# Remove two consecutive integers from 1 to N to make sum equal to S

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

• 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`
```Time complexity: O(n)
Space complexity: O(n)
```
• 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 ` `using` `namespace` `std; ` ` `  `// Function to find the numbers ` `// to be removed ` `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; ` `} ` ` `  `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 code ` `int` `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 S  ` ` `  `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``)  ` `            ``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 removed  ` `def` `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 code  ` `if` `__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 S  ` `using` `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  `

Output:

```Yes: 3, 4
No
```

Time complexity: O(1)
Space complexity: O(1)

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