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.
Java
// Java implementation of simple method to find count of// pairs with given sum.public class find { public static void main(String args[]) { int[] arr = { 1, 5, 7, -1, 5 }; int sum = 6; getPairsCount(arr, sum); } // Prints number of pairs in arr[0..n-1] with sum equal // to 'sum' public static void getPairsCount(int[] arr, int sum) { int count = 0; // Initialize result // Consider all possible pairs and check their sums for (int i = 0; i < arr.length; i++) for (int j = i + 1; j < arr.length; j++) if ((arr[i] + arr[j]) == sum) count++; System.out.printf("Count of pairs is %d", count); }}// This program is contributed by Jyotsna |
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 :
Java
/* Java implementation of simple method to find count ofpairs with given sum*/import java.util.HashMap;class Test { static int arr[] = new int[] { 1, 5, 7, -1, 5 }; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' static int getPairsCount(int n, int sum) { HashMap<Integer, Integer> hm = new HashMap<>(); // Store counts of all elements in map hm for (int i = 0; i < n; i++) { // initializing value to 0, if key not found if (!hm.containsKey(arr[i])) hm.put(arr[i], 0); hm.put(arr[i], hm.get(arr[i]) + 1); } int twice_count = 0; // iterate through each element and increment the // count (Notice that every pair is counted twice) for (int i = 0; i < n; i++) { if (hm.get(sum - arr[i]) != null) twice_count += hm.get(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--; } // return the half of twice_count return twice_count / 2; } // Driver method to test the above function public static void main(String[] args) { int sum = 6; System.out.println( "Count of pairs is " + getPairsCount(arr.length, sum)); }}// This code is contributed by Gaurav Miglani |
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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