Given an array arr[] of N integers, the task is to count the number of pairs with positive sum.
Examples:
Input: arr[] = {-7, -1, 3, 2}
Output: 3
Explanation:
The pairs with positive sum are: {-1, 3}, {-1, 2}, {3, 2}.Input: arr[] = {-4, -2, 5}
Output: 2
Explanation:
The pairs with positive sum are: {-4, 5}, {-2, 5}.
Naive Approach:
A naive approach is to traverse each element and check if there is another number in the array arr[] which can be added to it to give the positive-sum or not.
Below is the implementation of the above approach:
C++
// Naive approach to count pairs // with positive sum. #include <bits/stdc++.h> using namespace std; // Returns number of pairs in // arr[0..n-1] with positive sum int CountPairs(int arr[], int n) { // Initialize result int count = 0; // Consider all possible pairs // and check their sums for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If arr[i] & arr[j] // form valid pair if (arr[i] + arr[j] > 0) count += 1; } } return count; } // Driver's Code int main() { int arr[] = { -7, -1, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call to find the // count of pairs cout << CountPairs(arr, n); return 0; } |
Java
// Java implementation of the above approach class GFG { // Naive approach to count pairs // with positive sum. // Returns number of pairs in // arr[0..n-1] with positive sum static int CountPairs(int arr[], int n) { // Initialize result int count = 0; // Consider all possible pairs // and check their sums for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If arr[i] & arr[j] // form valid pair if (arr[i] + arr[j] > 0) count += 1; } } return count; } // Driver's Code public static void main (String[] args) { int []arr = { -7, -1, 3, 2 }; int n = arr.length; // Function call to find the // count of pairs System.out.println(CountPairs(arr, n)); } } // This code is contributed by Yash_R |
Python3
# Naive approach to count pairs # with positive sum. # Returns number of pairs in # arr[0..n-1] with positive sum def CountPairs(arr, n) : # Initialize result count = 0; # Consider all possible pairs # and check their sums for i in range(n) : for j in range( i + 1, n) : # If arr[i] & arr[j] # form valid pair if (arr[i] + arr[j] > 0) : count += 1; return count; # Driver's Code if __name__ == "__main__" : arr = [ -7, -1, 3, 2 ]; n = len(arr); # Function call to find the # count of pairs print(CountPairs(arr, n)); # This code is contributed by Yash_R |
Output:
3
Time Complexity: O(N2)
Efficient Approach:
The idea is to use the Two Pointers Technique. Following are the steps:
- Sort the given array arr[] in increasing order.
- Take two pointers. one representing the first element and second representing the last element of the sorted array.
- If the sum of elements at these pointers is greater than 0, then the difference between the pointers will give the count of pairs with positive-sum for the element at second pointers. Decrease the second pointer by 1.
- Else increase the first pointers by 1.
- Repeat the above steps untill both pointers converge towards each other.
Below is the implementation of the above approach:
C++
// C++ program to count the // pairs with positive sum #include <bits/stdc++.h> using namespace std; // Returns number of pairs // in arr[0..n-1] with // positive sum int CountPairs(int arr[], int n) { // Sort the array in // increasing order sort(arr, arr + n); // Intialise result int count = 0; // Intialise first and // second pointer int l = 0, r = n - 1; // Till the pointers // doesn't converge // traverse array to // count the pairs while (l < r) { // If sum of arr[i] && // arr[j] > 0, then the // count of pairs with // positive sum is the // difference between // the two pointers if (arr[l] + arr[r] > 0) { // Increase the count count += (r - l); r--; } else { l++; } } return count; } // Driver's Code int main() { int arr[] = { -7, -1, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call to count // the pairs with positive // sum cout << CountPairs(arr, n); return 0; } |
Python3
# Python3 program to count the # pairs with positive sum # Returns number of pairs # in arr[0..n-1] with # positive sum def CountPairs(arr, n) : # Sort the array in # increasing order arr.sort() # Intialise result count = 0; # Intialise first and # second pointer l = 0; r = n - 1; # Till the pointers # doesn't converge # traverse array to # count the pairs while (l < r) : # If sum of arr[i] && # arr[j] > 0, then the # count of pairs with # positive sum is the # difference between # the two pointers if (arr[l] + arr[r] > 0) : # Increase the count count += (r - l); r -= 1; else : l += 1; return count; # Driver's Code if __name__ == "__main__" : arr = [ -7, -1, 3, 2 ]; n = len(arr); # Function call to count # the pairs with positive # sum print(CountPairs(arr, n)); # This code is contributed by Yash_R |
Output:
3
Time Complexity: O(N*log N)
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