## Interview Question for Software Engineers

Country: United States
Interview Type: Phone Interview

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0
of 0 vote

We do following

``````public static int BinaryGap(int n)
{
int maxGap = 0;
int currentGap = 0;
while(n > 0)
{
if(n % 2 == 0)
{
currentGap++;
if(maxGap < currentGap)
{
maxGap = currentGap;
}
}
else if(n % 2 == 1)
{
currentGap = 0;
}
n = n/2;
}
return maxGap;
}``````

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0

it wont work if you pass 20 rupees

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0
of 0 vote

In a loop, divide the number by 2. Keep count of the number of times, n%2==0 is found continuously

``````public int maxGap(int n){
if(n<0){ return -1;}
int maxgap = 0;
int gap=0;
while(n>0){
if(n%2==0){
gap++;
if(gap>maxgap){ maxgap = gap; }
} else {
gap=0;
}
}
}``````

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0
of 0 vote

ujyujy

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0
of 0 vote

public int solution(int N) {
String binary = Integer.toBinaryString(N);
int compte = 0, comptegap = 0, result = 0;
char[] value = binary.toCharArray();

System.out.println(value);

for (char v : value) {
if (v == '1') {
compte++;
}

if (v == '0') {

if (compte == 2) {

if (result < comptegap) {
result = comptegap;
}

comptegap = 0;
} else {
result = 0;
}

comptegap++;
}
}

return result;
}

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0
of 0 vote

``````public int solution(int N) {
// write your code in Java SE 8
String binary = Integer.toBinaryString(N);
int compte = 0, comptegap = 0, result = 0;
char[] value = binary.toCharArray();

System.out.println(value);

for (char v : value) {
if (v == '1') {
compte++;
}

if (v == '0') {

if (compte == 2) {

if (result < comptegap) {
result = comptegap;
}

comptegap = 0;
} else {
result = 0;
}

comptegap++;
}
}

return result;
}``````

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