Number expressed as sum of five consecutive integers
Given an integer n, the task is to find whether n can be expressed as sum of five consecutive integer. If yes, find the five consecutive integers, else print “-1”.
Examples:
Input : n = 15 Output : 1 2 3 4 5 15 = 1 + 2 + 3 + 4 + 5 Input : n = 18 Output : -1
Method 1: (Brute Force)
The idea is to run a loop from i = 0 to n – 4, check if (i + i+1 + i+2 + i+3 + i+4) is equal to n. Also, check if n is positive or negative and accordingly increment or decrement i by 1.
Below is the implementation of this approach:
C++
// CPP Program to check if a number can // be expressed as sum of five consecutive // integers. #include <bits/stdc++.h> using namespace std; // function to check if a number can be expressed as // sum of five consecutive integers. void checksum( int n) { // if n is 0 if (n == 0) { cout << "-2 -1 0 1 2" << endl; return ; } int inc; // if n is positive, increment loop by 1. if (n > 0) inc = 1; // if n is negative, decrement loop by 1. else inc = -1; // Running loop from 0 to n - 4 for ( int i = 0; i <= n - 4; i += inc) { // check if sum of five consecutive // integer is equal to n. if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) { cout << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << " " << i + 4; return ; } } cout << "-1" ; } // Driver Program int main() { int n = 15; checksum(n); return 0; } |
Java
// Java Program to check if a number can // be expressed as sum of five consecutive // integers. import java.io.*; class GFG { // function to check if a number can be // expressed as sum of five consecutive // integers. static void checksum( int n) { // if n is 0 if (n == 0 ) { System.out.println( "-2 -1 0 1 2" ); return ; } int inc; // if n is positive, increment loop by 1. if (n > 0 ) inc = 1 ; // if n is negative, decrement loop by 1. else inc = - 1 ; // Running loop from 0 to n - 4 for ( int i = 0 ; i <= n - 4 ; i += inc) { // check if sum of five consecutive // integer is equal to n. if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) { System.out.print( (i ) + " " + (i + 1 ) + " " + (i + 2 ) + " " + (i + 3 ) + " " + (i + 4 )); return ; } } System.out.println( "-1" ); } // Driver Program public static void main (String[] args) { int n = 15 ; checksum(n); } } // This code is contributed by anuj_67 |
Python3
# Python3 code to check if a number # can be expressed as sum of five # consecutive integers. # function to check if a number # can be expressed as sum of five # consecutive integer. def checksum(n): # if n is 0 if n = = 0 : print ( "-2 -1 0 1 2" ) return 0 inc = 0 # if n is positive, # increment loop by 1. if n > 0 : inc = 1 # if n is negative, # decrement loop by 1. else : inc = - 1 # Running loop from 0 to n - 4 for i in range ( 0 , n - 3 , inc): # check if sum of five consecutive # integer is equal to n. if i + i + 1 + i + 2 + i + 3 + i + 4 = = n: print (i, " " , i + 1 , " " , i + 2 , " " , i + 3 , " " , i + 4 ) return 0 print ( "-1" ) # Driver Code n = 15 checksum(n) |
C#
// C# Program to check if a number can // be expressed as sum of five consecutive // integers. using System; class GFG { // function to check if a number can be // expressed as sum of five consecutive // integers. static void checksum( int n) { // if n is 0 if (n == 0) { Console.Write( "-2 -1 0 1 2" ); return ; } int inc; // if n is positive, increment loop by 1. if (n > 0) inc = 1; // if n is negative, decrement loop by 1. else inc = -1; // Running loop from 0 to n - 4 for ( int i = 0; i <= n - 4; i += inc) { // check if sum of five consecutive // integer is equal to n. if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) { Console.Write( (i ) + " " + (i + 1) + " " + (i + 2) + " " + (i + 3) + " " + (i + 4)); return ; } } Console.WriteLine( "-1" ); } // Driver Program public static void Main () { int n = 15; checksum(n); } } // This code is contributed by anuj_67 |
PHP
<?php // PHP Program to check // if a number can be // expressed as sum of // five consecutive integers. // function to check if a // number can be expressed // as sum of five consecutive // integers. function checksum( $n ) { // if n is 0 if ( $n == 0) { echo "-2 -1 0 1 2" , "\n" ; return ; } $inc ; // if n is positive, // increment loop by 1. if ( $n > 0) $inc = 1; // if n is negative, // decrement loop by 1. else $inc = -1; // Running loop from // 0 to n - 4 for ( $i = 0; $i <= $n - 4; $i += $inc ) { // check if sum of five // consecutive integer // is equal to n. if ( $i + $i + 1 + $i + 2 + $i + 3 + $i + 4 == $n ) { echo $i , " " , $i + 1, " " , $i + 2, " " , $i + 3, " " , $i + 4; return ; } } echo "-1" ; } // Driver Code $n = 15; checksum( $n ); // This code is contributed // by ajit ?> |
Javascript
<script> // JavaScript program to check if a number can // be expressed as sum of five consecutive // integers. // function to check if a number can be // expressed as sum of five consecutive // integers. function checksum(n) { // if n is 0 if (n == 0) { document.write( "-2 -1 0 1 2" ); return ; } var inc; // if n is positive, increment loop by 1. if (n > 0) inc = 1; // if n is negative, decrement loop by 1. else inc = -1; // Running loop from 0 to n - 4 for (i = 0; i <= n - 4; i += inc) { // check if sum of five consecutive // integer is equal to n. if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) { document.write((i) + " " + (i + 1) + " " + (i + 2) + " " + (i + 3) + " " + (i + 4)); return ; } } document.write( "-1" ); } // Driver Program var n = 15; checksum(n); // This code contributed by gauravrajput1 </script> |
Output :
1 2 3 4 5
Method 2: (Efficient Approach)
The idea is to check if n is multiple of 5 or not.
Let n is sum of five consecutive integer of k – 2, k-1, k, k + 1, k+2. Therefore,
k-2 + k-1 + k + k+1 + k+2= n
5*k = n
The five numbers will be n/5 – 2, n/5 – 1, n/5, n/5 + 1, n/5 + 2.
Below is the implementation of this approach:
C++
// CPP Program to check if a number can be // expressed as sum of five consecutive integer. #include <bits/stdc++.h> using namespace std; // function to check if a number can be // expressed as sum of five consecutive // integers. void checksum( int n) { // if n is multiple of 5 if (n % 5 == 0) cout << n / 5 - 2 << " " << n / 5 - 1 << " " << n / 5 << " " << n / 5 + 1 << " " << n / 5 + 2; // else print "-1". else cout << "-1" ; } // Driver Program int main() { int n = 15; checksum(n); return 0; } |
Java
// Java Program to check if a number can // be expressed as sum of five consecutive // integer. import java.io.*; class GFG { // function to check if a number can // be expressed as sum of five // consecutive integers. static void checksum( int n) { // if n is multiple of 5 if (n % 5 == 0 ) System.out.println( (n / 5 - 2 ) + " " + (n / 5 - 1 ) + " " + (n / 5 ) + " " + (n / 5 + 1 ) + " " + (n / 5 + 2 )); // else print "-1". else System.out.println( "-1" ); } // Driver Program public static void main (String[] args) { int n = 15 ; checksum(n); } } // This code is contributed by vt_m. |
Python3
# Python3 code to check if a number # can be expressed as sum of five # consecutive integer. # function to check if a number # can be expressed as sum of five # consecutive integers. def checksum(n): n = int (n) # if n is multiple of 5 if n % 5 = = 0 : print ( int (n / 5 - 2 ), " " , int (n / 5 - 1 ), " " , int (n / 5 ), " " , int (n / 5 + 1 ), " " , int (n / 5 + 2 )) # else print "-1". else : print ( "-1" ) # Driver Code n = 15 checksum(n) |
C#
// C# Program to check if a number can // be expressed as sum of five consecutive // integer. using System; class GFG { // function to check if a number can // be expressed as sum of five // consecutive integers. static void checksum( int n) { // if n is multiple of 5 if (n % 5 == 0) Console.WriteLine( (n / 5 - 2) + " " + (n / 5 - 1) + " " + (n / 5) + " " + (n / 5 + 1 ) + " " + (n / 5 + 2)); // else print "-1". else Console.WriteLine( "-1" ); } // Driver Program public static void Main () { int n = 15; checksum(n); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP Program to check if // a number can be expressed // as sum of five consecutive // integer. // function to check if a // number can be expressed // as sum of five consecutive // integers. function checksum( $n ) { // if n is multiple of 5 if ( $n % 5 == 0) echo $n / 5 - 2 , " " , $n / 5 - 1 , " " , $n / 5 , " " , $n / 5 + 1 , " " , $n / 5 + 2; // else print "-1". else echo "-1" ; } // Driver Code $n = 15; checksum( $n ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // javascript Program to check if a number can // be expressed as sum of five consecutive // integer. // function to check if a number can // be expressed as sum of five // consecutive integers. function checksum(n) { // if n is multiple of 5 if (n % 5 == 0) document.write((n / 5 - 2) + " " + (n / 5 - 1) + " " + (n / 5) + " " + (n / 5 + 1) + " " + (n / 5 + 2)); // else print "-1". else document.write( "-1" ); } // Driver Program var n = 15; checksum(n); // This code is contributed by todaysgaurav </script> |
Output:
1 2 3 4 5
Time Complexity : O(1)
Auxilitary Space Complexity : O(1)
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