int divideTwoInteger(int dividend, int divisor) {
// Write your code here.
int s = dividend / divisor;
return s;
}
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Divide Two Integers
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Problem statement
You are given two integers, ‘dividend’ and ‘divisor’.
You are required to divide the integers without using multiplication, division, and modular operators.
Your task is to return the quotient after dividing ‘dividend’ by ‘divisor’.
Note :
In cases where the dividend is not perfectly divisible by the divisor, you are required to return the integer value of the quotient which is closer to zero.
For example - If the answer is '8.34', we return the integer part, i.e., '8'. Also, If the answer is '-2.67', we return the integer part, i.e., '-2'.
Assume we're dealing with a system that can only store integers in the 32-bit signed integer range: [2^31, 2^31-1]. If the quotient is higher than 2^31 - 1, return 2^31 - 1; if it is less than -2^31, return -2^31.
For example :
If the answer is ‘9.333’, then return ‘9’, or if the answer is ‘-8.123’, then return ‘-8’.
Detailed explanation ( Input/output format, Notes, Images )
Input format:
The first input line contains two space-separated integers, ‘dividend’ and ‘divisor’.
The only line contains the result after division, following the above rules.
You do not need to print anything. It has already been taken care of. Just implement the given function.
Sample Input 1:
10 3
Sample Output 1:
3
Explanation of sample input 1:
Given ‘dividend = 10’ and ‘divisor = 3’
So the division is ‘10/3 = 3.3333……’.
Return only integer part ‘3’
Sample Input 2:
7 -3
Sample Output 2:
-2
Constraints:
-10^9 <= dividend, divisor <= 10^9
divisor != 0
Time limit: 1 second
Hint
Hint 1
Hint 2
Change multiplicative property as a subtraction.
Approaches (2)
Approach 1
Approach 2
Using Subtraction
We can subtract the divisor from the dividend until the dividend is greater than the divisor. The quotient will be equal to the number of total subtractions performed.
Below is the detailed algorithm:
- Store the ‘IS_DIVIDEND_NEGATIVE = false’ and ‘IS_DIVISOR_NEGATIVE = false’.
- Check if the ‘DIVIDEND’ and ‘DIVISOR’ are positive or negative and update the values of ‘IS_DIVIDEND_NEGATIVE’ and ‘IS_DIVISOR_NEGATIVE’.
- Iterate a while loop with ‘quotient = 0’
- if ( ‘DIVIDEND’ > divisor) then:
- (‘DIVIDEND’ = ‘DIVIDEND’ - ‘DIVISOR’) and (‘QUOTIENT’ = ‘QUOTIENT’ + 1).
- if ( ‘DIVIDEND’ > divisor) then:
- If both ‘IS_DIVIDEND_NEGATIVE’ and ‘IS_DIVISOR_NEGATIVE’ are of the same sign
- Return ‘QUOTIENT’.
- Else, we need to make ‘quotient’ negative
- Return ‘-1 * QUOTIENT’.
Time Complexity
0(N / M), where ‘N’ is ‘dividend’ and ‘M’ is ‘divisor’.
We are using a while loop, at max, it can iterate upto ‘dividend/divisor’ times.
Space Complexity
O(1).
Constant space is used.
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Divide Two Integers
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Interview problems
Easy python solution
def divideTwoInteger(a: int, b: int) -> int:
sign = 1
if a < 0 and b < 0:
sign = 1
elif a < 0 or b < 0:
sign = -1
a, b, quotient = abs(a), abs(b), 0
while a >= b:
a -= b
quotient += 1
return sign * quotient
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1 upvote
Interview problems
easy sol
only return statement I return d/s;
int divideTwoInteger(int d, int s) {
// Write your code here.
return d/s;
}
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0 upvotes
Interview problems
Easy approach
In this I used the basic logic like subtracting the number till it give less than 0 from that I will get the answer in easy way.
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0 upvotes
Interview problems
Optimal Solution PYT Code using Bit Manipulation | Time: O(log n * log n)
from typing import List
def divideTwoInteger(dividend: int, divisor: int) -> int:
if dividend==divisor:
return 1
sign=True
if (dividend>=0 and divisor<0) or (dividend<=0 and divisor>0):
sign=False
quo=0
dividend=abs(dividend)
divisor=abs(divisor)
while dividend>=divisor:
n=0
while (dividend>=(divisor<<(n+1))):
n+=1
quo+=(1<<n)
dividend-=(divisor<<n)
if quo==(1<<31) and sign:
return maxsize
if quo==(1<<31) and not sign:
return minsize
return quo if sign else -quo
51 views
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1 upvote
Interview problems
C++ Optimised Approach Using Binary Search
int divideTwoInteger(int dividend, int divisor) {
long int count = 0;
long int sum = 0;
while((sum + abs(divisor)) <= abs(dividend)){
sum += abs(divisor);
count++;
}
if((divisor > 0 && dividend > 0) || (divisor < 0 && dividend < 0))
return count;
else
return -count;
};
136 views
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0 upvotes
Interview problems
:)
int divideTwoInteger(int dividend, int divisor) {
// Handle negative numbers
int sign = 1;
if (dividend < 0) {
dividend = -1*dividend;
sign = (-1)*sign;
}
if (divisor < 0) {
divisor = -1*divisor;
sign = -1*sign;
}
int count = 0;
while (dividend >= divisor) {
dividend -= divisor;
count++;
}
return sign * count;
}
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0 upvotes
Interview problems
C++ Solution
int divideTwoInteger(int dividend, int divisor) {
int dd=abs(dividend);
int dr=abs(divisor);
int count=0;
while(dd>=dr){
dd-=dr;
count++;
}
if(divisor*dividend < 0) return (-count);
else return count;
}
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0 upvotes
Interview problems
Java Code
public class Solution{
public static int divideTwoInteger(int dividend, int divisor) {
// Write your code here.
int sign = ((dividend<0) ^ (divisor<0)) ? -1 : 1;
dividend = Math.abs(dividend);
divisor = Math.abs(divisor);
int res = 0;
while(dividend>=divisor){
int count = 0;
while(dividend >= (divisor<<count+1)){
count++;
}
res+= 1<<count;
dividend -= divisor<<(count);
}
return sign*res;
}
}
//Works on the logic of basic maths, that division means subs of a number from another multiple times. //Here we check that while divisor's multiple of 2 is less than the dividend. we inc the count, at last we add till what multiple of divisor is divisible. Later we subs the dividend with 3*multiple of 2.
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Interview problems
What they want is this 👇 ( without using long or / )
#include<bits/stdc++.h>
int divideTwoInteger(int dividend, int divisor) {
// Write your code here.
if (dividend == INT_MIN && divisor==-1) return INT_MAX;
if(dividend==divisor)return 1;
unsigned int n=abs(dividend),d=abs(divisor);
bool neg=false;
if(dividend <0 && divisor>=0)neg=true;
else if(dividend >=0 && divisor<0)neg=true;
unsigned int ans=0;
while(n>=d){
short count=0;
while(n > (d<<count+1)){
count++;
}
ans+=1<<count;
n-=(d*(1<<count));
}
if(neg)ans*=-1;
return ans;
}
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1 upvote
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