Given two integers say a and b. Find the quotient after dividing a by b without using multiplication, division and mod operator.
Examples:
Input: a = 10, b = 3
Output: 3Input: a = 43, b = -8
Output: -5
This problem has been already discussed here. In this post, a different approach is discussed.
Approach :
- Let a/b = c.
- Take log on both sides
- log(a) – log(b) = log(c)
- Now Log of RHS can be written as exp in LHS
- Final formula is : exp(log(a) – log(b)) = c
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Returns the quotient of dividend/divisor.void Divide(int a, int b){ long long dividend = (long long)a; long long divisor = (long long)b; // Calculate sign of divisor i.e., // sign will be negative only if // either one of them is negative // otherwise it will be positive long long sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1; // Remove signs of dividend and divisor dividend = abs(dividend); divisor = abs(divisor); // Zero division Exception. if (divisor == 0) { cout << "Cannot Divide by 0" << endl; return; } if (dividend == 0) { cout << a << " / " << b << " is equal to : " << 0 << endl; return; } if (divisor == 1) { cout << a << " / " << b << " is equal to : " << sign * dividend << endl; return; } // Using Formula derived above. cout << a << " / " << b << " is equal to : " << sign * exp(log(dividend) - log(divisor)) << endl;}// Drivers codeint main(){ int a = 10, b = 5; Divide(a, b); a = 49, b = -7; Divide(a, b); return 0;} |
Java
// Java program for// above approachimport java.io.*;class GFG{static void Divide(int a, int b){ long dividend = (long)a; long divisor = (long)b; // Calculate sign of divisor i.e., // sign will be negative only if // either one of them is negative // otherwise it will be positive long sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1; // Remove signs of // dividend and divisor dividend = Math.abs(dividend); divisor = Math.abs(divisor); // Zero division Exception. if (divisor == 0) { System.out.println("Cannot Divide by 0"); return; } if (dividend == 0) { System.out.println(a + " / " + b + " is equal to : " + 0); return; } if (divisor == 1) { System.out.println(a + " / " + b + " is equal to : " + sign * dividend); return; } // Using Formula // derived above. System.out.println(a + " / " + b + " is equal to : " + Math.floor(sign * (Math.exp(Math.log(dividend) - Math.log(divisor)))));}// Driver codepublic static void main (String[] args) {int a = 10, b = 5;Divide(a, b);a = 49; b = -7;Divide(a, b);}}// This code is contributed// by shiv_bhakt. |
Python3
# Python3 program# for above approachimport mathdef Divide(a, b): dividend = a; divisor = b; # Calculate sign of divisor # i.e., sign will be negative # only if either one of them # is negative otherwise it # will be positive sign = -1 if ((dividend < 0) ^ (divisor < 0)) else 1; # Remove signs of # dividend and divisor dividend = abs(dividend); divisor = abs(divisor); # Zero division Exception. if (divisor == 0): print("Cannot Divide by 0"); if (dividend == 0): print(a, "/", b, "is equal to :", 0); if (divisor == 1): print(a, "/", b, "is equal to :", (sign * dividend)); # Using Formula # derived above. print(a, "/", b, "is equal to :", math.floor(sign * math.exp(math.log(dividend) - math.log(divisor))));# Driver codea = 10;b = 5;Divide(a, b);a = 49;b = -7;Divide(a, b);# This code is contributed# by mits |
C#
// C# program for// above approachusing System;class GFG{static void Divide(int a, int b){ long dividend = (long)a; long divisor = (long)b; // Calculate sign of divisor // i.e., sign will be negative // only if either one of them // is negative otherwise it // will be positive long sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1; // Remove signs of // dividend and divisor dividend = Math.Abs(dividend); divisor = Math.Abs(divisor); // Zero division Exception. if (divisor == 0) { Console.WriteLine("Cannot Divide by 0"); return; } if (dividend == 0) { Console.WriteLine(a + " / " + b + " is equal to : " + 0); return; } if (divisor == 1) { Console.WriteLine(a + " / " + b + " is equal to : " + sign * dividend); return; } // Using Formula // derived above. Console.WriteLine(a + " / " + b + " is equal to : " + Math.Floor(sign * (Math.Exp(Math.Log(dividend) - Math.Log(divisor)))));}// Driver codepublic static void Main () { int a = 10, b = 5; Divide(a, b); a = 49; b = -7; Divide(a, b);}}// This code is contributed// by shiv_bhakt. |
PHP
<?php// PHP program for above approachfunction Divide($a, $b){ $dividend = $a; $divisor = $b; // Calculate sign of divisor // i.e., sign will be negative // only if either one of them // is negative otherwise it // will be positive $sign = ($dividend < 0) ^ ($divisor < 0) ? -1 : 1; // Remove signs of // dividend and divisor $dividend = abs($dividend); $divisor = abs($divisor); // Zero division Exception. if ($divisor == 0) { echo "Cannot Divide by 0"; echo""; } if ($dividend == 0) { echo $a , " / " , $b , " is equal to : " , 0 ; echo ""; } if ($divisor == 1) { echo $a , " / " , $b , " is equal to : ", $sign * $dividend. "\n"; echo ""; } // Using Formula // derived above. echo $a , " / " , $b , " is equal to : " , $sign * exp(log($dividend) - log($divisor)). "\n"; echo "";}// Driver code$a = 10;$b = 5;Divide($a, $b);$a = 49;$b = -7;Divide($a, $b);// This code is contributed// by shiv_bhakt.?> |
Javascript
<script>// Javascript program for// above approach function Divide(a, b){ var dividend = a; var divisor = b; // Calculate sign of divisor i.e., // sign will be negative only if // either one of them is negative // otherwise it will be positive var sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1; // Remove signs of // dividend and divisor dividend = Math.abs(dividend); divisor = Math.abs(divisor); // Zero division Exception. if (divisor == 0) { document.write("Cannot Divide by 0"); return; } if (dividend == 0) { document.write(a + " / " + b + " is equal to : " + 0 + "<br>"); return; } if (divisor == 1) { System.out.println(a + " / " + b + " is equal to : " + sign * dividend + "<br>"); return; } // Using Formula // derived above. document.write(a + " / " + b + " is equal to : " + Math.floor(sign * (Math.exp(Math.log(dividend) - Math.log(divisor)))) + "<br>");}// Driver Codevar a = 10, b = 5;Divide(a, b);a = 49; b = -7;Divide(a, b); // This code is contributed by Kirti</script> |
Output:
10 / 5 is equal to : 2 49 / -7 is equal to : -7
Time complexity: O(1)
Auxiliary space: O(1)
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