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Find pairs with given sum such that elements of pair are in different rows in Python
Suppose we have a matrix of unique elements and a sum; we have to find all the pairs from the matrix whose sum is equal to given sum. Here, each element of pair will be taken from different rows.
So, if the input is like −
| 2 | 4 | 3 | 5 |
| 6 | 9 | 8 | 7 |
| 10 | 11 | 14 | 12 |
| 13 | 1 | 15 | 16 |
sum = 13, then the output will be [(2, 11), (4, 9), (3, 10), (5, 8), (12, 1)]
To solve this, we will follow these steps −
res := a new list
n := size of matrix
for i in range 0 to n, do
sort the list matrix[i]
for i in range 0 to n - 1, do
for j in range i + 1 to n, do
low := 0, high := n - 1
while low < n and high >= 0, do
if (matrix[i, low] + matrix[j, high]) is same as sum, then
pair := make pair using (matrix[i, low],matrix[j, high])
insert pair at the end of res
low := low + 1
high := high - 1
otherwise,
if (matrix[i][low] + matrix[j][high]) < sum, then
low := low + 1
otherwise,
high := high - 1
return res
Example (Python)
Let us see the following implementation to get better understanding −
MAX = 100 def sum_pair(matrix, sum): res = [] n = len(matrix) for i in range(n): matrix[i].sort() for i in range(n - 1): for j in range(i + 1, n): low = 0 high = n - 1 while (low < n and high >= 0): if ((matrix[i][low] + matrix[j][high]) == sum): pair = (matrix[i][low],matrix[j][high]) res.append(pair) low += 1 high -= 1 else: if ((matrix[i][low] + matrix[j][high]) < sum): low += 1 else: high -= 1 return res sum = 13 matrix = [ [2, 4, 3, 5], [6, 9, 8, 7], [10, 11, 14, 12], [13, 1, 15, 16]] print(sum_pair(matrix, sum))
Input
[[2, 4, 3, 5], [6, 9, 8, 7], [10, 11, 14, 12], [13, 1, 15, 16]] sum = 13
Output
[(4, 9), (5, 8), (2, 11), (3, 10), (12, 1)]
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