Given two integers N and K, the task is to find any K distinct odd integers such that their sum is equal to N. If no such integers exists, print -1.
Examples:
Input: N = 10, K = 2
Output: 1, 9
Explanation:
There are two possible distinct odd integers, such that their sum is equal to N.
Possible K integers can be – {(1, 9), (3, 7)}Input: N = 5, K = 4
Output: -1
Explanation:
There are no such 4 distinct odd integers such that their sum is 5.
Approach:
- The key observation in this problem is if N and K have different parity then it is not possible to find K such distinct integers such that their sum is equal to N,
- Otherwise such K – 1 integers will consist of first K-1 odd positive integers
- The K^{th} odd number will be equal to (N – the sum of first (K-1) odd integers)
K^{th} Odd number = N - sum of first K-1 integer
Below is the implementation of the above approach:
// C++ implementation to find k // odd integers such that their sum is N #include <bits/stdc++.h> using namespace std;
// Function to find K odd integers // such that their sum is N void oddIntegers( int n, int k)
{ // Condition to check if there
// exist such K integers
if (n % 2 != k % 2) {
cout << "-1"
<< "\n" ;
return ;
}
int sum = 0;
int i = 1;
int j = 1;
// Loop to find first K-1
// distinct odd integers
while (j < k) {
sum = sum + i;
cout << i << " " ;
i = i + 2;
j++;
}
// Final Kth odd number
int finalOdd = n - sum;
cout << finalOdd << "\n" ;
} // Driver code int main()
{ int n = 10;
int k = 2;
oddIntegers(n, k);
return 0;
} |
// Java implementation to find k // odd integers such that their sum is N class GFG
{ // Function to find K odd integers // such that their sum is N static void oddIntegers( int n, int k)
{ // Condition to check if there
// exist such K integers
if (n % 2 != k % 2 ) {
System.out.println( "-1" );
return ;
}
int sum = 0 ;
int i = 1 ;
int j = 1 ;
// Loop to find first K-1
// distinct odd integers
while (j < k) {
sum = sum + i;
System.out.print(i+ " " );
i = i + 2 ;
j++;
}
// Final Kth odd number
int finalOdd = n - sum;
System.out.println(finalOdd);
} // Driver code public static void main (String[] args)
{ int n = 10 ;
int k = 2 ;
oddIntegers(n, k);
} } // This code is contributed by shubhamsingh |
# Python3 implementation to find k # odd integers such that their sum is N # Function to find K odd integers # such that their sum is N def oddIntegers(n, k) :
# Condition to check if there
# exist such K integers
if (n % 2 ! = k % 2 ) :
print ( "-1" );
return ;
sum = 0 ;
i = 1 ;
j = 1 ;
# Loop to find first K-1
# distinct odd integers
while (j < k) :
sum + = i;
print (i,end = " " );
i + = 2 ;
j + = 1 ;
# Final Kth odd number
finalOdd = n - sum ;
print (finalOdd);
# Driver code if __name__ = = "__main__" :
n = 10 ;
k = 2 ;
oddIntegers(n, k);
# This code is contributed by AnkitRai01 |
// C# implementation to find k // odd integers such that their sum is N using System;
class GFG
{ // Function to find K odd integers // such that their sum is N static void oddints( int n, int k)
{ // Condition to check if there
// exist such K integers
if (n % 2 != k % 2) {
Console.WriteLine( "-1" );
return ;
}
int sum = 0;
int i = 1;
int j = 1;
// Loop to find first K-1
// distinct odd integers
while (j < k) {
sum = sum + i;
Console.Write(i+ " " );
i = i + 2;
j++;
}
// Final Kth odd number
int finalOdd = n - sum;
Console.WriteLine(finalOdd);
} // Driver code public static void Main(String[] args)
{ int n = 10;
int k = 2;
oddints(n, k);
} } // This code is contributed by PrinciRaj1992 |
1 9
Performance Analysis:
- Time Complexity: As in the above approach, There is a loop to find such K odd integers which takes O(K) time in worst case. Hence the Time Complexity will be O(K).
- Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Possible values of Q such that, for any value of R, their product is equal to X times their sum
- Find the first N integers such that the sum of their digits is equal to 10
- Find three integers less than or equal to N such that their LCM is maximum
- Count of pairs in a given range with sum of their product and sum equal to their concatenated number
- Find K distinct positive odd integers with sum N
- Find K consecutive integers such that their sum is N
- Count pairs from 1 to N such that their Sum is divisible by their XOR
- Find two distinct numbers such that their LCM lies in given range
- Choose two elements from the given array such that their sum is not present in any of the arrays
- Find the pair (a, b) with minimum LCM such that their sum is equal to N
- Check if the sum of distinct digits of two integers are equal
- Find the number of positive integers less than or equal to N that have an odd number of digits
- Count of integers in a range which have even number of odd digits and odd number of even digits
- Minimum number of primes required such that their sum is equal to N
- Find two numbers such that difference of their squares equal to N
- Find K numbers with sum equal to N and sum of their squares maximized
- Find N distinct integers with zero sum
- Find N distinct integers with sum N
- Sum of product of all integers upto N with their count of divisors
- Distinct powers of a number N such that the sum is equal to K
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.