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 implementation of simple method # to find count of pairs with given sum. # Returns number of pairs in arr[0..n-1] # with sum equal to 'sum' def getPairsCount(arr, n, sum ):
count = 0 # Initialize result
# Consider all possible pairs
# and check their sums
for i in range ( 0 , n):
for j in range (i + 1 , n):
if arr[i] + arr[j] = = sum :
count + = 1
return count
# Driver function arr = [ 1 , 5 , 7 , - 1 , 5 ]
n = len (arr)
sum = 6
print ( "Count of pairs is" ,
getPairsCount(arr, n, sum ))
# This code is contributed by Smitha Dinesh Semwal |
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 :
# Python 3 implementation of simple method # to find count of pairs with given sum. import sys
# Returns number of pairs in arr[0..n-1] # with sum equal to 'sum' def getPairsCount(arr, n, sum ):
m = [ 0 ] * 1000
# Store counts of all elements in map m
for i in range ( 0 , n):
m[arr[i]] + = 1
twice_count = 0
# Iterate through each element and increment
# the count (Notice that every pair is
# counted twice)
for i in range ( 0 , n):
twice_count + = m[ sum - arr[i]]
# if (arr[i], arr[i]) pair satisfies the
# condition, then we need to ensure that
# the count is decreased by one such
# that the (arr[i], arr[i]) pair is not
# considered
if ( sum - arr[i] = = arr[i]):
twice_count - = 1
# return the half of twice_count
return int (twice_count / 2 )
# Driver function arr = [ 1 , 5 , 7 , - 1 , 5 ]
n = len (arr)
sum = 6
print ( "Count of pairs is" , getPairsCount(arr,
n, sum ))
# This code is contributed by # Smitha Dinesh Semwal |
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!