Minimum count of numbers needed from 1 to N that yields the sum as K Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given two positive integers N and K, the task is to print the minimum count of numbers needed to get the sum equal to K, adding any element from the first N natural numbers only once. If it is impossible to get the sum equal to K, then print "-1". Examples: Input: N = 5, K = 10Output: 3Explanation: The most optimal way is to choose number {1, 4, 5} to which will sum up to 10.Therefore, print 3 which is the minimum count of elements needed. Input: N = 5, K = 1000Output: -1Explanation: It is impossible to get the sum equal to 1000. Approach: The problem can be solved using the greedy algorithm. Follow the steps below to solve this problem: If K is greater than the sum of first N natural numbers, then the print "-1" and then return.If K is less than equal to N, then print 1 and then return.Otherwise, Initialize the variable say sum, and count as 0 to store the sum and minimum count of numbers needed.Iterate until N is greater than equal to 1 and sum is less than K and perform the following steps:Incrementing count by 1, sum by N, and then decrement the N by 1.Finally, if none of the above cases satisfy then print the count as the answer. Below is the implementation of the above approach: C++ // C++ program for the above approach #include <iostream> using namespace std; // Function to find minimum number of // elements required to obtain sum K int Minimum(int N, int K) { // Stores the maximum sum that // can be obtained int sum = N * (N + 1) / 2; // If K is greater than the // Maximum sum if (K > sum) return -1; // If K is less than or equal // to N if (K <= N) return 1; // Stores the sum sum = 0; // Stores the count of numbers // needed int count = 0; // Iterate until N is greater than // or equal to 1 and sum is less // than K while (N >= 1 && sum < K) { // Increment count by 1 count += 1; // Increment sum by N sum += N; // Update the sum N -= 1; } // Finally, return the count return count; } // Driver Code int main() { // Given Input int N = 5, K = 10; // Function Call cout << (Minimum(N, K)); return 0; } // This code is contributed by Potta Lokesh Java // Java program for the above approach import java.io.*; // Function to find minimum number of // elements required to obtain sum K class GFG { static int Minimum(int N, int K) { // Stores the maximum sum that // can be obtained int sum = N * (N + 1) / 2; // If K is greater than the // Maximum sum if (K > sum) return -1; // If K is less than or equal // to N if (K <= N) return 1; // Stores the sum sum = 0; // Stores the count of numbers // needed int count = 0; // Iterate until N is greater than // or equal to 1 and sum is less // than K while (N >= 1 && sum < K) { // Increment count by 1 count += 1; // Increment sum by N sum += N; // Update the sum N -= 1; } // Finally, return the count return count; } // Driver Code public static void main(String[] args) { // Given Input int N = 5, K = 10; // Function Call System.out.println(Minimum(N, K)); } } Python3 # Python3 program for the above approach # Function to find minimum number of # elements required to obtain sum K def Minimum(N, K): # Stores the maximum sum that # can be obtained sum = N * (N + 1) // 2 # If K is greater than the # Maximum sum if (K > sum): return -1 # If K is less than or equal # to N if (K <= N): return 1 # Stores the sum sum = 0 # Stores the count of numbers # needed count = 0 # Iterate until N is greater than # or equal to 1 and sum is less # than K while (N >= 1 and sum < K): # Increment count by 1 count += 1 # Increment sum by N sum += N # Update the sum N -= 1 # Finally, return the count return count # Driver Code if __name__ == '__main__': # Given Input N = 5 K = 10 # Function Call print(Minimum(N, K)) # This code is contributed by mohit kumar 29 C# // C# program for the above approach using System; // Function to find minimum number of // elements required to obtain sum K class GFG{ static int Minimum(int N, int K) { // Stores the maximum sum that // can be obtained int sum = N * (N + 1) / 2; // If K is greater than the // Maximum sum if (K > sum) return -1; // If K is less than or equal // to N if (K <= N) return 1; // Stores the sum sum = 0; // Stores the count of numbers // needed int count = 0; // Iterate until N is greater than // or equal to 1 and sum is less // than K while (N >= 1 && sum < K) { // Increment count by 1 count += 1; // Increment sum by N sum += N; // Update the sum N -= 1; } // Finally, return the count return count; } // Driver Code static public void Main() { // Given Input int N = 5, K = 10; // Function Call Console.Write(Minimum(N, K)); } } // This code is contributed by sanjoy_62 JavaScript <script> // JavaScript implementation of // the above approach function Minimum(N, K) { // Stores the maximum sum that // can be obtained let sum = N * (N + 1) / 2; // If K is greater than the // Maximum sum if (K > sum) return -1; // If K is less than or equal // to N if (K <= N) return 1; // Stores the sum sum = 0; // Stores the count of numbers // needed let count = 0; // Iterate until N is greater than // or equal to 1 and sum is less // than K while (N >= 1 && sum < K) { // Increment count by 1 count += 1; // Increment sum by N sum += N; // Update the sum N -= 1; } // Finally, return the count return count; } // Driver Code // Given Input let N = 5, K = 10; // Function Call document.write(Minimum(N, K)); </script> Output3 Time Complexity: O(N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms S shubhamharan69 Follow Improve Article Tags : Greedy Searching Mathematical DSA Practice Tags : GreedyMathematicalSearching Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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