Given an integer N. Find the maximum length sequence containing distinct even integers with sum N. In case, such a sequence does not exist, return an empty sequence.
Examples:
Input: N = 12
Output : {2, 6, 4}
Explanation: 2, 6 and 4 are even and 2+6+4=12. Note that {4, 8} sequence also contains even numbers whose sum is 12 . But we need the sequence of maximum length.Input: N = 25
Output: {}
Explanation: No sequence of distinct positive even integers is possible so that they add up to 25.
Approach: This problem can be solved using the greedy approach.
Start inserting even numbers from 2 into a list with a sum less than or equal to N. If a subsequence of even numbers with a sum equal to N is not possible simply add the remaining difference after all iterations to the last number in the list.
The following approach can be followed to solve the problem:
- If N is odd return the empty vector, as an odd number cannot be represented as the sum of even numbers.
- Initialize a variable (say sum) by 0, to store the sum of elements in a sequence, and a variable (say curr), to store the current even number.
- Until the value of sum is less than N, iteratively keep pushing values of curr in the answer and keep adding these values to sum. Also, keep incrementing the value of curr by 2 at each iteration.
- In end, add the value of remainder (N-sum) to the last element of the sequence.
Below is the implementation of the above approach:
C++
// C++ program to find Maximum length// sequence containing distinct// even integers with sum N#include <bits/stdc++.h>using namespace std;
// Function to find maximum// length sequence of distinct// positive even integers// that adds up to Nvector<int> maxSequence(int N)
{ // Initializing a vector
// to store the sequence
vector<int> v;
// Initializing sum and curr
// that stores the
// sum of sequence and current
// even integer respectively
int curr = 2, sum = 0;
// If N is odd, just return
// the empty vector
// as the required sequence cannot
// be deduced for an odd integer
if (N % 2 == 1) {
return v;
}
// While sum of sequence remains
// less than or equal to N, we keep
// inserting the value of curr in
// vector v as well as adding into sum
// also, increment curr by 2 at each
// iteration to get next even number
while (sum + curr <= N) {
v.push_back(curr);
sum += curr;
curr += 2;
}
int sz = v.size();
// We add the difference of N and
// sum to the last digit of sequence
v[sz - 1] += (N - sum);
// Returning the sequence
return v;
}// Driver Codeint main()
{ int N = 12;
vector<int> max_sequence = maxSequence(N);
// Printing the required sequence
for (int i = 0; i < max_sequence.size(); i++) {
cout << max_sequence[i] << " ";
}
} |
Java
// Java program to find Maximum length// sequence containing distinct// even integers with sum Nimport java.io.*;
import java.util.*;
class GFG
{ // Function to find maximum
// length sequence of distinct
// positive even integers
// that adds up to N
public static ArrayList<Integer> maxSequence(int N)
{
// Initializing a arraylist
// to store the sequence
ArrayList<Integer> v = new ArrayList<Integer>();
// Initializing sum and curr
// that stores the
// sum of sequence and current
// even integer respectively
int curr = 2, sum = 0;
// If N is odd, just return
// the empty arraylist
// as the required sequence cannot
// be deduced for an odd integer
if (N % 2 == 1) {
return v;
}
// While sum of sequence remains
// less than or equal to N, we keep
// inserting the value of curr in
// arraylist v as well as adding into sum
// also, increment curr by 2 at each
// iteration to get next even number
while (sum + curr <= N) {
v.add(curr);
sum += curr;
curr += 2;
}
int sz = v.size();
// We add the difference of N and
// sum to the last digit of sequence
v.set(sz - 1,v.get(sz-1)+(N-sum));
// Returning the sequence
return v;
}
public static void main(String[] args)
{
int N = 12;
ArrayList<Integer> max_sequence = maxSequence(N);
// Printing the required sequence
for (int i = 0; i < max_sequence.size(); i++) {
System.out.print(max_sequence.get(i) + " ");
}
}
}// This code is contributed by Rohit Pradhan |
Python3
# Python3 program to find Maximum length# sequence containing distinct# even integers with sum N# Function to find maximum# length sequence of distinct# positive even integers# that adds up to Ndef maxSequence(N):
# Initializing a vector
# to store the sequence
v = []
# Initializing sum and curr
# that stores the
# sum of sequence and current
# even integer respectively
curr, sum = 2, 0
# If N is odd, just return
# the empty vector
# as the required sequence cannot
# be deduced for an odd integer
if (N % 2 == 1):
return v
# While sum of sequence remains
# less than or equal to N, we keep
# inserting the value of curr in
# vector v as well as adding into sum
# also, increment curr by 2 at each
# iteration to get next even number
while (sum + curr <= N):
v.append(curr)
sum += curr
curr += 2
sz = len(v)
# We add the difference of N and
# sum to the last digit of sequence
v[sz - 1] += (N - sum)
# Returning the sequence
return v
# Driver Codeif __name__ == "__main__":
N = 12
max_sequence = maxSequence(N)
# Printing the required sequence
for i in range(0, len(max_sequence)):
print(max_sequence[i], end=" ")
# This code is contributed by rakeshsahni
|
C#
// C# program to find Maximum length// sequence containing distinct// even integers with sum Nusing System;
using System.Collections;
using System.Collections.Generic;
public class GFG{
// Function to find maximum
// length sequence of distinct
// positive even integers
// that adds up to N
public static ArrayList maxSequence(int N)
{
// Initializing a arraylist
// to store the sequence
ArrayList v = new ArrayList();
// Initializing sum and curr
// that stores the
// sum of sequence and current
// even integer respectively
int curr = 2, sum = 0;
// If N is odd, just return
// the empty arraylist
// as the required sequence cannot
// be deduced for an odd integer
if (N % 2 == 1) {
return v;
}
// While sum of sequence remains
// less than or equal to N, we keep
// inserting the value of curr in
// arraylist v as well as adding into sum
// also, increment curr by 2 at each
// iteration to get next even number
while (sum + curr <= N) {
v.Add(curr);
sum += curr;
curr += 2;
}
int sz = v.Count;
// We add the difference of N and
// sum to the last digit of sequence
int temp = sz - 1;
v[temp]= (int)v[temp] + N - sum;
// Returning the sequence
return v;
}
static public void Main (){
int N = 12;
ArrayList max_sequence = maxSequence(N);
// Printing the required sequence
foreach (int i in max_sequence) {
Console.Write(i + " ");
}
}
}// This code is contributed by hrithikgarg03188. |
Javascript
<script> // JavaScript code for the above approach
// Function to find maximum
// length sequence of distinct
// positive even integers
// that adds up to N
function maxSequence(N)
{
// Initializing a vector
// to store the sequence
let v = [];
// Initializing sum and curr
// that stores the
// sum of sequence and current
// even integer respectively
let curr = 2, sum = 0;
// If N is odd, just return
// the empty vector
// as the required sequence cannot
// be deduced for an odd integer
if (N % 2 == 1) {
return v;
}
// While sum of sequence remains
// less than or equal to N, we keep
// inserting the value of curr in
// vector v as well as adding into sum
// also, increment curr by 2 at each
// iteration to get next even number
while (sum + curr <= N) {
v.push(curr);
sum += curr;
curr += 2;
}
let sz = v.length;
// We add the difference of N and
// sum to the last digit of sequence
v[sz - 1] += (N - sum);
// Returning the sequence
return v;
}
// Driver Code
let N = 12;
let max_sequence = maxSequence(N);
// Printing the required sequence
for (let i = 0; i < max_sequence.length; i++) {
document.write(max_sequence[i] + " ");
}
// This code is contributed by Potta Lokesh
</script>
|
Output
2 4 6
Time Complexity: O(N)
Auxiliary Space: O(N)