Pair with largest sum which is less than K in the array
Last Updated :
31 Oct, 2022
Given an array arr of size N and an integer K. The task is to find the pair of integers such that their sum is maximum and but less than K
Examples:
Input : arr = {30, 20, 50} , K = 70
Output : 30, 20
30 + 20 = 50, which is the maximum possible sum which is less than K
Input : arr = {5, 20, 110, 100, 10}, K = 85
Output : 20, 10
Approach :
An efficient approach is to sort the given array and find the element which is greater than or equal to K. If found at index p, we have to find pairs only between arr[0, ..., p-1]. Run nested loops. One to take care of the first element in the pair and the other to take care of the second element in the pair. Maintain a variable maxsum and two other variables a and b to keep track of the possible solution. Initialize maxsum to 0. Find A[i] + A[j] (assuming i runs in the outer loop and j in the inner loop). If it is greater than maxsum then update maxsum with this sum and change a and b to the i'th and j'th elements of this array.
Below is the implementation of the above approach :
C++
// CPP program to find pair with largest
// sum which is less than K in the array
#include <bits/stdc++.h>
using namespace std;
// Function to find pair with largest
// sum which is less than K in the array
void Max_Sum(int arr[], int n, int k)
{
// To store the break point
int p = n;
// Sort the given array
sort(arr, arr + n);
// Find the break point
for (int i = 0; i < n; i++)
{
// No need to look beyond i'th index
if (arr[i] >= k) {
p = i;
break;
}
}
int maxsum = 0, a, b;
// Find the required pair
for (int i = 0; i < p; i++)
{
for (int j = i + 1; j < p; j++)
{
if (arr[i] + arr[j] < k and arr[i] + arr[j] >
maxsum)
{
maxsum = arr[i] + arr[j];
a = arr[i];
b = arr[j];
}
}
}
// Print the required answer
cout << a << " " << b;
}
// Driver code
int main()
{
int arr[] = {5, 20, 110, 100, 10}, k = 85;
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
Max_Sum(arr, n, k);
return 0;
}
// Java program to find pair with largest
// sum which is less than K in the array
import java.util.Arrays;
class GFG
{
// Function to find pair with largest
// sum which is less than K in the array
static void Max_Sum(int arr[], int n, int k)
{
// To store the break point
int p = n;
// Sort the given array
Arrays.sort(arr);
// Find the break point
for (int i = 0; i < n; i++)
{
// No need to look beyond i'th index
if (arr[i] >= k)
{
p = i;
break;
}
}
int maxsum = 0, a = 0, b = 0;
// Find the required pair
for (int i = 0; i < p; i++)
{
for (int j = i + 1; j < p; j++)
{
if (arr[i] + arr[j] < k &&
arr[i] + arr[j] > maxsum)
{
maxsum = arr[i] + arr[j];
a = arr[i];
b = arr[j];
}
}
}
// Print the required answer
System.out.print( a + " " + b);
}
// Driver code
public static void main (String[] args)
{
int []arr = {5, 20, 110, 100, 10};
int k = 85;
int n = arr.length;
// Function call
Max_Sum(arr, n, k);
}
}
// This code is contributed by anuj_67..
# Python3 program to find pair with largest
# sum which is less than K in the array
# Function to find pair with largest
# sum which is less than K in the array
def Max_Sum(arr, n, k):
# To store the break point
p = n
# Sort the given array
arr.sort()
# Find the break point
for i in range(0, n):
# No need to look beyond i'th index
if (arr[i] >= k):
p = i
break
maxsum = 0
a = 0
b = 0
# Find the required pair
for i in range(0, p):
for j in range(i + 1, p):
if(arr[i] + arr[j] < k and
arr[i] + arr[j] > maxsum):
maxsum = arr[i] + arr[j]
a = arr[i]
b = arr[j]
# Print the required answer
print(a, b)
# Driver code
arr = [5, 20, 110, 100, 10]
k = 85
n = len(arr)
# Function call
Max_Sum(arr, n, k)
# This code is contributed by Sanjit_Prasad
// C# program to find pair with largest
// sum which is less than K in the array
using System;
class GFG
{
// Function to find pair with largest
// sum which is less than K in the array
static void Max_Sum(int []arr, int n, int k)
{
// To store the break point
int p = n;
// Sort the given array
Array.Sort(arr);
// Find the break point
for (int i = 0; i < n; i++)
{
// No need to look beyond i'th index
if (arr[i] >= k)
{
p = i;
break;
}
}
int maxsum = 0, a = 0, b = 0;
// Find the required pair
for (int i = 0; i < p; i++)
{
for (int j = i + 1; j < p; j++)
{
if (arr[i] + arr[j] < k &&
arr[i] + arr[j] > maxsum)
{
maxsum = arr[i] + arr[j];
a = arr[i];
b = arr[j];
}
}
}
// Print the required answer
Console.WriteLine( a + " " + b);
}
// Driver code
public static void Main ()
{
int []arr = {5, 20, 110, 100, 10};
int k = 85;
int n = arr.Length;
// Function call
Max_Sum(arr, n, k);
}
}
// This code is contributed by anuj_67..
<script>
// Javascript program to find pair with largest
// sum which is less than K in the array
// Function to find pair with largest
// sum which is less than K in the array
function Max_Sum(arr, n, k)
{
// To store the break point
let p = n;
// Sort the given array
arr.sort(function(a, b){return a - b});
// Find the break point
for (let i = 0; i < n; i++)
{
// No need to look beyond i'th index
if (arr[i] >= k)
{
p = i;
break;
}
}
let maxsum = 0, a = 0, b = 0;
// Find the required pair
for (let i = 0; i < p; i++)
{
for (let j = i + 1; j < p; j++)
{
if (arr[i] + arr[j] < k &&
arr[i] + arr[j] > maxsum)
{
maxsum = arr[i] + arr[j];
a = arr[i];
b = arr[j];
}
}
}
// Print the required answer
document.write( a + " " + b + "</br>");
}
let arr = [5, 20, 110, 100, 10];
let k = 85;
let n = arr.length;
// Function call
Max_Sum(arr, n, k);
</script>
Time complexity: O(N^2)
Auxiliary Space: O(1)
Efficient Approach: Two Pointer Method
After sorting, we can place two pointers at the left and right extremes of the array. The desired pair can be found by the following steps:
- Find the current sum of the values at both the pointers.
- If the sum is lesser than k:
- update the answer with the maximum of the previous answer and the current sum.
- increase the left pointer.
- Else Decrease the right pointer.
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find max sum less than k
int maxSum(vector<int> arr, int k)
{
// Sort the array
sort(arr.begin(), arr.end());
int n = arr.size(), l = 0, r = n - 1, ans = 0;
// While l is less than r
while (l < r) {
if (arr[l] + arr[r] < k) {
ans = max(ans, arr[l] + arr[r]);
l++;
}
else {
r--;
}
}
return ans;
}
// Driver Code
int main()
{
vector<int> A = { 20, 10, 30, 100, 110 };
int k = 85;
// Function Call
cout << maxSum(A, k) << endl;
}
// Java program for the above approach
import java.util.Arrays;
class GFG
{
// Function to find max sum less than k
static int maxSum(int arr[], int k)
{
// Sort the array
Arrays.sort(arr);
int n = arr.length, l = 0, r = n - 1, ans = 0;
// While l is less than r
while (l < r) {
if (arr[l] + arr[r] < k) {
ans = Math.max(ans, arr[l] + arr[r]);
l++;
}
else {
r--;
}
}
return ans;
}
// Driver Code
public static void main (String[] args)
{
int []A = { 20, 10, 30, 100, 110 };
int k = 85;
// Function Call
System.out.print(maxSum(A, k));
}
}
// This code is contributed by Aman Kumar.
# Python program for above approach
# Function to find max sum less than k
def maxSum(arr, k):
# Sort the array
arr.sort()
n,l = len(arr),0
r,ans = n - 1,0
# While l is less than r
while (l < r):
if (arr[l] + arr[r] < k):
ans = max(ans, arr[l] + arr[r])
l += 1
else:
r -= 1
return ans
# Driver Code
A = [20, 10, 30, 100, 110]
k = 85
# Function Call
print(maxSum(A, k))
# This code is contributed by shinjanpatra
// C# implementation of the approach
using System;
class GFG
{
// Function to find max sum less than k
static int maxSum(int[] arr, int k)
{
// Sort the array
Array.Sort(arr);
int n = arr.Length, l = 0, r = n - 1, ans = 0;
// While l is less than r
while (l < r) {
if (arr[l] + arr[r] < k) {
ans = Math.Max(ans, arr[l] + arr[r]);
l++;
}
else {
r--;
}
}
return ans;
}
// Driver code
public static void Main ()
{
int[] A = { 20, 10, 30, 100, 110 };
int k = 85;
// Function Call
Console.WriteLine(maxSum(A, k));
}
}
// This code is contributed by target_2.
<script>
// Javascript program for the above approach
// Function to find max sum less than k
function maxSum(arr, k)
{
// Sort the array
arr.sort((a,b)=>a-b);
var n = arr.length, l = 0, r = n - 1, ans = 0;
// While l is less than r
while (l < r) {
if (arr[l] + arr[r] < k) {
ans = Math.max(ans, arr[l] + arr[r]);
l++;
}
else {
r--;
}
}
return ans;
}
// Driver Code
var A = [20, 10, 30, 100, 110];
var k = 85;
// Function Call
document.write( maxSum(A, k));
</script>
Time complexity: O(NlogN)
Auxiliary Space: O(1)
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