CareerCup

int i = 0,j=9;
while(i<j){

if(a[i]==0 && a[j]==1){
a[j]=0;
a[i] =1;
i++;
j--;
}else{
if(a[i]==0)
j--;
else
i++;
}
}

int i = p = 0, j= q = 9;
int b[10];
while(i < j) {
if(a[i] == 0){
     b[q] = 0;
     q--;
} else {
     b[p] = 1;
     p++;
}
if(a[j] == 0){
     b[q] = 0;
     q--;
} else {
     b[p] = 1;
     p++;
}
i++;
j--;
}

<pre lang="" line="1" title="CodeMonkey5559" class="run-this">public void seregate(int[] arr) {
for (int i = 0, j = arr.length - 1; i < j;) {
if (arr[i] == 1) {
++i;
continue;
}
if (arr[j] == 0) {
--j;
continue;
}
arr[i++] = 1;
arr[j--] = 0;
}
}
</pre>

public void seregate(int[] arr) {
for (int i = 0, j = arr.length - 1; i < j;) {
while (arr[i] == 1) {
++i;
}
while (arr[j] == 0) {
--j;
}
if (i < j) {
arr[i++] = 1;
arr[j--] = 0;
}
}
}

public int[] ar (String a[]) {
if(a == null)
return null;
if(a.length == 1)
return a;

int sum = 0;
for(int i = 0; i < a.length; i++)
sum =+ a[i]; a[i] = 0;

for(int i = 0; i < sum; i++)
a[i] = 1;

return a;


}

int a[10] = {0,1,1,1,0,1,0,0,1,0};
int y = 0;
int l = 4;
for(int i = 0, int n = 9;i < n;)
{
cout << "Execution " << y++ << endl;
if((a[i] == 0) && (a[n] == 1))
{
l = a[i];
a[i] = a[n];
a[n] = l;
i++;
n--;
// k--;
}
else if((a[i] == 1) && (a[n] == 1))
{
i++;
l = a[i];
a[i] = a[n];
a[n] = l;
i++;


}
else if((a[i] == 0) && (a[n] == 0))
{
n--;
l=a[n];
a[n]=a[i];
a[i]=l;
n--;

}
else
{
n--;
i++;
}

}

Even this code will work:

i=0;
j=n-1;

while (i<j){

while (((a[i]==a[j])&&a[i]==0)||(a[i]<a[j]))
      i++;

while ((a[i]==a[j])&&(a[i]==1))
      j--;

 if (i!=j)
  {
    int temp=a[i];
    a[i]=a[j];
    a[j]=temp;
    i++; j--:
  }
}

int A[10] = {0,0,0,0,0,0,0,0,0,1};
int i = 0;
int j = sizeof(A)/sizeof(int)-1;
while(i < j)
{
if(A[i] == 1)
i++;
if(A[j] == 0)
j--;
if(A[i] == 0 && A[j] == 1)
{
A[i] = 1; A[j] = 0;
i++; j--;
}
}

void X::sort()
{
int zero=-1,one=-1;

for(int i=0,j=length-1;i<length,j>-1;i++,j--)
{

if(a[i]==0)
{
zero = i;
cout<<endl<<zero;
}
if(a[j]==1)
one = j;
if(zero < one && zero!=-1 && one != -1)
{
cout<<"here";
swap(zero,one);
zero=one=-1;
}


}

}

The idea is to keep a pointer to the beginning(i) and to the end of the array(j).

Run a loop until i < j
Case 1: if i = 1 and j = 0, i++, j--
Case 2: if j = 1 => swap i+1th element and j, i++
Case 3: if j = 0 => j--;
Case 4: if i = 0 => Swap j-1th element and i, j--

int[] input = new int[] { 1,0,0,1,1,0,0,1};
int i = 0, j = input.Length - 1;
while (i <= j)
{
if (input[i] == 1 && input[j] == 0)
{
i++; j--;
}
else if (input[i] == 0 && input[j] == 1)
{
input[i] = 1;
input[j] = 0;
i++; j--;
}
else if (input[i] == 0 && input[j] == 0)
{
j--;
}
else if (input[i] == 1 && input[j] == 1)
{
i++;
}
}

this is just like the in-place partition in quick sort

//considering no. of 0 and 1s in array are same it will take o(n/2) time

int[] a = {0, 1, 1, 0, 0, 0, 0, 1, 1, 1};
        int i = 0;
        int j = a.length - 1;
        int temp = 0;
        do {
            if (a[i] == 0 && a[j] == 1) {
                temp = a[i];
                a[i] = a[j];
                a[j] = temp;
                i++;
                j--;
            }
            if (a[i] == 1) {
                i++;
            } 
            if (a[j] == 0) {
                j--;
            }
        } while (i < j);

1.Count the no. of 1s and 0s , store it in two variables(x and y).
2. For index 0 to x-1 in the array , put 1s. For index x to x+y-1 put 0s.

Funny question.

it can be simply done by using two pointers.

same as function of finding pivot in quick sort

Generally speaking it would always take O(1) time for a[10].