Given two integers N and K, the task is to find K distinct positive odd integers such that their sum is equal to the given number N.
Examples:
Input: N = 10, K = 2
Output: 1 9
Explanation:
Two odd positive integers such that their sum is 10 can be (1, 9) or (3, 7).
Input: N = 10, K = 4
Output: NO
Explanation:
There does not exists four odd positive integers with sum 10.
Approach:
The number N can be represented as the sum of K positive odd integers only is the following two conditions satisfies:
- If the square of K is less than or equal to N and,
- If the sum of N and K is an even number.
If these conditions are satisfied then there exist K positive odd integers whose sum is N.
To generate K such odd numbers:
- Print first K-1 odd numbers starting from 1, i.e. 1, 3, 5, 7, 9…….
- The last odd number will be : N – (Sum of first K-1 odd positive integers)
Below is the implementation of the above approach:
C++
// C++ implementation to find K // odd positive integers such that// their sum is equal to given number#include <bits/stdc++.h>using namespace std;#define ll long long int// Function to find K odd positive// integers such that their sum is Nvoid findDistinctOddSum(ll n, ll k){ // Condition to check if there // are enough values to check if ((k * k) <= n && (n + k) % 2 == 0){ int val = 1; int sum = 0; for(int i = 1; i < k; i++){ cout << val << " "; sum += val; val += 2; } cout << n - sum << endl; } else cout << "NO \n";}// Driver Codeint main(){ ll n = 100; ll k = 4; findDistinctOddSum(n, k); return 0;} |
Java
// Java implementation to find K // odd positive integers such that// their sum is equal to given numberimport java.util.*;class GFG{ // Function to find K odd positive// integers such that their sum is Nstatic void findDistinctOddSum(int n, int k){ // Condition to check if there // are enough values to check if ((k * k) <= n && (n + k) % 2 == 0){ int val = 1; int sum = 0; for(int i = 1; i < k; i++){ System.out.print(val+ " "); sum += val; val += 2; } System.out.print(n - sum +"\n"); } else System.out.print("NO \n");} // Driver Codepublic static void main(String[] args){ int n = 100; int k = 4; findDistinctOddSum(n, k);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation to find K # odd positive integers such that# their sum is equal to given number# Function to find K odd positive# integers such that their summ is Ndef findDistinctOddsumm(n, k): # Condition to check if there # are enough values to check if ((k * k) <= n and (n + k) % 2 == 0): val = 1 summ = 0 for i in range(1, k): print(val, end = " ") summ += val val += 2 print(n - summ) else: print("NO")# Driver Coden = 100k = 4findDistinctOddsumm(n, k)# This code is contributed by shubhamsingh10 |
C#
// C# implementation to find K // odd positive integers such that// their sum is equal to given numberusing System;public class GFG{ // Function to find K odd positive// integers such that their sum is Nstatic void findDistinctOddSum(int n, int k){ // Condition to check if there // are enough values to check if ((k * k) <= n && (n + k) % 2 == 0){ int val = 1; int sum = 0; for(int i = 1; i < k; i++){ Console.Write(val+ " "); sum += val; val += 2; } Console.Write(n - sum +"\n"); } else Console.Write("NO \n");} // Driver Codepublic static void Main(String[] args){ int n = 100; int k = 4; findDistinctOddSum(n, k);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Function to find K odd positive// integers such that their sum is Nfunction findDistinctOddSum(n,k){ // Condition to check if there // are enough values to check if ((k * k) <= n && (n + k) % 2 == 0){ var val = 1; var sum = 0; for(var i = 1; i < k; i++){ document.write( val+ " "); sum += val; val += 2; } document.write( n - sum) ; } else document.write( "NO \n");}// Driver Codevar n = 100; var k = 4; findDistinctOddSum(n, k);</script> |
Output
1 3 5 91
Performance Analysis:
- Time Complexity: In the above-given approach, there is one loop which takes O(K) time in the worst case. Therefore, the time complexity for this approach will be O(K).
- Auxiliary Space: O(1)
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