
            auto it = hm.find(remSum); // it is an iterator
                                       // of unordered_map type
  
            // If an element with value equal to remaining sum exists
            if (it != hm.end())
            {
                // Find row and column numbers of element with
                // value equal to remaining sum.
                pair<int, int> 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;
}

chevron_right



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.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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



My Personal Notes arrow_drop_up

Improved By : chitranayal



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.