Count of pairs of Array elements with average at least K
Last Updated :
28 Mar, 2023
Given an array A[] of size N consisting of N integers, the task is to count the number of pairs such that their average is greater or equal to K.
Example:
Input: N = 4, K = 3, A = {5, 1, 3, 4}
Output: 4
Explanation: (5, 1), (5, 3), (5, 4) and (3, 4) are the required pairs with average greater or equal to K = 3.
Input: N = 3, K = 3, A = {1, 2, 3}
Output: 0
Explanation: No pairs exist with average greater or equal to K = 3.
Approach: This problem can be solved using binary search for the first occurrence of an element in based on the following observation:
- Just need to find 2*K - A[i] because average of two numbers X and Y is K ≤ (X+Y)/2. Now replace X with the current element that we are traversing i.e A[i] then equation becomes Y ≥ 2*K-A[i].
- So for any element A[i] in the array A[] total number of pairs formed are all the numbers in A[] that are greater than or equal to 2*K-A[i] i.e size of array 'A' -index of 2*K-A[i].
- So go as left as possible in A[] and for that find the first occurrence of 2*K-A[i]. If 2*K-A[i] is not found in A[] then return the index of next greater element of 2*K-A[i] because if average ≤ (X+Y)/2 for any two integers then also average ≤ (X+Z)/2 for all Z ≥ Y.
Follow the below steps to solve this problem:
- Sort the array A[].
- Traverse the array A[].
- Find the first occurrence of 2*k-A[i] for every element A[i] in A.
- If 2*k-A[i] does not exist then find the first occurrence of an element just greater than 2*k-A[i] in the array A and store its result in a variable (say ind).
- If ind is not -1, then add N-ind in the answer.
- Return the final answer after the traversal is over.
Below is the implementation of the above approach:
C++14
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the index of 2*K-A[i]
int findElement(int A[], int low,
int high, int key)
{
int ans = -1;
// Binary search
while (low <= high) {
int mid = low + (high - low) / 2;
if (key <= A[mid]) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Count the number of pairs
int countPairs(int A[], int& N, int& k)
{
sort(A, A + N);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int index
= findElement(A, i + 1, N - 1,
2 * k - A[i]);
if (index != -1)
count += N - index;
}
return count;
}
// Driver Code
int main()
{
int A[] = { 5, 1, 3, 4 };
int N = sizeof(A) / sizeof(A[0]);
int K = 3;
// Function call
cout << countPairs(A, N, K);
return 0;
}
// Java code to implement the approach
import java.util.*;
class GFG {
// Function to return the index of 2*K-A[i]
static int findElement(int A[], int low,
int high, int key)
{
int ans = -1;
// Binary search
while (low <= high) {
int mid = low + (high - low) / 2;
if (key <= A[mid]) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Count the number of pairs
static int countPairs(int A[], int N, int k)
{
Arrays.sort(A);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int index
= findElement(A, i + 1, N - 1,
2 * k - A[i]);
if (index != -1)
count += N - index;
}
return count;
}
// Driver Code
public static void main (String[] args) {
int A[] = { 5, 1, 3, 4 };
int N = A.length;
int K = 3;
// Function call
System.out.print(countPairs(A, N, K));
}
}
// This code is contributed by hrithikgarg03188.
# Python3 program for above approach
# Function to return the index of 2*K-A[i]
def findElement(A, low, high, key):
ans = -1
# binary search
while (low <= high):
mid = low + (high - low)//2
if key <= A[mid]:
ans = mid
high = mid - 1
else:
low = mid + 1
return ans
# Count the number of pairs
def countPairs(A, N, k):
A.sort()
count = 0
# Loop to count the number of pairs
for i in range(N):
index = findElement(A, i + 1, N - 1, 2 * k - A[i])
if index != -1:
count += N - index
return count
# Driver code
A = [5, 1, 3, 4]
N = len(A)
K = 3
# Function call
print(countPairs(A, N, K))
# this code is contributed by phasing17
// C# code to implement the approach
using System;
public class GFG {
// Function to return the index of 2*K-A[i]
static int findElement(int[] A, int low, int high,
int key)
{
int ans = -1;
// Binary search
while (low <= high) {
int mid = low + (high - low) / 2;
if (key <= A[mid]) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Count the number of pairs
static int countPairs(int[] A, int N, int k)
{
Array.Sort(A);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int index = findElement(A, i + 1, N - 1,
2 * k - A[i]);
if (index != -1)
count += N - index;
}
return count;
}
// Driver Code
static public void Main()
{
int[] A = { 5, 1, 3, 4 };
int N = A.Length;
int K = 3;
// Function call
Console.Write(countPairs(A, N, K));
}
}
// This code is contributed by Rohit Pradhan
<script>
// JavaScript code for the above approach
// Function to return the index of 2*K-A[i]
function findElement(A, low,
high, key) {
let ans = -1;
// Binary search
while (low <= high) {
let mid = low + Math.floor((high - low) / 2);
if (key <= A[mid]) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Count the number of pairs
function countPairs(A, N, k) {
A.sort(function (a, b) { return a - b });
let count = 0;
// Loop to count the number of pairs
for (let i = 0; i < N; i++) {
let index
= findElement(A, i + 1, N - 1,
2 * k - A[i]);
if (index != -1)
count += N - index;
}
return count;
}
// Driver Code
let A = [5, 1, 3, 4];
let N = A.length;
let K = 3;
// Function call
document.write(countPairs(A, N, K));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N log N)
Auxiliary Space: O(N)
Another Approach:
- Sort the array A[] in non-decreasing order using std::sort() function.
- Initialize a count variable to zero which will keep track of the number of pairs that have an average greater than or equal to K.
- Loop through the array A[] from left to right.
- For each element A[i], calculate the target value using the formula 2K - A[i]. This is because for two numbers X and Y, their average is greater than or equal to K if and only if (X + Y)/2 >= K, which simplifies to X + Y >= 2K.
- Use binary search to find the index of the first element in the array A[] that is greater than or equal to the target value. Initialize left to i + 1 and right to N - 1, and repeat until left > right. Inside the loop, calculate mid as the average of left and right, and compare A[mid] with the target value. If A[mid] is greater than or equal to the target, set right to mid - 1, otherwise set left to mid + 1.
- After the binary search, the index variable should contain the index of the first element in A[] that is greater than or equal to the target value. If index is less than N, then there are N - index pairs (A[i], A[j]), where i < j and A[j] >= target, that have an average greater than or equal to K. Add this count to the overall count variable.
- After the loop, return the count variable as the final answer.
Example:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the index of 2*K-A[i]
int countPairs(int A[], int& N, int& K)
{
sort(A, A + N);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int target = 2*K - A[i];
int left = i + 1, right = N - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (A[mid] >= target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
int index = left;
if (index != N)
count += N - index;
}
return count;
}
// Driver Code
int main()
{
int A[] = { 5, 1, 3, 4 };
int N = sizeof(A) / sizeof(A[0]);
int K = 3;
// Function call
cout << countPairs(A, N, K);
return 0;
}
// C# code to implement the approach
using System;
public class Program {
// Function to return the index of 2*K-A[i]
public static int CountPairs(int[] A, int N, int K) {
Array.Sort(A);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int target = 2 * K - A[i];
int left = i + 1, right = N - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (A[mid] >= target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
int index = left;
if (index != N)
count += N - index;
}
return count;
}
public static void Main() {
int[] A = { 5, 1, 3, 4 };
int N = A.Length;
int K = 3;
int result = CountPairs(A, N, K);
Console.WriteLine(result);
}
}
# Python code to implement the approach
def countPairs(A, N, K):
A.sort()
count = 0
# Loop to count the number of pairs
for i in range(N):
target = 2 * K - A[i]
left, right = i + 1, N - 1
while left <= right:
mid = left + (right - left) // 2
if A[mid] >= target:
right = mid - 1
else:
left = mid + 1
index = left
if index != N:
count += N - index
return count
# Driver code
A = [5, 1, 3, 4]
N = len(A)
K = 3
result = countPairs(A, N, K)
print(result)
// Java code to implement the approach
import java.util.Arrays;
class Main {
// Function to return the index of 2*K-A[i]
static int countPairs(int[] A, int N, int K)
{
Arrays.sort(A);
int count = 0;
// Loop to count the number of pairs
for (int i = 0; i < N; i++) {
int target = 2 * K - A[i];
int left = i + 1, right = N - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (A[mid] >= target) {
right = mid - 1;
}
else {
left = mid + 1;
}
}
int index = left;
if (index != N)
count += N - index;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int[] A = { 5, 1, 3, 4 };
int N = A.length;
int K = 3;
// Function call
System.out.println(countPairs(A, N, K));
}
}
function countPairs(A, N, K) {
A.sort(function(a, b) { return a - b; });
let count = 0;
for (let i = 0; i < N; i++) {
let target = 2 * K - A[i];
let left = i + 1, right = N - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (A[mid] >= target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
let index = left;
if (index != N)
count += N - index;
}
return count;
}
let A = [5, 1, 3, 4];
let N = A.length;
let K = 3;
let result = countPairs(A, N, K);
console.log(result);
Time Complexity: O(N*logN)
Auxiliary Space: O(1)
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