Given a sum S and an integer N, The task is to remove two consecutive integers from 1 to N to make sum equal to S. Examples:
Input: N = 4, S = 3 Output: Yes sum(1, 2, 3, 4) = 10, remove 3 and 4 from 1 to N now sum(1, 2) = 3 = S Hence by removing 3 and 4 we got sum = S Input: N = 5, S =3 Output: No Its not possible to remove two elements from 1 to N such that new sum is 3
- Method 1:
- Create an array with elements from 1 to N.
- Each time remove two consecutive elements and check the difference between sum of N natural numbers and sum of two removed elements is S.
- sum of N natural numbers can be calculated using formula
sum = (n)(n + 1) / 2
- Method 2:
- We know that, sum of N natural numbers can be calculated using formula
sum = (n)(n + 1) / 2
- According to the problem statement, we need to remove two integers from 1 to N such that the sum of remaining integers is S.
- Let the two consecutive integers to be removed be i and i + 1.
- Therefore,
required sum = S = (n)(n + 1) / 2 - (i) - (i + 1)
S = (n)(n + 1) / 2 - (2i - 1)
Therefore,
i = ((n)(n + 1) / 4) - ((S + 1) / 2)
- Hence find i using above formula
- if i is an integer, then answer is Yes else No
Below is the implementation of the above approach:
C++
// C++ program remove two consecutive integers// from 1 to N to make sum equal to S#include <bits/stdc++.h>using namespace std;// Function to find the numbers// to be removedfloat findNumber(int N, int S){ // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i;}void check(int N, int S){ float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) cout << "Yes: " << integerI << ", " << integerI + 1 << endl; else cout << "No" << endl;}// Driver codeint main(){ int N = 4, S = 3; check(N, S); N = 5, S = 3; check(N, S); return 0;} |
Java
// Java program remove two consecutive integers// from 1 to N to make sum equal to Sclass GFG{ // Function to find the numbers // to be removed static float findNumber(int N, int S) { // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i; } static void check(int N, int S) { float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) System.out.println("Yes: " + integerI + ", " + (integerI + 1)) ; else System.out.println("No"); } // Driver code public static void main (String[] args) { int N = 4, S = 3; check(N, S); N = 5; S = 3; check(N, S); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 program remove two consecutive integers# from 1 to N to make sum equal to S# Function to find the numbers# to be removeddef findNumber(N, S) : # typecast appropriately # so that answer is float i = (((N) * (N + 1)) / 4) - ((S + 1) / 2); # return the obtained result return i;def check(N, S) : i = findNumber(N, S); # Convert i to integer integerI = int(i); # If i is an integer is 0 # then answer is Yes if (i - integerI == 0) : print("Yes:", integerI, ",", integerI + 1); else : print("No");# Driver codeif __name__ == "__main__" : N = 4; S = 3; check(N, S); N = 5; S = 3; check(N, S);# This code is contributed by AnkitRai01 |
C#
// C# program remove two consecutive integers// from 1 to N to make sum equal to Susing System;class GFG{ // Function to find the numbers // to be removed static float findNumber(int N, int S) { // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i; } static void check(int N, int S) { float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) Console.WriteLine("Yes: " + integerI + ", " + (integerI + 1)) ; else Console.WriteLine("No"); } // Driver code public static void Main() { int N = 4, S = 3; check(N, S); N = 5; S = 3; check(N, S); }}// This code is contributed by AnkitRai01 |
Javascript
// JS program remove two consecutive integers// from 1 to N to make sum equal to S// Function to find the numbers// to be removedfunction findNumber(N, S){ // typecast appropriately // so that answer is float let i = (((N) * (N + 1)) / 4) - ((S + 1) / 2); // return the obtained result return i;}function check(N, S){ let i = findNumber(N, S); // Convert i to integer let integerI = parseInt(i); // If i is an integer is 0 // then answer is Yes if (i - integerI == 0){ let value = integerI + 1; console.log("Yes: " + integerI + "," + value); } else console.log("No")}// Driver codelet N = 4, S = 3;check(N, S);N = 5, S = 3;check(N, S);// This code is contributed by Yash Agarwal |
Output
Yes: 3, 4 No
Time Complexity: O(1)
Auxiliary Space: O(1)
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