Given an integer N (N ≠ 0), the task is to find a range [L, R] (−10⁻¹⁸ < L < R < 10¹⁸) such that sum of all integers in this range is equal to N.
L + (L+1) + … + (R−1) + R = N
Examples:
Input : N = 3
Output: -2 3
Explanation: For L = -2 and R = -3 the sum becomes -2 + (-1) + 0 + 1 + 2 + 3 = 3Input : N = -6
Output: -6 5
Explanation: The sum for this range [-6, 5] is -6 + (-5) + (-4) + (-3) + (-2) + (-1) + 0 + 1+ 2 + 3 + 4 + 5 = -6
Naive Approach: For every value of L try to find a value R which satisfies the condition L + (L+1) + . . . + (R-1) + R = N, using nested loop.
Time Complexity: O(N2)
Auxiliary space: O(1)
Efficient Approach: Since L and R are integers and can be negative numbers as well, the above problem can be solved in O(1) efficiently. Consider the below observation:
- For N being a positive integer we can consider:
[−(N – 1)] + [−(N – 2)] + . . . -1 + 0 + 1 + . . . + (N − 1) + N =
-(N – 1) + (N – 1) – (N – 2) + (N – 2) + . . . + 1 – 1 + 0 + N = N
So, L = -(N – 1) and R = N
- Similarly for N being a negative, we can consider:
N + (N + 1) + . . . -1 + 0 + 1 + . . . + [-(N + 2)] + [-(N + 1)] =
(N + 1) – (N + 1) + (N + 2) – (N + 2) + . . . -1 + 1 + 0 + N = N
So L = N and R = -(N + 1)
Therefore, the solution to this problem in unit time complexity is:
L = -(N – 1) and R = N, when N is a positive integer.
L = N and R = -(N + 1), when N is a negative integer.
Note: This is the longest possible range, (i.e. R – L has the highest value) which satisfies the problem requirement.
Below is the implementation of the approach:
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Function to find two integersvoid Find_Two_Intergers(long long int N){ // Variable to store value of L and R long long int L, R; // When N is positive if (N > 0) { L = -(N - 1); R = N; } // When N is negative else { L = N; R = -(N + 1); } cout << L << " " << R;}// Driver Codeint main(){ long long int N = 3; Find_Two_Integers(N); return 0;} |
C
// C code to implement above approach#include <stdio.h>// Function to find two integersvoid Find_Two_Intergers(long long int N){ // Variable to store L and R long long int L, R; // When N is positive if (N > 0) { L = -(N - 1); R = N; } // When N is negative else { L = N; R = -(N + 1); } printf("%lld %lld", L, R);}// Driver codeint main(){ long long int N = 3; Find_Two_Integers(N); return 0;} |
Java
// Java code for the above approachimport java.io.*;class GFG{ // Function to find two integers static void Find_Two_Intergers(long N) { // Variable to store value of L and R long L, R; // When N is positive if (N > 0) { L = -(N - 1); R = N; } // When N is negative else { L = N; R = -(N + 1); } System.out.print( L + " " + R); } // Driver Code public static void main (String[] args) { long N = 3; Find_Two_Integers(N); }}// This code is contributed by Potta Lokesh |
Python3
# Python code to implement above approach# Function to find two integersdef Find_Two_Intergers(N): # variable to store L and R L = 0 R = 0 # When N is positive if N > 0: L = -(N-1) R = N # When N is negative else: L = N R = -(N+1) print(L, R)# Driver codeN = 3Find_Two_Integers(N) |
C#
// C# code for the above approachusing System;class GFG{ // Function to find two integers static void Find_Two_Intergers(long N) { // Variable to store value of L and R long L, R; // When N is positive if (N > 0) { L = -(N - 1); R = N; } // When N is negative else { L = N; R = -(N + 1); } Console.Write( L + " " + R); } // Driver Code public static void Main (String[] args) { long N = 3; Find_Two_Integers(N); }}// This code is contributed by Saurabh Jaiswal |
Javascript
<script>// Javascript code to implement above approach// Function to find two integersfunction Find_Two_Intergers(N) { // Variable to store value of L and R let L, R; // When N is positive if (N > 0) { L = -(N - 1); R = N; } // When N is negative else { L = N; R = -(N + 1); } document.write(L + " " + R);}// Driver Codelet N = 3;Find_Two_Integers(N);// This code is contributed by gfgking.</script> |
-2 3
Time Complexity: O(1)
Auxiliary Space: O(1)
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