Python Program for Count pairs with given sum
Last Updated :
15 Feb, 2023
Given an array of integers, and a number ‘sum’, find the number of pairs of integers in the array whose sum is equal to ‘sum’.
Examples:
Input : arr[] = {1, 5, 7, -1},
sum = 6
Output : 2
Pairs with sum 6 are (1, 5) and (7, -1)
Input : arr[] = {1, 5, 7, -1, 5},
sum = 6
Output : 3
Pairs with sum 6 are (1, 5), (7, -1) &
(1, 5)
Input : arr[] = {1, 1, 1, 1},
sum = 2
Output : 6
There are 3! pairs with sum 2.
Input : arr[] = {10, 12, 10, 15, -1, 7, 6,
5, 4, 2, 1, 1, 1},
sum = 11
Output : 9
Expected time complexity O(n)
Naive Solution – A simple solution is to traverse each element and check if there’s another number in the array which can be added to it to give sum.
Python3
def getPairsCount(arr, n, sum ):
count = 0
for i in range ( 0 , n):
for j in range (i + 1 , n):
if arr[i] + arr[j] = = sum :
count + = 1
return count
arr = [ 1 , 5 , 7 , - 1 , 5 ]
n = len (arr)
sum = 6
print ( "Count of pairs is" ,
getPairsCount(arr, n, sum ))
|
Output
Count of pairs is 3
Time Complexity: O(n2)
Auxiliary Space: O(1)
Efficient solution –
A better solution is possible in O(n) time. Below is the Algorithm –
- Create a map to store frequency of each number in the array. (Single traversal is required)
- In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
- After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.
Below is the implementation of above idea :
Python3
import sys
def getPairsCount(arr, n, sum ):
m = [ 0 ] * 1000
for i in range ( 0 , n):
m[arr[i]] + = 1
twice_count = 0
for i in range ( 0 , n):
twice_count + = m[ sum - arr[i]]
if ( sum - arr[i] = = arr[i]):
twice_count - = 1
return int (twice_count / 2 )
arr = [ 1 , 5 , 7 , - 1 , 5 ]
n = len (arr)
sum = 6
print ( "Count of pairs is" , getPairsCount(arr,
n, sum ))
|
Output
Count of pairs is 3
Time Complexity: O(n)
Auxiliary Space: O(n)
The extra space is used to store the elements in the map.
Please refer complete article on Count pairs with given sum for more details!
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